Chapter VI

Asymptotics

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Section VI.1Asymptotics of Integrals[27/96%]
Problem VI.1.1

Prove that (p,q∈Rp, q \in \mathbb {R})

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(a)

∫1xtpetdt∼x→+∞xpex\int_{1}^{x} t^{p} e^{t} d t \underset {x \rightarrow +\infty }{\sim } x^{p} e^{x};

(b)

∫x+∞tpe−tdt∼x→+∞xpe−x\int_{x}^{+\infty } t^{p} e^{-t} d t \underset {x \rightarrow +\infty }{\sim } x^{p} e^{-x};

(c)

∫2xtp−1ln⁡qtdt∼x→+∞xppln⁡qx(p>0)\int_{2}^{x} t^{p-1} \ln^{q} t d t \underset {x \rightarrow +\infty }{\sim } \frac{x^{p}}{p} \ln^{q} x \quad (p>0);

(d)

∫0xtp−1∣ln⁡t∣qdt∼x→0xpp∣ln⁡x∣q(p>0)\int_{0}^{x} t^{p-1}\left|\ln t\right|^{q} d t \underset {x \rightarrow 0}{\sim } \frac{x^{p}}{p}\left|\ln x\right|^{q} \quad (p>0).

Problem VI.1.2

Compute the integral ∫010xxdx\int_{0}^{10} x^{x} d x with a relative error of at most 5.

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Problem VI.1.3

Let −∞<a<b⩽+∞;f,g,h∈C([a,b))-\infty <a<b \leqslant +\infty ; f, g, h \in C([a, b)), with f>0f>0 on [a,b)[a, b) and g(t)=o(f(t))g(t)=o(f(t)), h(t)∼f(t)h(t) \sim f(t) as t→b−0t \rightarrow b-0. Prove that as t→b−0t \rightarrow b-0 the following relations hold:

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(a)

if ∫abf(t)dt=+∞\int_{a}^{b} f(t) d t=+\infty, then

∫axg(t)dt=o(∫axf(t)dt),∫axh(t)dt∼∫axf(t)dt \int _{a}^{x} g(t) d t=o\left(\int _{a}^{x} f(t) d t\right), \quad \int _{a}^{x} h(t) d t \sim \int _{a}^{x} f(t) d t
(b)

if ∫abf(t)dt<+∞\int_{a}^{b} f(t) d t<+\infty, then

∫xbg(t)dt=o(∫xbf(t)dt),∫xbh(t)dt∼∫xbf(t)dt \int _{x}^{b} g(t) d t=o\left(\int _{x}^{b} f(t) d t\right), \quad \int _{x}^{b} h(t) d t \sim \int _{x}^{b} f(t) d t
Problem VI.1.4

Find the asymptotics as t→1−0t \rightarrow 1-0 of the integrals

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(a)

J(t)=∫0tdx1−xa(a≠0)\mathscr {J}(t)=\int_{0}^{t} \frac{d x}{1-x^{a}} \quad (a \neq 0);

(b)

K(t)=∫01dx(1−x2)(1−t2x2)\mathcal{K}(t)=\int_{0}^{1} \frac{d x}{\sqrt{\left(1-x^{2}\right)\left(1-t^{2} x^{2}\right)}};

(c)

L(t)=∫0πdx1−2tcos⁡x+t2\mathscr {L}(t)=\int_{0}^{\pi } \frac{d x}{1-2 t \cos x+t^{2}}.

Problem VI.1.5

Find the asymptotics as A→+∞A \rightarrow +\infty of the integrals

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(a)

∫A2Adxln⁡x\int_{A}^{2 A} \frac{d x}{\ln x};

(b)

∫A2Axxdx\int_{A}^{2 A} x^{x} d x;

(c)

∫0Aee−xdx\int_{0}^{A} e^{e^{-x}} d x;

(d)

∫πA∣sin⁡x∣ea/xdx(a∈R)\int_{\pi }^{A}\left|\sin x\right| e^{a / x} d x \quad (a \in \mathbb {R});

(e)

∫A+∞e−x2dx\int_{A}^{+\infty } e^{-x^{2}} d x;

(f)

∫A+∞xae−xdx(a∈R)\int_{A}^{+\infty } x^{a} e^{-x} d x \quad (a \in \mathbb {R}).

Problem VI.1.6

Find the asymptotics as A→+∞A \rightarrow +\infty of the integrals

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(a)

∫01eAx2dx\int_{0}^{1} e^{A x^{2}} d x;

(b)

∫1+∞e−xp/Adxx(p>0)\int_{1}^{+\infty } e^{-x^{p} / A} \frac{d x}{x} \quad (p>0);

(c)

∫0+∞ex−x2/Adx\int_{0}^{+\infty } e^{x-x^{2} / A} d x;

(d)

∫2+∞e−x/Aln⁡pxdx(p∈R)\int_{2}^{+\infty } e^{-x / A} \ln^{p} x d x \quad (p \in \mathbb {R}).

Problem VI.1.7

Find the asymptotics as n→+∞n \rightarrow +\infty of the integrals (n∈N)(n \in \mathbb {N})

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(a)

∫πnπ(n+2)sin⁡xxpdx(p≠0)\quad \int_{\pi n}^{\pi (n+2)} \frac{\sin x}{x^{p}} d x \quad (p \neq 0);

(b)

∫πn+∞sin⁡xxpdx(p>0)\int_{\pi n}^{+\infty } \frac{\sin x}{x^{p}} d x \quad (p>0);

(c)

∫πnπ(n+1)cos⁡xxpdx(p≠0)\quad \int_{\pi n}^{\pi (n+1)} \frac{\cos x}{x^{p}} d x \quad (p \neq 0);

(d)

∫πn+∞cos⁡xxpdx(p>0)\int_{\pi n}^{+\infty } \frac{\cos x}{x^{p}} d x \quad (p>0).

Problem VI.1.8

Prove that as ε→+0\varepsilon \rightarrow +0

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(a)

∫1+∞sin⁡2xx1+εdx∼12ε\int_{1}^{+\infty } \frac{\sin^{2} x}{x^{1+\varepsilon }} d x \sim \frac{1}{2 \varepsilon };

(b)

∫1+∞cos⁡2xx1+εdx∼12ε\int_{1}^{+\infty } \frac{\cos^{2} x}{x^{1+\varepsilon }} d x \sim \frac{1}{2 \varepsilon };

(c)

∫0+∞sin⁡xxεdx→1\int_{0}^{+\infty } \frac{\sin x}{x^{\varepsilon }} d x \rightarrow 1;

(d)

∫0+∞cos⁡xxεdx∼π2ε\int_{0}^{+\infty } \frac{\cos x}{x^{\varepsilon }} d x \sim \frac{\pi }{2} \varepsilon.

Problem VI.1.9

Find the asymptotics as ε→+0\varepsilon \rightarrow +0 of the integrals

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(a)

∫0∞e−εx2dxx+ε\int_{0}^{\infty } \frac{e^{-\varepsilon x^{2}} d x}{x+\varepsilon };

(b)

∫0∞sin⁡εx2dxx2+εx\int_{0}^{\infty } \frac{\sin \varepsilon x^{2} d x}{x^{2}+\varepsilon \sqrt{x}};

(c)

∫0∞sin⁡(x2+ε)dxx3+εx\int_{0}^{\infty } \frac{\sin \left(x^{2}+\varepsilon \right) d x}{x^{3}+\varepsilon \sqrt{x}}.

Problem VI.1.10

Prove that as A→+∞A \rightarrow +\infty

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(a)

∫0∞sin⁡(xA)dx∼π2A\int_{0}^{\infty } \sin \left(x^{A}\right) d x \sim \frac{\pi }{2 A};

(b)

∫0∞cos⁡(xA)dx→1\int_{0}^{\infty } \cos \left(x^{A}\right) d x \rightarrow 1.

Problem VI.1.11

Let Cx=∫xx+1cos⁡t2dtC_{x}=\int_{x}^{x+1} \cos t^{2} d t and Sx=∫xx+1sin⁡t2dtS_{x}=\int_{x}^{x+1} \sin t^{2} d t for x>0x>0. Prove that

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(a)

Cx2+Sx2⩽1x\sqrt{C_{x}^{2}+S_{x}^{2}} \leqslant \frac{1}{x} for all x>0x>0;

(b)

Cx2+Sx2=1x∣sin⁡(x+12)∣+O(1x2)\sqrt{C_{x}^{2}+S_{x}^{2}}=\frac{1}{x}\left|\sin \left(x+\frac{1}{2}\right)\right|+O\left(\frac{1}{x^{2}}\right) as x→+∞x \rightarrow +\infty.

Problem VI.1.12

Let the function φ\varphi be continuous on R\mathbb {R} and have period T>0T>0, with Cφ=1T∫0Tφ(t)dt≠0C_{\varphi }=\frac{1}{T} \int_{0}^{T} \varphi (t) d t \neq 0. Prove that

∫1+∞φ(t)t1+εdt∼Cφε as ε→+0 \int _{1}^{+\infty } \frac{\varphi (t)}{t^{1+\varepsilon }} d t \sim \frac{C_{\varphi }}{\varepsilon } \quad \text{ as } \quad \varepsilon \rightarrow +0
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Problem VI.1.13

Find the asymptotics as A→+∞A \rightarrow +\infty of the integrals

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(a)

∫0Asin⁡2xxpdx(p⩽1)\int_{0}^{A} \frac{\sin^{2} x}{x^{p}} d x \quad (p \leqslant 1);

(b)

∫1A∣sin⁡x∣p∣cos⁡x∣xdx(p>−1)\int_{1}^{A} \frac{\left|\sin x\right|^{p}\left|\cos x\right|}{x} d x \quad (p>-1);

(c)

∫0A∣sin⁡x∣xpdx(p⩽1)\int_{0}^{A} \frac{\left|\sin x\right|}{x^{p}} d x \quad (p \leqslant 1);

(d)

∫A+∞∣sin⁡x∣xpdx(p>1)\int_{A}^{+\infty } \frac{\left|\sin x\right|}{x^{p}} d x \quad (p>1).

Problem VI.1.14

Prove that as A→+∞A \rightarrow +\infty

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(a)

∫0π/2(sin⁡Axsin⁡x)2dx=π2A+O(1)\int_{0}^{\pi / 2}\left(\frac{\sin A x}{\sin x}\right)^{2} d x=\frac{\pi }{2} A+O(1);

(b)

∫01cos⁡Axxdx=π2A+O(1A)\int_{0}^{1} \frac{\cos A x}{\sqrt{x}} d x=\sqrt{\frac{\pi }{2 A}}+O\left(\frac{1}{A}\right);

(c)

∫0π/2cos⁡2xdx1+cos⁡2Ax→π42\int_{0}^{\pi / 2} \frac{\cos^{2} x d x}{1+\cos^{2} A x} \rightarrow \frac{\pi }{4 \sqrt{2}};

(d)

∫0π/2(sin⁡Axsin⁡x)4dx=π3A3+O(A)\int_{0}^{\pi / 2}\left(\frac{\sin A x}{\sin x}\right)^{4} d x=\frac{\pi }{3} A^{3}+O(A).

Problem VI.1.15

Using the Poisson integral representation for the Bessel functions

Jν(A)=2π(A2)νΓ(ν+12)∫0π/2cos⁡(Asin⁡φ)cos⁡2νφdφ(ν>−12) J_{\nu }(A)=\frac{2}{\sqrt{\pi }} \frac{\left(\frac{A}{2}\right)^{\nu }}{\Gamma \left(\nu +\frac{1}{2}\right)} \int _{0}^{\pi / 2} \cos (A \sin \varphi ) \cos ^{2 \nu } \varphi d \varphi \quad \left(\nu >-\frac{1}{2}\right)

prove that as A→+∞\boldsymbol {A} \rightarrow +\infty

J0(A)=sin⁡A+cos⁡AπA+O(A−3/2),J1(A)=sin⁡A−cos⁡AπA+O(A−3/2) J_{0}(A)=\frac{\sin A+\cos A}{\sqrt{\pi A}}+O\left(A^{-3 / 2}\right), \quad J_{1}(A)=\frac{\sin A-\cos A}{\sqrt{\pi A}}+O\left(A^{-3 / 2}\right)
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Problem VI.1.16

Find the asymptotics as A→+∞A \rightarrow +\infty of the following integrals:

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(a)

∫0∞cos⁡Axdxxp(1+x2)(−2<p<1)\int_{0}^{\infty } \frac{\cos A x d x}{x^{p}\left(1+x^{2}\right)} \quad (-2<p<1);

(b)

∫0∞sin⁡Axdxxp(1+x2)(∣p∣<2)\int_{0}^{\infty } \frac{\sin A x d x}{x^{p}\left(1+x^{2}\right)} \quad (\left|p\right|<2);

(c)

∫0∞cos⁡Axexpdx(0<p<2)\int_{0}^{\infty } \frac{\cos A x}{e^{x^{p}}} d x \quad (0<p<2);

(d)

∫0∞sin⁡Axexpdx(0<p<2)\int_{0}^{\infty } \frac{\sin A x}{e^{x^{p}}} d x \quad (0<p<2).

Problem VI.1.17

Let the function ff decrease to zero on [a,+∞)[a,+\infty ), and let the function φ\varphi be continuous on R\mathbb {R} and have period T>0T>0, with ∫0Tφ(t)dt=0\int_{0}^{T} \varphi (t) d t=0. Prove that

∣∫A+∞f(t)φ(t)dt∣⩽f(A)∫0T∣φ(t)∣dt for A⩾a \left|\int _{A}^{+\infty } f(t) \varphi (t) d t\right| \leqslant f(A) \int _{0}^{T}\left|\varphi (t)\right| d t \quad \text{ for } \quad A \geqslant a
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Problem VI.1.18

Let the function ff decrease to zero on [a,+∞)[a,+\infty ), and let the function φ\varphi be continuous on R\mathbb {R} and have period T>0T>0. Prove that

∫aAf(t)φ(t)dt=Cφ∫aAf(t)dt+I+O(f(A)) \int _{a}^{A} f(t) \varphi (t) d t=C_{\varphi } \int _{a}^{A} f(t) d t+I+O(f(A))

where

Cφ=1T∫0Tφ(t)dt,I=∫a+∞(φ(t)−Cφ)f(t)dt C_{\varphi }=\frac{1}{T} \int _{0}^{T} \varphi (t) d t, \quad I=\int _{a}^{+\infty }\left(\varphi (t)-C_{\varphi }\right) f(t) d t
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Problem VI.1.19

Let the function ff be nonnegative and monotonic on [a,+∞)[a,+\infty ), and let the function φ\varphi be continuous on R\mathbb {R} and have period T>0T>0, with Cφ=1T∫0Tφ(t)dt≠0C_{\varphi }=\frac{1}{T} \int_{0}^{T} \varphi (t) d t \neq 0. Prove that

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(a)

if ∫a∞f(t)dt=+∞\int_{a}^{\infty } f(t) d t=+\infty and ∫AA+1f(t)dt=o(∫aAf(t)dt)\int_{A}^{A+1} f(t) d t=o\left(\int_{a}^{A} f(t) d t\right), then

∫aAf(t)φ(t)dt∼Cφ∫aAf(t)dt as A→+∞ \int _{a}^{A} f(t) \varphi (t) d t \sim C_{\varphi } \int _{a}^{A} f(t) d t \quad \text{ as } \quad A \rightarrow +\infty
(b)

if ∫a∞f(t)dt<+∞\int_{a}^{\infty } f(t) d t<+\infty and ∫AA+1f(t)dt=o(∫A∞f(t)dt)\int_{A}^{A+1} f(t) d t=o\left(\int_{A}^{\infty } f(t) d t\right), then

∫A+∞f(t)φ(t)dt∼Cφ∫A+∞f(t)dt as A→+∞ \int _{A}^{+\infty } f(t) \varphi (t) d t \sim C_{\varphi } \int _{A}^{+\infty } f(t) d t \quad \text{ as } \quad A \rightarrow +\infty
Problem VI.1.20

Let f∈C([a,b]),φ∈C(R)f \in C(\left[a, b\right]), \varphi \in C(\mathbb {R}). Find the limits

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(a)

lim⁡A→+∞∫abf(t)φ(At)dt\lim_{A \rightarrow +\infty } \int_{a}^{b} f(t) \varphi (A t) d t, if φ\varphi has period T>0T>0;

(b)

lim⁡A→+∞A∫abf(t)φ(At)dt\lim_{A \rightarrow +\infty } A \int_{a}^{b} f(t) \varphi (A t) d t, if ∫R∣φ(t)∣dt<∞\int_{\mathbb {R}}\left|\varphi (t)\right| d t<\infty.

Problem VI.1.21

Let f∈C([a,b]×[−1,1])f \in C(\left[a, b\right] \times [-1,1]). Prove that

I(A)=∫abf(x,sin⁡Ax)dx⟶A→+∞12π∫ab(∫02πf(x,sin⁡t)dt)dx I(A)=\int _{a}^{b} f(x, \sin A x) d x \underset {A \rightarrow +\infty }{\longrightarrow } \frac{1}{2 \pi } \int _{a}^{b}\left(\int _{0}^{2 \pi } f(x, \sin t) d t\right) d x
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Problem VI.1.22
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(a)

Let p(A)⩾0p(A) \geqslant 0 and p(A)=o(A2)p(A)=o\left(A^{2}\right) as A→+∞A \rightarrow +\infty. Prove that

IA=∫0π/2sin⁡Axsin⁡xcos⁡p(A)xdx→π2 as A→+∞ I_{A}=\int _{0}^{\pi / 2} \frac{\sin A x}{\sin x} \cos ^{p(A)} x d x \rightarrow \frac{\pi }{2} \quad \text{ as } \quad A \rightarrow +\infty
(b)

Compute the integral

LA=∫0π/2sin⁡Axsin⁡xcos⁡A−1xdx(A>0) L_{A}=\int _{0}^{\pi / 2} \frac{\sin A x}{\sin x} \cos ^{A-1} x d x \quad (A>0)
Problem VI.1.23

Find the asymptotics as ε→+0\varepsilon \rightarrow +0 of the integrals

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(a)

∫0+∞e−x(1−x−ε)dx\int_{0}^{+\infty } e^{-x\left(1-x^{-\varepsilon }\right)} d x;

(b)

∫0+∞e−x(xε−1)dx\int_{0}^{+\infty } e^{-x\left(x^{\varepsilon }-1\right)} d x.

Problem VI.1.24

Prove that as ε→+0\varepsilon \rightarrow +0

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(a)

∫01xεxdx=1−ε4+ε227+O(ε3)\int_{0}^{1} x^{\varepsilon x} d x=1-\frac{\varepsilon }{4}+\frac{\varepsilon^{2}}{27}+O\left(\varepsilon^{3}\right);

(b)

∫01e−ε/xdx=1−2ε−ε2ln⁡ε+O(ε2)\int_{0}^{1} e^{-\varepsilon / \sqrt{x}} d x=1-2 \varepsilon -\varepsilon^{2} \ln \varepsilon +O\left(\varepsilon^{2}\right);

(c)

∫01exp⁡(−εxln⁡2(x/e))dx=1−ε−εln⁡ε+o(εln⁡ε)\int_{0}^{1} \exp \left(-\frac{\varepsilon }{x \ln^{2}(x / e)}\right) d x=1-\varepsilon -\frac{\varepsilon }{\ln \varepsilon }+o\left(\frac{\varepsilon }{\ln \varepsilon }\right).

Problem VI.1.25

Let the function ff be positive and integrable on the interval (0,1)(0,1), and suppose that for some number pp in the interval (1,2](1,2] the integral ∫01∣ln⁡f(x)∣pdx\int_{0}^{1}\left|\ln f(x)\right|^{p} d x is finite. Prove that as ε→+0\varepsilon \rightarrow +0

∫01fε(x)dx=1+ε∫01ln⁡f(x)dx+O(εp) \int _{0}^{1} f^{\varepsilon }(x) d x=1+\varepsilon \int _{0}^{1} \ln f(x) d x+O\left(\varepsilon ^{p}\right)
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Problem VI.1.26

Find the asymptotics as n→∞n \rightarrow \infty of the following integrals:

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(a)

∫0π∣sin⁡nx∣dxx\int_{0}^{\pi }\left|\sin n x\right| \frac{d x}{x};

(b)

∫0πmax⁡1⩽k⩽n∣sin⁡kx∣dxx\int_{0}^{\pi } \max_{1 \leqslant k \leqslant n}\left|\sin k x\right| \frac{d x}{x};

(c)

∫0πmax⁡2⩽k⩽n∣sin⁡kx∣ln⁡kdxx\int_{0}^{\pi } \max_{2 \leqslant k \leqslant n} \frac{\left|\sin k x\right|}{\ln k} \frac{d x}{x};

(d)

∫0πmax⁡1⩽k⩽n∣sin⁡kx∣kdxx\int_{0}^{\pi } \max_{1 \leqslant k \leqslant n} \frac{\left|\sin k x\right|}{k} \frac{d x}{x}.

Problem VI.1.27

Let P(x)=a1x+…+anxnP(x)=a_{1} x+\ldots +a_{n} x^{n} be a polynomial with real coefficients, I(P)=∫01eiP(x)dxI(P)=\int_{0}^{1} e^{i P(x)} d x. Prove that ∣I(P)∣⩽CnAn\left|I(P)\right| \leqslant \frac{C_{n}}{\sqrt[n]{A}}, where A=∣a1∣+∣a2∣+…+∣an∣A=\left|a_{1}\right|+\left|a_{2}\right|+\ldots +\left|a_{n}\right|, and the coefficient CnC_{n} depends only on nn.

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Section VI.2Laplace's Method[23/100%]
Problem VI.2.1

Let f∈C([0,π2])f \in C\left(\left[0, \frac{\pi }{2}\right]\right). Find the limit

lim⁡n→+∞n∫0π/2xf(x)cos⁡nxdx \lim _{n \rightarrow +\infty } n \int _{0}^{\pi / 2} x f(x) \cos ^{n} x d x
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Problem VI.2.2

Let c>0c>0 and Ic=∫0ce−t2dtI_{c}=\int_{0}^{c} e^{-t^{2}} d t. Prove that as A→+∞A \rightarrow +\infty

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(a)

∫0c/A(1−x2)Adx∼IcA\int_{0}^{c / \sqrt{A}}\left(1-x^{2}\right)^{A} d x \sim \frac{I_{c}}{\sqrt{A}};

(b)

∫0c/Adx(1+x2)A∼IcA\int_{0}^{c / \sqrt{A}} \frac{d x}{\left(1+x^{2}\right)^{A}} \sim \frac{I_{c}}{\sqrt{A}}.

Problem VI.2.3

Let c>0c>0. Find the asymptotics as A→+∞A \rightarrow +\infty of the integral

∫0c/Acos⁡Axdx \int _{0}^{c / \sqrt{A}} \cos ^{A} x d x
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Problem VI.2.4

Verify that the asymptotics as A→+∞A \rightarrow +\infty of the following integrals do not depend on the parameter c>0c>0:

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(a)

∫0c(1−x2)Adx(c⩽1)\int_{0}^{c}\left(1-x^{2}\right)^{A} d x \quad (c \leqslant 1);

(b)

∫0cdx(1+x2)A\int_{0}^{c} \frac{d x}{\left(1+x^{2}\right)^{A}};

(c)

∫0ccos⁡Axdx(c⩽π2)\int_{0}^{c} \cos^{A} x d x \quad \left(c \leqslant \frac{\pi }{2}\right);

(d)

∫0cdx(1+x+x2)A\int_{0}^{c} \frac{d x}{\left(1+x+x^{2}\right)^{A}};

(e)

∫0c(ln⁡(1+x)x)Adx\int_{0}^{c}\left(\frac{\ln (1+x)}{x}\right)^{A} d x;

(f)

∫0c(sin⁡xx)Adx(c⩽π)\int_{0}^{c}\left(\frac{\sin x}{x}\right)^{A} d x \quad (c \leqslant \pi );

(g)

∫11+cx−Axdx\int_{1}^{1+c} x^{-A x} d x;

(h)

∫0cln⁡A(e−x)dx(c⩽e−1)\int_{0}^{c} \ln^{A}(e-x) d x \quad (c \leqslant e-1).

Remark.
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A characteristic feature of Problems 2.1−2.42.1-2.4 is the phenomenon of localization — the asymptotics of the integral depend on the behavior of the integrand only in a neighborhood of a single point. This effect manifests itself distinctly also in the more general case, in the study of the important integrals of the form

Φ(A)=∫abf(x)φA(x)dx \Phi (A)=\int _{a}^{b} f(x) \varphi ^{A}(x) d x

where the functions ff and φ\varphi are nonnegative and φ\varphi is piecewise monotone. For large values of the parameter AA, the graph of the function φA\varphi^{A} has sharply pronounced "humps" in neighborhoods of those points at which the function φ\varphi has a strict local maximum. By splitting the interval [a,b]\left[a, b\right] into several intervals if necessary, we may assume that φ\varphi is monotone on [a,b]\left[a, b\right]. In doing so it suffices to consider only the case where φ\varphi is decreasing. Then the values of the function φA\varphi^{A} at points far from the point aa are negligibly small compared with its values at points close to aa, which give the main contribution to the integral Φ(A)\Phi (A). To find the principal part of Φ(A)\Phi (A) it remains to approximate the functions ff and φ\varphi near the point aa by simpler ones and to compute the resulting integral. The implementation of the scheme just described constitutes the Laplace method for studying integrals of the form Φ(A)\Phi (A), as well as their modifications. The most frequently encountered case is one in which the difference φ(a)−φ(x)\varphi (a)-\varphi (x) is an infinitesimal of power type, i.e.,

φ(a)−φ(x)∼x→a+0C(x−a)p(C,p>0). \varphi (a)-\varphi (x) \underset {x \rightarrow a+0}{\sim } C(x-a)^{p} \quad (C, p>0) .

This is investigated in Problem 2.7. The result obtained in this way is called Laplace's asymptotic formula (in Problem 2.7, for brevity of statement, it is assumed that φ(a)=1\varphi (a)=1 and f≡1f \equiv 1). The reader will have no difficulty formulating the analogous statement if the function φ\varphi increases on [a,b]\left[a, b\right] and

φ(b)−φ(x)∼x→b−0C(b−x)p. \varphi (b)-\varphi (x) \underset {x \rightarrow b-0}{\sim } C(b-x)^{p} .

From this Laplace's asymptotic formula for a piecewise monotone function follows immediately.

By a suitable change of variable in the integral Φ(A)\Phi (A), the case of a nonconstant nonnegative function ff can be reduced to the case f≡1f \equiv 1 (as the new variable one should take the antiderivative ∫axf(t)dt\int_{a}^{x} f(t) d t). The result of Problem 2.8 makes it possible to substantially reduce the technical difficulties that arise in doing so.

Problem VI.2.5

We assume that φ\varphi is a positive nonincreasing function on the interval [a,b)(−∞<a<b⩽+∞)[a, b)(-\infty <a<b \leqslant +\infty ), with φ(a)=lim⁡x→a+0φ(x)=1\varphi (a)=\lim_{x \rightarrow a+0} \varphi (x)=1 a strict maximum of φ\varphi on [a,b)[a, b). We also assume that f∈C((a,b)),f⩾0f \in C((a, b)), f \geqslant 0 on (a,b)(a, b) and f≢0f \not\equiv 0 near the point aa, and that the integral ∫abf(x)φA(x)dx\int_{a}^{b} f(x) \varphi^{A}(x) d x is finite for sufficiently large AA. Prove the following properties of the integral Φ(A)=∫abf(x)φA(x)dx\Phi (A)=\int_{a}^{b} f(x) \varphi^{A}(x) d x as A→+∞A \rightarrow +\infty:

?
(a)

Φ(A)\boldsymbol {\Phi }(A) decreases to zero;

(b)

Φ(A)\Phi (A) cannot decrease exponentially: ln⁡Φ(A)=o(A)\ln \Phi (A)=o(A);

(c)

statement b) cannot, in general, be strengthened: for any function ω(A)\omega (A) decreasing to zero as A→+∞A \rightarrow +\infty and such that ln⁡ω(A)=o(A)\ln \omega (A)=o(A), there exists a nonnegative function φ\varphi, decreasing on [0,1)[0,1), with lim⁡x→0φ(x)=1\lim_{x \rightarrow 0} \varphi (x)=1, such that ∫01φA(x)dx=o(ω(A))\int_{0}^{1} \varphi^{A}(x) d x=o(\omega (A));

(d)

Φ(A)\Phi (A) can decrease arbitrarily slowly: for any function ω(A)\omega (A) decreasing to zero as A→+∞A \rightarrow +\infty, there exists a function φ\varphi such that ∫abφA(x)dx⩾ω(A)\int_{a}^{b} \varphi^{A}(x) d x \geqslant \omega (A) for sufficiently large AA;

(e)

the asymptotics of Φ(A)\Phi (A) is determined by the behavior of the functions φ\varphi and ff in an arbitrarily small neighborhood of the point aa:

Φ(A)∼∫aβf(x)φA(x)dx for any number β∈(a,b); \Phi (A) \sim \int _{a}^{\beta } f(x) \varphi ^{A}(x) d x \quad \text{ for any number } \beta \in (a, b) ;
(f)

Φ(A)∼Φ(A+H)\Phi (A) \sim \Phi (A+H) for any number H∈RH \in \mathbb {R}.

Remark.
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Note that statement f) allows one, without loss of generality, to assume that ∫abf(x)dx<∞\int_{a}^{b} f(x) d x<\infty, since otherwise, without changing the asymptotics of the integral Φ(A)\Phi (A), the function f(x)f(x) can be replaced by f(x)φH(x)f(x) \varphi^{H}(x) for sufficiently large HH.

Problem VI.2.6

We assume that φ\varphi is a positive nonincreasing function on the interval [a,b)(−∞<a<b⩽+∞)[a, b)(-\infty <a<b \leqslant +\infty ), with φ(a)=lim⁡x→a+0φ(x)=1\varphi (a)=\lim_{x \rightarrow a+0} \varphi (x)=1 a strict maximum of φ\varphi on [a,b)[a, b). We also assume that f∈C((a,b)),f⩾0f \in C((a, b)), f \geqslant 0 on (a,b)(a, b) and f≢0f \not\equiv 0 near the point aa, and that the integral ∫abf(x)φA(x)dx\int_{a}^{b} f(x) \varphi^{A}(x) d x is finite for sufficiently large AA. Let Φ(A)=∫abφA(x)dx\Phi (A)=\int_{a}^{b} \varphi^{A}(x) d x. Prove that

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(a)

if for some L,p>0L, p>0 the inequality φ(x)⩾1−L(x−a)p\varphi (x) \geqslant 1-L(x-a)^{p} holds near the point aa, then

A1/pΦ(A)⩾Γ(1+1p)L1/p+o(1) as A→+∞; A^{1 / p} \Phi (A) \geqslant \frac{\Gamma \left(1+\frac{1}{p}\right)}{L^{1 / p}}+o(1) \quad \text{ as } \quad A \rightarrow +\infty ;
(b)

if 1−φ(x)(x−a)p→0\frac{1-\varphi (x)}{(x-a)^{p}} \rightarrow 0 as x→a+0x \rightarrow a+0, then A1/pΦ(A)→+∞A^{1 / p} \Phi (A) \rightarrow +\infty as A→+∞A \rightarrow +\infty;

(c)

if for some l,p>0l, p>0 the inequality φ(x)⩽1−l(x−a)p\varphi (x) \leqslant 1-l(x-a)^{p} holds near the point aa, then

A1/pΦ(A)⩽Γ(1+1p)l1/p+o(1) as A→+∞; A^{1 / p} \Phi (A) \leqslant \frac{\Gamma \left(1+\frac{1}{p}\right)}{l^{1 / p}}+o(1) \quad \text{ as } \quad A \rightarrow +\infty ;
(d)

if 1−φ(x)(x−a)p→+∞\frac{1-\varphi (x)}{(x-a)^{p}} \rightarrow +\infty as x→a+0x \rightarrow a+0, then

A1/pΦ(A)→0 as A→+∞. A^{1 / p} \Phi (A) \rightarrow 0 \quad \text{ as } \quad A \rightarrow +\infty .
Problem VI.2.7

We assume that φ\varphi is a positive nonincreasing function on the interval [a,b)(−∞<a<b⩽+∞)[a, b)(-\infty <a<b \leqslant +\infty ), with φ(a)=lim⁡x→a+0φ(x)=1\varphi (a)=\lim_{x \rightarrow a+0} \varphi (x)=1 a strict maximum of φ\varphi on [a,b)[a, b). We also assume that f∈C((a,b)),f⩾0f \in C((a, b)), f \geqslant 0 on (a,b)(a, b) and f≢0f \not\equiv 0 near the point aa, and that the integral ∫abf(x)φA(x)dx\int_{a}^{b} f(x) \varphi^{A}(x) d x is finite for sufficiently large AA. Let 1−φ(x)∼x→a+0C(x−a)p(C,p>0)1-\varphi (x) \underset {x \rightarrow a+0}{\sim } C(x-a)^{p}(C, p>0). Prove the Laplace asymptotic formula

Φ(A)=∫abφA(x)dx∼A→+∞Γ(1+1p)(AC)1/p \Phi (A)=\int _{a}^{b} \varphi ^{A}(x) d x \underset {A \rightarrow +\infty }{\sim } \frac{\Gamma \left(1+\frac{1}{p}\right)}{(A C)^{1 / p}}

In particular,

Φ(A)∼1AC for p=1,Φ(A)∼12πAC for p=2. \Phi (A) \sim \frac{1}{A C} \quad \text{ for } p=1, \quad \Phi (A) \sim \frac{1}{2} \sqrt{\frac{\pi }{A C}} \quad \text{ for } p=2 .

Supplement the Laplace formula by proving that for any q>−1q>-1

∫ab(x−a)qφA(x)dx∼A→+∞1q+1Γ(1+q+1p)(AC)q+1p \int _{a}^{b}(x-a)^{q} \varphi ^{A}(x) d x \underset {A \rightarrow +\infty }{\sim } \frac{1}{q+1} \frac{\Gamma \left(1+\frac{q+1}{p}\right)}{(A C)^{\frac{q+1}{p}}}
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Problem VI.2.8

We assume that φ\varphi is a positive nonincreasing function on the interval [a,b)(−∞<a<b⩽+∞)[a, b)(-\infty <a<b \leqslant +\infty ), with φ(a)=lim⁡x→a+0φ(x)=1\varphi (a)=\lim_{x \rightarrow a+0} \varphi (x)=1 a strict maximum of φ\varphi on [a,b)[a, b). We also assume that f∈C((a,b)),f⩾0f \in C((a, b)), f \geqslant 0 on (a,b)(a, b) and f≢0f \not\equiv 0 near the point aa, and that the integral ∫abf(x)φA(x)dx\int_{a}^{b} f(x) \varphi^{A}(x) d x is finite for sufficiently large AA. Let g∈C((a,b))g \in C((a, b)) and let the function g(x)φA(x)g(x) \varphi^{A}(x) be absolutely integrable on (a,b)(a, b) for AA sufficiently large. Prove that

∫abf(x)φA(x)dx∼A→∞∫abg(x)φA(x)dx, if f(x)∼x→a+0g(x) \int _{a}^{b} f(x) \varphi ^{A}(x) d x \underset {A \rightarrow \infty }{\sim } \int _{a}^{b} g(x) \varphi ^{A}(x) d x, \quad \text{ if } f(x) \underset {x \rightarrow a+0}{\sim } g(x)

In particular, if the function ff is continuous at the point aa and f(a)≠0f(a) \neq 0, then for any β∈(a,b)\beta \in (a, b)

Φ(A)=∫abf(x)φA(x)dx∼A→+∞f(a)∫aβφA(x)dx \Phi (A)=\int _{a}^{b} f(x) \varphi ^{A}(x) d x \underset {A \rightarrow +\infty }{\sim } f(a) \int _{a}^{\beta } \varphi ^{A}(x) d x
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Problem VI.2.9

Find the asymptotics as A→+∞A \rightarrow +\infty of the following integrals:

?
(a)

∫01(1−xp)Adx(p>0)\int_{0}^{1}\left(1-x^{p}\right)^{A} d x(p>0);

(b)

∫0+∞dx(1+xp)A(p>0)\int_{0}^{+\infty } \frac{d x}{\left(1+x^{p}\right)^{A}}(p>0);

(c)

∫0π/2xpcos⁡Axdx(p>−1)\int_{0}^{\pi / 2} x^{p} \cos^{A} x d x(p>-1);

(d)

∫0π/2xpsin⁡Axdx(p∈R)\int_{0}^{\pi / 2} x^{p} \sin^{A} x d x(p \in \mathbb {R});

(e)

∫0+∞sin⁡x(1+x2)Adx\int_{0}^{+\infty } \frac{\sin x}{\left(1+x^{2}\right)^{A}} d x;

(f)

∫0+∞cos⁡x(1+x2)Adx\int_{0}^{+\infty } \frac{\cos x}{\left(1+x^{2}\right)^{A}} d x;

(g)

∫0Aex(A−x)Adx;\int_{0}^{A} e^{x}(A-x)^{A} d x ;

(h)

∫0+∞eA(x−xp)dx(p>1)\int_{0}^{+\infty } e^{A\left(x-x^{p}\right)} d x(p>1);

(i)

∫0+∞sin⁡xxp⋅dxeAx⋅(p<2)\int_{0}^{+\infty } \frac{\sin x}{x^{p}} \cdot \frac{d x}{e^{A x}} \cdot (p<2);

(j)

∫0π/2cos⁡xeAsin⁡2xdx\int_{0}^{\pi / 2} \frac{\cos x}{e^{A \sqrt{\sin 2 x}}} d x;

(k)

∫−∞+∞xcos⁡(π2x)1−x5⋅dxeAx2\int_{-\infty }^{+\infty } \frac{x \cos \left(\frac{\pi }{2} x\right)}{1-x^{5}} \cdot \frac{d x}{e^{A x^{2}}};

(l)

∫−∞+∞xsin⁡(πx)1−x1999⋅dxeAx2\int_{-\infty }^{+\infty } \frac{x \sin (\pi x)}{1-x^{1999}} \cdot \frac{d x}{e^{A x^{2}}};

(m)

∫01xAxdx\int_{0}^{1} x^{A x} d x;

(n)

∫01(Ax)pxdx(p∈R)\int_{0}^{1}(A x)^{p x} d x \quad (p \in \mathbb {R}).

Problem VI.2.10

Find the asymptotics as n→+∞(n∈N)n \rightarrow +\infty (n \in \mathbb {N}) of the following integrals:

?
(a)

∫0π(a+cos⁡x)ndx(a>0)\int_{0}^{\pi }(a+\cos x)^{n} d x(a>0);

(b)

∫0+∞(sin⁡xx)ndx\int_{0}^{+\infty }\left(\frac{\sin x}{x}\right)^{n} d x;

(c)

∫02(1−4x+2x2)ndxx\int_{0}^{2}\left(1-4 x+2 x^{2}\right)^{n} \frac{d x}{\sqrt{x}};

(d)

∫0+∞cos⁡nxdxex\int_{0}^{+\infty } \cos^{n} x \frac{d x}{e^{x}}.

Problem VI.2.11

Let p,q,C>0p, q, C>0 and r∈Rr \in \mathbb {R}. Prove that

I(A)=∫0+∞xre−C/xqe−Axpdx∼C0Aσe−γ(A) as A→+∞ I(A)=\int _{0}^{+\infty } x^{r} e^{-C / x^{q}} e^{-A x^{p}} d x \sim \frac{C_{0}}{A^{\sigma }} e^{-\gamma (A)} \quad \text{ as } A \rightarrow +\infty

where σ=1+r+q2p+q,γ(A)=(p+q)(Cp)pp+q(Aq)qp+q,C0=(Cqp)σ2πCq(p+q)\sigma =\frac{1+r+\frac{q}{2}}{p+q}, \gamma (A)=(p+q)\left(\frac{C}{p}\right)^{\frac{p}{p+q}}\left(\frac{A}{q}\right)^{\frac{q}{p+q}}, C_{0}=\left(\frac{C q}{p}\right)^{\sigma } \sqrt{\frac{2 \pi }{C q(p+q)}}. In particular, as A→+∞A \rightarrow +\infty

∫0∞xre−C/xpe−Axpdx∼πp(CA)1+r2pe−2ACAC4 \int _{0}^{\infty } x^{r} e^{-C / x^{p}} e^{-A x^{p}} d x \sim \frac{\sqrt{\pi }}{p}\left(\frac{C}{A}\right)^{\frac{1+r}{2 p}} \frac{e^{-2 \sqrt{A C}}}{\sqrt[4]{A C}}

Note that the exponential factor e−γ(A)e^{-\gamma (A)} does not depend on rr and, consequently, this parameter does not affect the asymptotics of ln⁡I(A)\ln I(A).

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Problem VI.2.12

Compute the integral

∫−∞+∞dx(x4+4x+4)100 \int _{-\infty }^{+\infty } \frac{d x}{\left(x^{4}+4 x+4\right)^{100}}

with relative error no more than 10%10 \%.

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Problem VI.2.13

Find the asymptotics as A→+∞A \rightarrow +\infty of the integrals

?
(a)

∫0∞(2x1+x2)AdxexA\int_{0}^{\infty }\left(\frac{2 x}{1+x^{2}}\right)^{A} \frac{d x}{e^{x \sqrt{A}}};

(b)

∫0Aexp(A−x)Adx(0<p<1)\int_{0}^{A} e^{x^{p}}(A-x)^{A} d x(0<p<1);

(c)

∫01/eeA/ln⁡xdx\int_{0}^{1 / e} e^{A / \ln x} d x;

(d)

∫01/e(1+1ln⁡x)Adx\int_{0}^{1 / e}\left(1+\frac{1}{\ln x}\right)^{A} d x.

Remark.
?

In this and the following problems we consider integrals Φ(A)=∫abfA(x)dx\Phi (A)=\int_{a}^{b} f_{A}(x) d x, where the function fAf_{A}, while no longer of the form considered in problems 2.52.5--2.82.8, nevertheless retains their characteristic feature: as AA grows, the graph of the function fAf_{A} has an increasingly sharply pronounced "hump" in a neighborhood of the point xAx_{A} at which fAf_{A} attains its maximum value. Although the result of problem 2.7 does not apply here, the idea of the solution is nevertheless preserved: writing the function fAf_{A} in the form eln⁡fAe^{\ln f_{A}} and replacing, in some neighborhood of the point xAx_{A}, the function ln⁡fA\ln f_{A} by its Taylor expansion, and the integral over the interval (a,b)(a, b) by the integral over a neighborhood of the point xAx_{A}, we find the principal part of Φ(A)\Phi (A). The main difficulty here is the choice of the neighborhood. On the one hand, it should not be too large, since otherwise the error caused by applying Taylor's formula will make itself felt. On the other hand, to keep down the error arising from replacing the integral over the interval (a,b)(a, b) by the integral over the neighborhood, it cannot be taken too small. A good choice of neighborhood, allowing both of the errors mentioned to be well estimated, constitutes the main content of the solution. Let us also note that in the case when xA→x0x_{A} \rightarrow x_{0} as A→+∞A \rightarrow +\infty, it may turn out to be more convenient to consider the Taylor expansion of the function ln⁡fA\ln f_{A} in a neighborhood of the point x0x_{0}.

Problem VI.2.14

Prove that as A→+∞A \rightarrow +\infty

?
(a)

∫01/2xAxdx∼1Aln⁡A\int_{0}^{1 / 2} x^{A x} d x \sim \frac{1}{A \ln A};

(b)

∫01∣ln⁡x∣p(1−x)Adx∼ln⁡pAA(p∈R)\int_{0}^{1}\left|\ln x\right|^{p}(1-x)^{A} d x \sim \frac{\ln^{p} A}{A} \quad (p \in \mathbb {R}).

Problem VI.2.15

Let the functions ψ,θ\psi , \theta be strictly increasing and not exceed 1 on [0,1],lim⁡x→0ψ(x)=lim⁡x→0θ(x)=0\left[0,1\right], \lim_{x \rightarrow 0} \psi (x)=\lim_{x \rightarrow 0} \theta (x)=0. Prove that if ψ−1(y)∼θ−1(y)\psi^{-1}(y) \sim \theta^{-1}(y) as y→+0y \rightarrow +0, then

∫01(1−ψ(x))Adx∼∫01(1−θ(x))Adx as A→+∞ \int _{0}^{1}(1-\psi (x))^{A} d x \sim \int _{0}^{1}(1-\theta (x))^{A} d x \quad \text{ as } \quad A \rightarrow +\infty
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Problem VI.2.16

Let

Ψ(A)=∫abf(x)e−Aψ(x)dx and Θ(A)=∫abf(x)e−Aθ(x)dx \Psi (A)=\int _{a}^{b} f(x) e^{-A \psi (x)} d x \quad \text{ and } \quad \Theta (A)=\int _{a}^{b} f(x) e^{-A \theta (x)} d x

where the function ff is nonnegative and integrable, and the functions ψ\psi and θ\theta are continuous and increasing on [a,b),ψ(a)=θ(a)=0[a, b), \psi (a)=\theta (a)=0. Suppose that

  1. ψ(x)−θ(x)=O(ψ2(x))\psi (x)-\theta (x)=O\left(\psi^{2}(x)\right) as x→ax \rightarrow a;

  2. the integral Ψ(A)\Psi (A) does not decrease to zero too quickly:

    ln⁡Ψ(A)=o(A) as A→+∞. \ln \Psi (A)=o(\sqrt{A}) \quad \text{ as } \quad A \rightarrow +\infty .

Prove that Ψ(A)∼Θ(A)\Psi (A) \sim \Theta (A) as A→+∞A \rightarrow +\infty. In particular, if ψ<1\psi <1, then, taking θ=−ln⁡(1−ψ)\theta =-\ln (1-\psi ), we obtain

∫abf(x)(1−ψ(x))Adx∼∫abf(x)e−Aψ(x)dx as A→+∞ \int _{a}^{b} f(x)(1-\psi (x))^{A} d x \sim \int _{a}^{b} f(x) e^{-A \psi (x)} d x \quad \text{ as } \quad A \rightarrow +\infty

Can condition 2) be weakened by replacing it with ln⁡Φ(A)=O(A)\ln \Phi (A)=O(\sqrt{A})?

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Problem VI.2.17

Let f,ψ∈C((a,b)),∫abf(x)dx<∞,0<ψ<1,ψf, \psi \in C((a, b)), \int_{a}^{b} f(x) d x<\infty , 0<\psi <1, \psi be increasing and ψ(x)=O((x−a)p)\psi (x)=O\left((x-a)^{p}\right) as x→a(p>0)x \rightarrow a(p>0). Prove that if the function ff is positive and not too small near the point aa, more precisely, ln⁡f(x)=o((x−a)−p)\ln f(x)=o\left((x-a)^{-p}\right) as x→ax \rightarrow a, then

∫abf(x)(1−ψ(x))Adx∼∫abf(x)e−Aψ(x)dx as A→+∞. \int _{a}^{b} f(x)(1-\psi (x))^{A} d x \sim \int _{a}^{b} f(x) e^{-A \psi (x)} d x \quad \text{ as } \quad A \rightarrow +\infty .
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Problem VI.2.18

Prove Stirling's formula

Γ(1+x)∼x→+∞2πx(xe)x \Gamma (1+x) \underset {x \rightarrow +\infty }{\sim } \sqrt{2 \pi x}\left(\frac{x}{e}\right)^{x}

Supplement this result by proving that for any r∈Nr \in \mathbb {N}

Γ(r)(1+x)=Γ(1+x)ln⁡rx(1+O(1xln⁡x)) as x→+∞. \Gamma ^{(r)}(1+x)=\Gamma (1+x) \ln ^{r} x\left(1+O\left(\frac{1}{\sqrt{x} \ln x}\right)\right) \quad \text{ as } \quad x \rightarrow +\infty .
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Remark.
?

Here Γ\Gamma is Euler's Γ\Gamma-function

Γ(x)=∫0+∞tx−1e−tdt,x>0. \Gamma (x)=\int _{0}^{+\infty } t^{x-1} e^{-t} d t, \quad x>0 .
Problem VI.2.19

Prove that Γ(x+c)∼x→+∞xcΓ(x)(c∈R)\Gamma (x+c) \underset {x \rightarrow +\infty }{\sim } x^{c} \Gamma (x)(c \in \mathbb {R}).

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Remark.
?

Here Γ\Gamma is Euler's Γ\Gamma-function

Γ(x)=∫0+∞tx−1e−tdt,x>0. \Gamma (x)=\int _{0}^{+\infty } t^{x-1} e^{-t} d t, \quad x>0 .
Problem VI.2.20

Prove that ∫0xtxe−tdt∼x→+∞12Γ(1+x)\int_{0}^{x} t^{x} e^{-t} d t \underset {x \rightarrow +\infty }{\sim } \frac{1}{2} \Gamma (1+x).

?
Remark.
?

Here Γ\Gamma is Euler's Γ\Gamma-function

Γ(x)=∫0+∞tx−1e−tdt,x>0. \Gamma (x)=\int _{0}^{+\infty } t^{x-1} e^{-t} d t, \quad x>0 .
Problem VI.2.21

Let φ\varphi be a positive function defined on (0,+∞)(0,+\infty ). Prove that as x→+∞x \rightarrow +\infty

?
(a)

φ(x)−xx→0⟺1Γ(1+x)∫0φ(x)txe−tdt→12\frac{\varphi (x)-x}{\sqrt{x}} \rightarrow 0 \quad \Longleftrightarrow \quad \frac{1}{\Gamma (1+x)} \int_{0}^{\varphi (x)} t^{x} e^{-t} d t \rightarrow \frac{1}{2};

(b)

φ(x)−xx→+∞⟺1Γ(1+x)∫0φ(x)txe−tdt→1\frac{\varphi (x)-x}{\sqrt{x}} \rightarrow +\infty \quad \Longleftrightarrow \quad \frac{1}{\Gamma (1+x)} \int_{0}^{\varphi (x)} t^{x} e^{-t} d t \rightarrow 1;

(c)

φ(x)−xx→−∞⟺1Γ(1+x)∫0φ(x)txe−tdt→0\frac{\varphi (x)-x}{\sqrt{x}} \rightarrow -\infty \quad \Longleftrightarrow \quad \frac{1}{\Gamma (1+x)} \int_{0}^{\varphi (x)} t^{x} e^{-t} d t \rightarrow 0.

Remark.
?

Here Γ\Gamma is Euler's Γ\Gamma-function

Γ(x)=∫0+∞tx−1e−tdt,x>0. \Gamma (x)=\int _{0}^{+\infty } t^{x-1} e^{-t} d t, \quad x>0 .
Problem VI.2.22

Let ff be the Cantor function (see problem III.3.14 and the text before it). Prove that

∫01f2n(t)dt∼C3n as n→+∞ \int _{0}^{1} f^{2^{n}}(t) d t \sim \frac{C}{3^{n}} \quad \text{ as } \quad n \rightarrow +\infty
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Remark.
?

The Cantor function is the function ff defined for x∈[0,1]x \in [0,1] by

f(x)=sup⁡{∑εk2k∣ ∑2εk3k⩽x,εk=0 or 1}. f(x)=\sup \left\{ \sum \frac{\varepsilon _{k}}{2^{k}} \left\lvert \, \sum \frac{2 \varepsilon _{k}}{3^{k}} \leqslant x\right., \varepsilon _{k}=0 \text{ or } 1\right\} .
Problem VI.2.23

Let ff be the Cantor function. Prove that

∫01t3ndf(t)∼C2n as n→+∞ \int _{0}^{1} t^{3^{n}} d f(t) \sim \frac{C}{2^{n}} \quad \text{ as } \quad n \rightarrow +\infty
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Section VI.3Asymptotics of Sums[41/88%]
Problem VI.3.1

Let Q>1,p,q∈RQ>1, p, q \in \mathbb {R}. Prove that as n→+∞n \rightarrow +\infty the following relations hold

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(a)

∑1⩽k⩽nkpln⁡qk∼np+1p+1ln⁡qn(p>−1)\sum_{1 \leqslant k \leqslant n} k^{p} \ln^{q} k \sim \frac{n^{p+1}}{p+1} \ln^{q} n \quad (p>-1);

(b)

∑k⩾nln⁡qkkp∼ln⁡qn(p−1)np−1(p>1)\sum_{k \geqslant n} \frac{\ln^{q} k}{k^{p}} \sim \frac{\ln^{q} n}{(p-1) n^{p-1}} \quad (p>1);

(c)

∑1⩽k⩽nkpQk∼QQ−1npQn\sum_{1 \leqslant k \leqslant n} k^{p} Q^{k} \sim \frac{Q}{Q-1} n^{p} Q^{n};

(d)

∑k⩾nkpQk∼QQ−1npQn\sum_{k \geqslant n} \frac{k^{p}}{Q^{k}} \sim \frac{Q}{Q-1} \frac{n^{p}}{Q^{n}}.

Problem VI.3.2

Let α>0\alpha >0 and xn=∑1⩽k⩽nα1k(1−1n)kx_{n}=\sum_{1 \leqslant k \leqslant n^{\alpha }} \frac{1}{k}\left(1-\frac{1}{n}\right)^{k}. Prove that xn∼min⁡(1,α)ln⁡nx_{n} \sim \min (1, \alpha ) \ln n.

?
Problem VI.3.3

Let an>0,bn=o(an),cn∼ana_{n}>0, b_{n}=o\left(a_{n}\right), c_{n} \sim a_{n}. Prove that

?
(a)

if ∑an=+∞\sum a_{n}=+\infty, then

∑1⩽k⩽nbk=o(∑1⩽k⩽nak),∑1⩽k⩽nck∼∑1⩽k⩽nak; \sum _{1 \leqslant k \leqslant n} b_{k}=o\left(\sum _{1 \leqslant k \leqslant n} a_{k}\right), \quad \sum _{1 \leqslant k \leqslant n} c_{k} \sim \sum _{1 \leqslant k \leqslant n} a_{k} ;
(b)

if ∑an<+∞\sum a_{n}<+\infty, then

∑k⩾nbk=o(∑k⩾nak),∑k⩾nck∼∑k⩾nak. \sum _{k \geqslant n} b_{k}=o\left(\sum _{k \geqslant n} a_{k}\right), \quad \sum _{k \geqslant n} c_{k} \sim \sum _{k \geqslant n} a_{k} .
Remark.
?

Recall that ∑an\sum a_{n} denotes the series ∑n⩾1an\sum_{n \geqslant 1} a_{n}.

Problem VI.3.4

Let an↓0,p⩾0,α∈Ra_{n} \downarrow 0, p \geqslant 0, \alpha \in \mathbb {R}. Prove that

?
(a)

if ∑1⩽k⩽nkαak=O(np)\sum_{1 \leqslant k \leqslant n} k^{\alpha } a_{k}=O\left(n^{p}\right), then an=O(np−α−1)a_{n}=O\left(n^{p-\alpha -1}\right);

(b)

if ∑1⩽k⩽nkαak=o(np)\sum_{1 \leqslant k \leqslant n} k^{\alpha } a_{k}=o\left(n^{p}\right), then an=o(np−α−1)a_{n}=o\left(n^{p-\alpha -1}\right);

(c)

if ∑k⩾nkαak=O(n−p)\quad \sum_{k \geqslant n} k^{\alpha } a_{k}=O\left(n^{-p}\right), then an=O(n−p−α−1)a_{n}=O\left(n^{-p-\alpha -1}\right);

(d)

if ∑k⩾nkαak=o(n−p)\quad \sum_{k \geqslant n} k^{\alpha } a_{k}=o\left(n^{-p}\right), then an=o(n−p−α−1)a_{n}=o\left(n^{-p-\alpha -1}\right).

Problem VI.3.5

Let p>0,α∈Rp>0, \alpha \in \mathbb {R} and ∑1⩽k⩽nkαxk∼np\sum_{1 \leqslant k \leqslant n} k^{\alpha } x_{k} \sim n^{p}. Prove that

?
(a)

if p>αp>\alpha, then Sn=x1+…+xn∼pp−αnp−αS_{n}=x_{1}+\ldots +x_{n} \sim \frac{p}{p-\alpha } n^{p-\alpha };

(b)

if p=αp=\alpha, then Sn=x1+…+xn∼pln⁡nS_{n}=x_{1}+\ldots +x_{n} \sim p \ln n;

(c)

if p<αp<\alpha, then the series ∑xn\sum x_{n} converges and σn=∑k⩾nxk∼pα−pnp−α\sigma_{n}=\sum_{k \geqslant n} x_{k} \sim \frac{p}{\alpha -p} n^{p-\alpha }.

Problem VI.3.6

Let p>0,α∈Rp>0, \quad \alpha \in \mathbb {R} \quad and \quad the series ∑kαxk\sum k^{\alpha } x_{k} \quad converge, with ∑k⩾nkαxk∼n−p\sum_{k \geqslant n} k^{\alpha } x_{k} \sim n^{-p}. Prove that

?
(a)

if p<−αp<-\alpha, then Sn=x1+…+xn∼p∣p+α∣n∣p+α∣S_{n}=x_{1}+\ldots +x_{n} \sim \frac{p}{\left|p+\alpha \right|} n^{\left|p+\alpha \right|};

(b)

if p=−αp=-\alpha, then Sn=x1+…+xn∼pln⁡nS_{n}=x_{1}+\ldots +x_{n} \sim p \ln n;

(c)

if p>−αp>-\alpha, then the series ∑xn\sum x_{n} converges and σn=∑k⩾nxk∼pp+αn−(p+α)\sigma_{n}=\sum_{k \geqslant n} x_{k} \sim \frac{p}{p+\alpha } n^{-(p+\alpha )}.

Problem VI.3.7

Let an>0a_{n}>0 and ann↑+∞\sqrt[n]{a_{n}} \uparrow +\infty. Prove that

?
(a)

∑1⩽k⩽nkpan∼npan\sum_{1 \leqslant k \leqslant n} k^{p} a_{n} \sim n^{p} a_{n};

(b)

∑k⩾nkpak∼npan\sum_{k \geqslant n} \frac{k^{p}}{a_{k}} \sim \frac{n^{p}}{a_{n}}.

Problem VI.3.8

Let an↓0,Sn=a1+…+an∼n1−p1−pa_{n} \downarrow 0, S_{n}=a_{1}+\ldots +a_{n} \sim \frac{n^{1-p}}{1-p}, where 0<p<10<p<1.

?
(a)

Prove that an∼n−pa_{n} \sim n^{-p}.

(b)

Can one claim that an∼1na_{n} \sim \frac{1}{n}, if Sn∼ln⁡nS_{n} \sim \ln n ?

Problem VI.3.9

Let an↓0a_{n} \downarrow 0 and ∑an<+∞\sum a_{n}<+\infty, with ∑k⩾nak∼n−p\sum_{k \geqslant n} a_{k} \sim n^{-p}, where p>0p>0. Prove that an∼pn−1−pa_{n} \sim p n^{-1-p}.

?
Problem VI.3.10

Let an↓0a_{n} \downarrow 0. Prove that

?
(a)

lim‾⁡ε⋅→+01ln⁡(1/ε)∑an1+ε=lim‾⁡N→∞1ln⁡ln⁡(1/aN)∑1⩽n⩽Nan\varlimsup_{\varepsilon \cdot \rightarrow +0} \frac{1}{\ln (1 / \varepsilon )} \sum a_{n}^{1+\varepsilon }=\varlimsup_{N \rightarrow \infty } \frac{1}{\ln \ln \left(1 / a_{N}\right)} \sum_{1 \leqslant n \leqslant N} a_{n};

(b)

lim⁡‾ε→+0ε∑an1+ε⩽lim⁡‾N→∞1ln⁡(1/aN)∑1⩽n⩽Nan⩽elim⁡‾ε→+0ε∑an1+ε\overline{\lim }_{\varepsilon \rightarrow +0} \varepsilon \sum a_{n}^{1+\varepsilon } \leqslant \overline{\lim }_{N \rightarrow \infty } \frac{1}{\ln \left(1 / a_{N}\right)} \sum_{1 \leqslant n \leqslant N} a_{n} \leqslant e \overline{\lim }_{\varepsilon \rightarrow +0} \varepsilon \sum a_{n}^{1+\varepsilon }.

Can the coefficient ee in inequality b) be decreased?

Problem VI.3.11

Let αn>0,∑αn<+∞\alpha_{n}>0, \sum \alpha_{n}<+\infty and f(t)=t∑αn1+t2αnf(t)=t \sum \frac{\alpha_{n}}{1+t^{2} \alpha_{n}} for t⩾0t \geqslant 0. Prove that the behavior of the function ff as t→+∞t \rightarrow +\infty is closely related to the behavior of the sequence {nαn}\left\{ n \sqrt{\alpha_{n}}\right\} :

π2lim⁡nαn⩽lim⁡t→+∞f(t)⩽lim‾⁡t→+∞f(t)⩽π2lim⁡‾nαn \frac{\pi }{2} \lim n \sqrt{\alpha _{n}} \leqslant \lim _{t \rightarrow +\infty } f(t) \leqslant \varlimsup _{t \rightarrow +\infty } f(t) \leqslant \frac{\pi }{2} \overline{\lim } n \sqrt{\alpha _{n}}

Moreover,

sup⁡t⩾0f(t)⩽π2sup⁡nαn \sup _{t \geqslant 0} f(t) \leqslant \frac{\pi }{2} \sup n \sqrt{\alpha _{n}}

and the coefficient π2\frac{\pi }{2} cannot be decreased.

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Problem VI.3.12

Let f:[0,+∞)→Rf:[0,+\infty ) \rightarrow \mathbb {R} be a nonnegative nonincreasing function, h>0h>0. Prove that

lim⁡h→+0h∑f(kh)=∫0+∞f(t)dt \lim _{h \rightarrow +0} h \sum f(k h)=\int _{0}^{+\infty } f(t) d t

Is this true if the function ff is continuous but not monotonic?

?
Problem VI.3.13

Determine the asymptotics of the following sums as A→+∞A \rightarrow +\infty :

?
(a)

∑k⩾11k2+A2\sum_{k \geqslant 1} \frac{1}{k^{2}+A^{2}};

(b)

∑k⩾1e−k2/A2\sum_{k \geqslant 1} e^{-k^{2} / A^{2}};

(c)

∑k⩾1kp−1kq+Aq(0<p<q)\sum_{k \geqslant 1} \frac{k^{p-1}}{k^{q}+A^{q}}(0<p<q).

Problem VI.3.14

Let the function ff be nonnegative and decreasing on [1,+∞)[1,+\infty ). Prove that

?
(a)

if ∫1+∞f(t)dt=+∞\int_{1}^{+\infty } f(t) d t=+\infty, then

∑1⩽k⩽nf(k)=∫1n+1f(t)dt+C+o(1),0⩽C⩽f(1) \sum _{1 \leqslant k \leqslant n} f(k)=\int _{1}^{n+1} f(t) d t+C+o(1), \quad 0 \leqslant C \leqslant f(1)

in particular ∑1⩽k⩽nf(k)∼∫1nf(t)dt\sum_{1 \leqslant k \leqslant n} f(k) \sim \int_{1}^{n} f(t) d t;

(b)

if ∫1+∞f(t)dt<+∞\int_{1}^{+\infty } f(t) d t<+\infty, then

∑k⩾nf(k)=∫n+∞f(t)dt+O(f(n)) \sum _{k \geqslant n} f(k)=\int _{n}^{+\infty } f(t) d t+O(f(n))

in particular, if f(x)=o(∫x+∞f(t)dt)f(x)=o\left(\int_{x}^{+\infty } f(t) d t\right) as x→+∞x \rightarrow +\infty, then

∑k⩾nf(k)∼∫n+∞f(t)dt \sum _{k \geqslant n} f(k) \sim \int _{n}^{+\infty } f(t) d t
Problem VI.3.15

If the function ff is monotonic on [m,n](m,n∈Z)\left[m, n\right](m, n \in \mathbb {Z}), then

∣∑m⩽k⩽nf(k)−∫mnf(t)dt∣⩽max⁡(∣f(n)∣,∣f(m)∣) \left|\sum _{m \leqslant k \leqslant n} f(k)-\int _{m}^{n} f(t) d t\right| \leqslant \max (\left|f(n)\right|,\left|f(m)\right|)

In particular, if the function ff is monotonic on [1,∞),∫1∞f(t)dt=+∞[1, \infty ), \int_{1}^{\infty } f(t) d t=+\infty and

f(x)=o(∫1xf(t)dt) as x→+∞, then ∑1⩽k⩽nf(k)∼∫1nf(t)dt. f(x)=o\left(\int _{1}^{x} f(t) d t\right) \text{ as } x \rightarrow +\infty , \text{ then } \sum _{1 \leqslant k \leqslant n} f(k) \sim \int _{1}^{n} f(t) d t .
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Problem VI.3.16

Let M,N∈Z,M⩽NM, N \in \mathbb {Z}, M \leqslant N, let the function ff (in general, complex-valued) be integrable on the interval [M−12,N+12]\left[M-\frac{1}{2}, N+\frac{1}{2}\right], and let

Δ=∑M⩽k⩽Nf(k)−∫M−1/2N+1/2f(t)dt \Delta =\sum _{M \leqslant k \leqslant N} f(k)-\int _{M-1 / 2}^{N+1 / 2} f(t) d t
?
(a)

Prove that ∣Δ∣⩽12 VN+12M−12(f)\left|\Delta \right| \leqslant \frac{1}{2} \underset {M-\frac{1}{2}}{\stackrel{N+\frac{1}{2}}{\mathrm{~ V}}}(f). In particular, if ff is monotonic, then

∣Δ∣⩽12∣f(N+12)−f(M−12)∣. \left|\Delta \right| \leqslant \frac{1}{2}\left|f\left(N+\frac{1}{2}\right)-f\left(M-\frac{1}{2}\right)\right| .
(b)

If f∈C1([M−12,N+12])f \in C^{1}\left(\left[M-\frac{1}{2}, N+\frac{1}{2}\right]\right), then

Δ=∫M−1/2N+1/2(t−[t]−12)f′(t)dt \Delta =\int _{M-1 / 2}^{N+1 / 2}\left(t-\left[t\right]-\frac{1}{2}\right) f^{\prime }(t) d t
(c)

If f∈C2([M−12,N+12])f \in C^{2}\left(\left[M-\frac{1}{2}, N+\frac{1}{2}\right]\right), then

Δ=−12∫M−1/2N+1/2(t−[t]−12)2f′′(t)dt \Delta =-\frac{1}{2} \int _{M-1 / 2}^{N+1 / 2}\left(t-\left[t\right]-\frac{1}{2}\right)^{2} f^{\prime \prime }(t) d t

and consequently, ∣Δ∣⩽18∫M−1/2N+1/2∣f′′(t)∣dt\left|\Delta \right| \leqslant \frac{1}{8} \int_{M-1 / 2}^{N+1 / 2}\left|f^{\prime \prime }(t)\right| d t. In particular, if ff is convex or concave, then ∣Δ∣⩽18∣f′(N+12)−f′(M−12)∣\left|\Delta \right| \leqslant \frac{1}{8}\left|f^{\prime }\left(N+\frac{1}{2}\right)-f^{\prime }\left(M-\frac{1}{2}\right)\right|.

Problem VI.3.17

Let f(p)=∑n⩾1(−1)n+1npf(p)=\sum_{n \geqslant 1} \frac{(-1)^{n+1}}{n^{p}} for p>0p>0. Prove that

f(p)=12+O(p). f(p)=\frac{1}{2}+O(p) .
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Problem VI.3.18

Prove that as p→+0p \rightarrow +0

∑n⩾11np+1=1p+γ+o(1)(γ− Euler’s constant). \sum _{n \geqslant 1} \frac{1}{n^{p+1}}=\frac{1}{p}+\gamma +o(1) \quad (\gamma -\text{ Euler's constant}) .
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Problem VI.3.19

Prove that

∏1⩽k⩽n(1+1k)∼Aexp⁡(2n)n,∏1⩽k⩽n(1+1k3)∼Bn3exp⁡(32(n2/3−n1/3)) \begin{aligned} \prod _{1 \leqslant k \leqslant n}\left(1+\frac{1}{\sqrt{k}}\right) & \sim A \frac{\exp (2 \sqrt{n})}{\sqrt{n}}, \\ \prod _{1 \leqslant k \leqslant n}\left(1+\frac{1}{\sqrt[3]{k}}\right) & \sim B \sqrt[3]{n} \exp \left(\frac{3}{2}\left(n^{2 / 3}-n^{1 / 3}\right)\right) \end{aligned}

for some positive numbers AA and BB.

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Problem VI.3.20

Let α∈R\N\alpha \in \mathbb {R} \backslash \mathbb {N}. Prove that

(n−αn)=(n−α)(n−α−1)…(1−α)n!∼Cαnα(Cα≠0). \binom {n-\alpha }{n}=\frac{(n-\alpha )(n-\alpha -1) \ldots (1-\alpha )}{n!} \sim \frac{C_{\alpha }}{n^{\alpha }} \quad \left(C_{\alpha } \neq 0\right) .
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Problem VI.3.21

Prove that

?
(a)

1⋅22⋅33…nnn2∼ne4\sqrt[n^{2}]{1 \cdot 2^{2} \cdot 3^{3} \ldots n^{n}} \sim \frac{\sqrt{n}}{\sqrt[4]{e}};

(b)

1!⋅(2!)2⋅(3!)3…(n!)nn3∼n3e4/9\sqrt[n^{3}]{1!\cdot (2!)^{2} \cdot (3!)^{3} \ldots (n!)^{n}} \sim \frac{\sqrt[3]{n}}{e^{4 / 9}};

(c)

1⋅22⋅33…nnn∼(ne)n/2n\sqrt[n]{1 \cdot 2^{2} \cdot 3^{3} \ldots n^{n}} \sim \left(\frac{n}{\sqrt{e}}\right)^{n / 2} \sqrt{n}.

Problem VI.3.22

Let α>0\alpha >0. Prove that as n→+∞(n∈N)n \rightarrow +\infty (n \in \mathbb {N})

?
(a)

∑1⩽k⩽nkαk/n∼n1+ααln⁡n\sum_{1 \leqslant k \leqslant n} k^{\alpha k / n} \sim \frac{n^{1+\alpha }}{\alpha \ln n};

(b)

∑1⩽k⩽nk−αk/n∼nαln⁡n\sum_{1 \leqslant k \leqslant n} k^{-\alpha k / n} \sim \frac{n}{\alpha \ln n};

(c)

∑1⩽k⩽n(k!)−α/n∼nαln⁡n\sum_{1 \leqslant k \leqslant n}(k!)^{-\alpha / n} \sim \frac{n}{\alpha \ln n};

(d)

∑1⩽k⩽nkαn∼nαn1−e−α\sum_{1 \leqslant k \leqslant n} k^{\alpha n} \sim \frac{n^{\alpha n}}{1-e^{-\alpha }}.

Problem VI.3.23

Prove that ∑1⩽k⩽nnkk!∼12en\sum_{1 \leqslant k \leqslant n} \frac{n^{k}}{k!} \sim \frac{1}{2} e^{n}.

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Problem VI.3.24

Prove that

?
(a)

∑1⩽k⩽nCnk∼2πn42n/2\sum_{1 \leqslant k \leqslant n} \sqrt{C_{n}^{k}} \sim \sqrt[4]{2 \pi n} 2^{n / 2};

(b)

∑1⩽k⩽nCnk∣2k−n∣∼Γ(34)π2n42n\sum_{1 \leqslant k \leqslant n} C_{n}^{k} \sqrt{\left|2 k-n\right|} \sim \frac{\Gamma \left(\frac{3}{4}\right)}{\sqrt{\pi }} \sqrt[4]{2 n} 2^{n}.

Problem VI.3.25

Let N∈NN \in \mathbb {N} and x>0x>0. Prove that

∑n⩾Ne−xn=(2xN−12+2x2+O(xN))e−xN−1/2 \sum _{n \geqslant N} e^{-x \sqrt{n}}=\left(\frac{2}{x} \sqrt{N-\frac{1}{2}}+\frac{2}{x^{2}}+O\left(\frac{x}{\sqrt{N}}\right)\right) e^{-x \sqrt{N-1 / 2}}

where the constant in the OO-estimate does not depend on xx and NN.

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Problem VI.3.26

Let the sequence of numbers {xn}\left\{ x_{n}\right\} be such that xnn→C\frac{x_{n}}{\sqrt{n}} \rightarrow C, 0⩽C⩽+∞0 \leqslant C \leqslant +\infty. Find the asymptotics of the sum

αN=∑n⩾Ne−xNn \alpha _{N}=\sum _{n \geqslant N} e^{-x_{N} \sqrt{n}}

in the following cases:

?
(a)

C=0C=0;

(b)

C=+∞C=+\infty;

(c)

0<C<+∞0<C<+\infty.

Problem VI.3.27

Let p>0p>0 and S(x)=∑k=0[x](x−k)pS(x)=\sum_{k=0}^{\left[x\right]}(x-k)^{p}. Prove that for some C>0C>0, for all x>0x>0 the inequality holds

∣S(x)−1p+1xp+1∣⩽Cxp. \left|S(x)-\frac{1}{p+1} x^{p+1}\right| \leqslant C x^{p} .
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Problem VI.3.28

Let ψ∈C1([a,+∞)),ψ,ψ′>0\psi \in C^{1}([a,+\infty )), \psi , \psi^{\prime }>0 and ψ=O(ψ′)\psi =O\left(\psi^{\prime }\right) on [a,+∞)[a,+\infty ). Prove that

∫a+∞e−εψ(t)dt=ψ−1(1ε)+O(1) as ε→+0 \int _{a}^{+\infty } e^{-\varepsilon \psi (t)} d t=\psi ^{-1}\left(\frac{1}{\varepsilon }\right)+O(1) \quad \text{ as } \quad \varepsilon \rightarrow +0
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Problem VI.3.29

Find the asymptotics as t→1−0t \rightarrow 1-0 of the sums of the series

?
(a)

∑1ntln⁡pn(p>0)\sum \frac{1}{n} t^{\ln^{p} n}(p>0);

(b)

∑tln⁡pn(p>1)\sum t^{\ln^{p} n} \quad (p>1);

(c)

∑tnp(p>0)\sum t^{n^{p}} \quad (p>0);

(d)

∑tan(a>1)\sum t^{a^{n}} \quad (a>1);

(e)

∑t(an)n(a>0)\sum t^{(a n)^{n}}(a>0);

(f)

∑tn!\sum t^{n!}.

Problem VI.3.30

Find the asymptotics as t→1−0t \rightarrow 1-0 of the sums of the series

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(a)

∑nptn(p>−1)\sum n^{p} t^{n}(p>-1);

(b)

∑n⩾2(ln⁡n)ptn(p∈R)\sum_{n \geqslant 2}(\ln n)^{p} t^{n}(p \in \mathbb {R}).

Problem VI.3.31

Find the asymptotics as t→1−0t \rightarrow 1-0 of the sums of the series

?
(a)

∑tn1+tn\sum \frac{t^{n}}{1+t^{n}};

(b)

∑ntn1+tn\sum \frac{n t^{n}}{1+t^{n}};

(c)

∑tn(1+tn)2\quad \sum \frac{t^{n}}{\left(1+t^{n}\right)^{2}};

(d)

∑ntn(1+tn)2\quad \sum \frac{n t^{n}}{\left(1+t^{n}\right)^{2}};

(e)

∑tn1−tn\sum \frac{t^{n}}{1-t^{n}};

(f)

∑ntn1−tn\sum \frac{n t^{n}}{1-t^{n}};

(g)

∑tn(1−tn)2\sum \frac{t^{n}}{\left(1-t^{n}\right)^{2}};

(h)

∑ntn(1−tn)2\sum \frac{n t^{n}}{\left(1-t^{n}\right)^{2}};

(i)

∑(−1)ntn1−tn\quad \sum \frac{(-1)^{n} t^{n}}{1-t^{n}};

(j)

∑(−1)nntn1−tn\quad \sum \frac{(-1)^{n} n t^{n}}{1-t^{n}};

(k)

∑(−1)ntn(1−tn)2\quad \sum \frac{(-1)^{n} t^{n}}{\left(1-t^{n}\right)^{2}};

(l)

∑(−1)nntn(1−tn)2\quad \sum \frac{(-1)^{n} n t^{n}}{\left(1-t^{n}\right)^{2}}.

Problem VI.3.32

Find the asymptotics as t→1−0t \rightarrow 1-0 of the sums of the series

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(a)

∑tn1+t2n\sum \frac{t^{n}}{1+t^{2 n}};

(b)

∑tn(1+t2n)2\sum \frac{t^{n}}{\left(1+t^{2 n}\right)^{2}};

(c)

∑ntn1−t2n\quad \sum \frac{n t^{n}}{1-t^{2 n}};

(d)

∑tn1−t2n\quad \sum \frac{t^{n}}{1-t^{2 n}};

(e)
Problem VI.3.33

Let τ(k)\tau (k) be the number of divisors of the natural number kk. Prove that

∑1⩽k⩽nτ(k)=n(ln⁡n+2γ−1)+O(n). \sum _{1 \leqslant k \leqslant n} \tau (k)=n(\ln n+2 \gamma -1)+O(\sqrt{n}) .

Here γ\gamma is Euler's constant (see problem II.2.8).

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Problem VI.3.34

Find the asymptotics as t→1−0t \rightarrow 1-0 of the sums of the series

?
(a)

T(t)=∑τ(n)tnT(t)=\sum \tau (n) t^{n},

(b)

S(t)=∑σ(n)tnS(t)=\sum \sigma (n) t^{n}, (τ(n)(\tau (n) is the number of divisors, σ(n)\sigma (n) is the sum of divisors of the number nn).

Problem VI.3.35

Let a smooth function φ\varphi be convex or concave on [1,+∞)[1,+\infty ), lim⁡x→+∞φ(x)=+∞\lim_{x \rightarrow +\infty } \varphi (x)=+\infty. Prove that

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(a)

if ln⁡φ(x)=o(x)\ln \varphi (x)=o(x) as x→+∞x \rightarrow +\infty, then

∑φ(n)tn∼11−t∑φ′(n)tn(t→1−0); \sum \varphi (n) t^{n} \sim \frac{1}{1-t} \sum \varphi ^{\prime }(n) t^{n} \quad (t \rightarrow 1-0) ;
(b)

if φ(x)=O(x)\varphi (x)=O(x) as x→+∞x \rightarrow +\infty, then

∑φ(n)tn=11−t(∑φ′(n)tn+C+o(1))(t→1−0), \sum \varphi (n) t^{n}=\frac{1}{1-t}\left(\sum \varphi ^{\prime }(n) t^{n}+C+o(1)\right) \quad (t \rightarrow 1-0),

where C=lim⁡(φ(n)−(φ′(1)+…+φ′(n)))C=\lim \left(\varphi (n)-\left(\varphi^{\prime }(1)+\ldots +\varphi^{\prime }(n)\right)\right). In particular, taking φ(x)=ln⁡x\varphi (x)=\ln x, we obtain a refinement of the result of problem 3.30 b) with p=1p=1:

∑tnln⁡n=∣ln⁡(1−t)∣−γ+o(1) \sum t^{n} \ln n=\left|\ln (1-t)\right|-\gamma +o(1)
Problem VI.3.36

Find the asymptotics as A→+∞A \rightarrow +\infty of the sums (C>0C>0)

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(a)

∑e−Cn(1−1n2)A\sum e^{-C n}\left(1-\frac{1}{n^{2}}\right)^{A};

(b)

∑e−Cn(1−1n)A\sum e^{-C n}\left(1-\frac{1}{n}\right)^{A}.

Problem VI.3.37

Let 1<b<a1<b<a. Prove that as s→+∞s \rightarrow +\infty

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(a)

∑n⩾1bnan+as∼(ba)sφ(s)\sum_{n \geqslant 1} \frac{b^{n}}{a^{n}+a^{s}} \sim \left(\frac{b}{a}\right)^{s} \varphi (s);

(b)

∑n⩾11ansh⁡(bs−n)∼ψ(s)as\sum_{n \geqslant 1} \frac{1}{a^{n} \operatorname {sh}\left(b^{s-n}\right)} \sim \frac{\psi (s)}{a^{s}},

where φ,ψ−1\varphi , \psi -1-periodic positive smooth functions:

φ(s)=∑k∈Zbk−s1+ak−s,ψ(s)=∑k∈Zak−ssh⁡(bk−s). \varphi (s)=\sum _{k \in \mathbb {Z}} \frac{b^{k-s}}{1+a^{k-s}}, \quad \psi (s)=\sum _{k \in \mathbb {Z}} \frac{a^{k-s}}{\operatorname {sh}\left(b^{k-s}\right)} .

What is the mean value of the function ψ\psi when a=b2a=b^{2}?

Problem VI.3.38

Find the asymptotics of the sum

Sm=∑n⩾012n(1−12n)2m S_{m}=\sum _{n \geqslant 0} \frac{1}{2^{n}}\left(1-\frac{1}{2^{n}}\right)^{2^{m}}
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Problem VI.3.39

Prove that

max⁡x12+…+xn2=1∑k=1n(x1+x22+…+xkk)2∼n→+∞nln⁡n. \max _{x_{1}^{2}+\ldots +x_{n}^{2}=1} \sum _{k=1}^{n}\left(x_{1}+\frac{x_{2}}{\sqrt{2}}+\ldots +\frac{x_{k}}{\sqrt{k}}\right)^{2} \underset {n \rightarrow +\infty }{\sim } n \ln n .
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Problem VI.3.40

Let s>1,{pk}s>1,\left\{ p_{k}\right\} be the sequence of all prime numbers, numbered in increasing order. For an odd index 2n+12 n+1 set φ(2n+1)=(−1)n\varphi (2 n+1)=(-1)^{n}.

Arguing in the same way as when solving problems I.2.11 b) and I.2.12 b), prove that

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(a)

∏k⩾2(1−φ(pk)pks)−1=∑n⩾0(−1)n(2n+1)s\prod_{k \geqslant 2}\left(1-\frac{\varphi \left(p_{k}\right)}{p_{k}^{s}}\right)^{-1}=\sum_{n \geqslant 0} \frac{(-1)^{n}}{(2 n+1)^{s}};

(b)

∑k⩾1φ(pk)pks=O(1)\sum_{k \geqslant 1} \frac{\varphi \left(p_{k}\right)}{p_{k}^{s}}=O(1).

(c)

Using the result of problem I.2.12b), derive from this the asymptotic relations

∑pk=1 mod 41pks∼s→112ln⁡1s−1,∑pk=−1 mod 41pks∼s→112ln⁡1s−1. \sum _{p_{k}=1 \bmod 4} \frac{1}{p_{k}^{s}} \underset {s \rightarrow 1}{\sim } \frac{1}{2} \ln \frac{1}{s-1}, \quad \sum _{p_{k}=-1 \bmod 4} \frac{1}{p_{k}^{s}} \underset {s \rightarrow 1}{\sim } \frac{1}{2} \ln \frac{1}{s-1} .

From part c) it follows that each of the arithmetic progressions {4m±1}\left\{ 4 m \pm 1\right\} contains infinitely many primes. This is a special case of the celebrated Dirichlet theorem, according to which any arithmetic progression {am+b}\left\{ a m+b\right\} with coprime natural numbers aa and bb contains infinitely many primes (see [GL], [Ka], [Tr]). Interesting information related to this theorem can be found in [Kn][\mathrm{Kn}]. It is also apparent that the progressions {4m±1}\left\{ 4 m \pm 1\right\} contain "approximately equal numbers" of primes. This result (as well as problem I.2.15) can be supplemented by the following asymptotic relations.

Problem VI.3.41

Prove that

∑pk=1mod4pk⩽N1pk∼N→∞ln⁡ln⁡N2,∑pk=−1mod4pk⩽N1pk∼N→∞ln⁡ln⁡N2. \sum _{\substack {p_{k}=1 \bmod 4 \\ p_{k} \leqslant N}} \frac{1}{p_{k}} \underset {N \rightarrow \infty }{\sim } \frac{\ln \ln N}{2}, \quad \sum _{\substack {p_{k}=-1 \bmod 4 \\ p_{k} \leqslant N}}^{\frac{1}{p_{k}}} \underset {N \rightarrow \infty }{\sim } \frac{\ln \ln N}{2} .
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Section VI.4Asymptotics of Implicit Functions and Recurrent Sequences[19/89%]
Problem VI.4.1

Let y(x)=∫x+∞e−t2/2dt(x∈R),x(y)y(x)=\int_{x}^{+\infty } e^{-t^{2} / 2} d t(x \in \mathbb {R}), x(y) be the function inverse to y(x)y(x). Prove that x(y)∼y→+02ln⁡1yx(y) \underset {y \rightarrow +0}{\sim } \sqrt{2 \ln \frac{1}{y}}.

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Problem VI.4.2

Prove that each of the following equations defines, in some neighborhood of the point (x0,c0)\left(x_{0}, c_{0}\right), an infinitely differentiable implicit function yy, and find the first nn coefficients of the expansion y(x)=c0+c1(x−x0)+…+cn(x−x0)n+…y(x)=c_{0}+c_{1}\left(x-x_{0}\right)+\ldots +c_{n}\left(x-x_{0}\right)^{n}+\ldots

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(a)

yey−x=0y e^{y}-x=0, x0=c0=0,n=3;x_{0}=c_{0}=0, \quad n=3 ;

(b)

y2+ln⁡2y−x=0y^{2}+\ln^{2} y-x=0, x0=c0=1,n=3x_{0}=c_{0}=1, \quad n=3;

(c)

xy+exy−x=0x y+e^{x y}-x=0, x0=1,c0=0x_{0}=1, c_{0}=0, n=3;n=3 ;

(d)

yln⁡y−x=0y \ln y-x=0, x0=0,c0=1x_{0}=0, c_{0}=1, n=3;n=3 ;

(e)

exy+x2+y−1=0e^{x y}+x^{2}+y-1=0, x0=c0=0x_{0}=c_{0}=0, n=6;n=6 ;

(f)

arctg⁡(x+y)−x−2y=0\operatorname {arctg}(x+y)-x-2 y=0, x0=c0=0x_{0}=c_{0}=0, n=6n=6.

Problem VI.4.3

Verify that the domain of definition of the implicit functions considered in problem 4.2 contains the half-line [x0,+∞)\left[x_{0},+\infty \right), and prove that as x→+∞x \rightarrow +\infty

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(a)

y(x)=ln⁡x−ln⁡ln⁡x+c1a1(x)+c2a2(x)+O(a2(x)ln⁡ln⁡x)y(x)=\ln x-\ln \ln x+c_{1} a_{1}(x)+c_{2} a_{2}(x)+O\left(\frac{a_{2}(x)}{\ln \ln x}\right);

(b)

y(x)=x(c0+c1a1(x)+c2a2(x)+O(a2(x)ln⁡x))y(x)=\sqrt{x}\left(c_{0}+c_{1} a_{1}(x)+c_{2} a_{2}(x)+O\left(\frac{a_{2}(x)}{\ln x}\right)\right);

(c)

y(x)=ln⁡xx+1x(c1a1(x)+c2a2(x)+O(a2(x)ln⁡x))y(x)=\frac{\ln x}{x}+\frac{1}{x}\left(c_{1} a_{1}(x)+c_{2} a_{2}(x)+O\left(\frac{a_{2}(x)}{\ln x}\right)\right);

(d)

y(x)=xln⁡x(c0+c1a1(x)+c2a2(x)+O(a2(x)ln⁡ln⁡x))y(x)=\frac{x}{\ln x}\left(c_{0}+c_{1} a_{1}(x)+c_{2} a_{2}(x)+O\left(\frac{a_{2}(x)}{\ln \ln x}\right)\right), where c0,c1,c2c_{0}, c_{1}, c_{2} are constants, and aka_{k} are functions of the form (ln⁡x)mxn\frac{(\ln x)^{m}}{x^{n}} or (ln⁡ln⁡x)mln⁡nx\frac{(\ln \ln x)^{m}}{\ln^{n} x} for suitable m,n∈Nm, n \in \mathbb {N};

(e)

y(x)=−x2+1−ex−x3+O(xe2(x−x3))y(x)=-x^{2}+1-e^{x-x^{3}}+O\left(x e^{2\left(x-x^{3}\right)}\right);

(f)

y(x)=−x2+π4−1x+O(1x2)y(x)=-\frac{x}{2}+\frac{\pi }{4}-\frac{1}{x}+O\left(\frac{1}{x^{2}}\right).

Problem VI.4.4

Prove that the equation yx+y=1y^{x}+y=1 defines, for all x>0x>0, a unique implicit function yy and that

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(a)

y∈C∞((0,∞)),0<y<1,y′>0y \in C^{\infty }((0, \infty )), 0<y<1, y^{\prime }>0;

(b)

y(x)=x∣ln⁡x∣+xln⁡∣ln⁡x∣+o(x)y(x)=x\left|\ln x\right|+x \ln \left|\ln x\right|+o(x) as x→+0x \rightarrow +0;

(c)

y(x)=1−ln⁡xx+ln⁡ln⁡xx+o(1x)y(x)=1-\frac{\ln x}{x}+\frac{\ln \ln x}{x}+o\left(\frac{1}{x}\right) as x→+∞x \rightarrow +\infty.

Problem VI.4.5

How many continuous solutions yy does the equation x5+x2y=y3x^{5}+x^{2} y=y^{3} have that tend to zero as x→0x \rightarrow 0? What is their asymptotic behavior as x→0x \rightarrow 0?

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Problem VI.4.6

Prove that Kepler's equation y=M+xsin⁡yy=M+x \sin y, where MM is a fixed positive parameter, defines in a neighborhood of the point (0,M)(0, M) an infinitely differentiable function yy, and moreover

y(x)=M+xsin⁡M+12x2sin⁡2M+O(x3) as x→0. y(x)=M+x \sin M+\frac{1}{2} x^{2} \sin 2 M+O\left(x^{3}\right) \quad \text{ as } \quad x \rightarrow 0 .
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Problem VI.4.7

The problem of "the globe in a string bag." The Earth's globe was girdled with an inextensible thread. Then the thread was lengthened by 1 cm and pulled taut by tugging at the knot. Compute the height to which it rose, with a relative error of no more than one ten-thousandth of a percent (take the radius of the globe to be 6400 km6400\ \text{km}).

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Problem VI.4.8

Find the derivatives with respect to ε\varepsilon of the lengths of the semi-axes of the ellipsoid x2+y2+z2+xy+yz+zx=1+εxyx^{2}+y^{2}+z^{2}+x y+y z+z x=1+\varepsilon x y at ε=0\varepsilon =0.

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Problem VI.4.9

Let xnx_{n} be the root of the equation x=tg⁡xx=\operatorname {tg} x lying in the interval (πn,πn+π\pi n, \pi n+\pi), n∈Nn \in \mathbb {N}. Prove that

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(a)

xn=π(n+12)−1πn+O(1n2)x_{n}=\pi \left(n+\frac{1}{2}\right)-\frac{1}{\pi n}+O\left(\frac{1}{n^{2}}\right);

(b)

−1π(n+12)−1(πn)3<xn−π(n+12)<−1π(n+12)-\frac{1}{\pi \left(n+\frac{1}{2}\right)}-\frac{1}{(\pi n)^{3}}<x_{n}-\pi \left(n+\frac{1}{2}\right)<-\frac{1}{\pi \left(n+\frac{1}{2}\right)}.

Problem VI.4.10

Let xn=h(xn−1)x_{n}=h\left(x_{n-1}\right), where h(x)=x(1+axp)1/ph(x)=\frac{x}{\left(1+a x^{p}\right)^{1 / p}} for x⩾0x \geqslant 0 (a,pa, p are positive constants, n∈N,x0>0n \in \mathbb {N}, x_{0}>0). Prove that xn∼(an)−1/px_{n} \sim (a n)^{-1 / p}.

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Problem VI.4.11

Let x0=10−3,xn=xn−1−xn−12(n∈N)x_{0}=10^{-3}, x_{n}=x_{n-1}-x_{n-1}^{2} \quad (n \in \mathbb {N}). Prove that x1000≈1210−3x_{1000} \approx \frac{1}{2} 10^{-3} with accuracy 10−610^{-6}.

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Problem VI.4.12

Find the asymptotics of the recurrent sequences xn=f(xn−1)(n∈N,x0>0)x_{n}=f\left(x_{n-1}\right)\left(n \in \mathbb {N}, x_{0}>0\right) in the following cases:

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(a)

f(x)=x(1−x),x0<1f(x)=x(1-x), x_{0}<1;

(b)

f(x)=sin⁡x,x0<πf(x)=\sin x, x_{0}<\pi;

(c)

f(x)=x1+x3f(x)=\frac{x}{1+x^{3}};

(d)

f(x)=arctg⁡xf(x)=\operatorname {arctg} x;

(e)

f(x)=ln⁡(1+x)f(x)=\ln (1+x);

(f)

f(x)=1−e−xf(x)=1-e^{-x};

(g)

f(x)=x1+xf(x)=\frac{x}{1+\sqrt{x}};

(h)

f(x)=x1+x4f(x)=\frac{x}{\sqrt{1+x^{4}}}.

Problem VI.4.13

Let C,p,x0>0C, p, x_{0}>0 and xn=xn−1+Cxn−11−p(n∈N)x_{n}=x_{n-1}+C x_{n-1}^{1-p}(n \in \mathbb {N}). Prove that

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(a)

xn∼(Cpn)1/px_{n} \sim (C p n)^{1 / p};

(b)

xn=(Cpn)1/p(1+p−12p2ln⁡nn+O(1n))x_{n}=(C p n)^{1 / p}\left(1+\frac{p-1}{2 p^{2}} \frac{\ln n}{n}+O\left(\frac{1}{n}\right)\right).

Problem VI.4.14

Construct a function ff, positive on (0,+∞0,+\infty), such that the sequence xn=f(xn−1)(n∈N,x0>0)x_{n}=f\left(x_{n-1}\right)\left(n \in \mathbb {N}, x_{0}>0\right) satisfies the relation xn∼1ln⁡nx_{n} \sim \frac{1}{\ln n}.

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Problem VI.4.15

Find the asymptotics of the recurrent sequences in the following cases (n∈N,x0>0n \in \mathbb {N}, x_{0}>0):

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(a)

xn=xn−1+ln⁡(n+xn−12)x_{n}=x_{n-1}+\ln \left(n+x_{n-1}^{2}\right);

(b)

xn=xn−1+ln⁡nxn−12x_{n}=x_{n-1}+\frac{\ln n}{x_{n-1}^{2}}.

Problem VI.4.16

Let x0∈(−1,2),xn=xn−1(xn−1−1)(n∈N)\quad x_{0} \in (-1,2), \quad x_{n}=x_{n-1}\left(x_{n-1}-1\right) \quad (n \in \mathbb {N}). Find the asymptotics of the sequence {xn}\left\{ x_{n}\right\}.

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Problem VI.4.17

Let x0=1,xn=xn−1+(∑0⩽k<nxk)−1(n∈N)x_{0}=1, x_{n}=x_{n-1}+\left(\sum_{0 \leqslant k<n} x_{k}\right)^{-1}(n \in \mathbb {N}). Prove that xn∼2ln⁡nx_{n} \sim \sqrt{2 \ln n}.

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Problem VI.4.18

Let ∣p∣<1,x0=1,xn=xn−1+(∑0⩽k<nxk)−p(n∈N)\left|p\right|<1, x_{0}=1, x_{n}=x_{n-1}+\left(\sum_{0 \leqslant k<n} x_{k}\right)^{-p}(n \in \mathbb {N}). Find the asymptotics of the sequence {xn}\left\{ x_{n}\right\}.

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Problem VI.4.19

Let p∈R,x0=1,xn=xn−1+(∑0⩽k<nxkp)−1(n∈N)p \in \mathbb {R}, x_{0}=1, x_{n}=x_{n-1}+\left(\sum_{0 \leqslant k<n} x_{k}^{p}\right)^{-1}(n \in \mathbb {N}). Find the asymptotics of the sequence {xn}\left\{ x_{n}\right\}.

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