Section VI.2

Laplace's Method

[23/100%]
Show
LaTeX
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
?
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

?
(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
?
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:

?
(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.
?

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.
?

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

?
(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}}}
?
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
?
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).

?
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 \%.

?
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
?
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})?

?
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 .
?
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 .
?
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}).

?
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
?
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
?