Subsection 3.13.2

Problems

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Problem 3.13.1

Prove that the function I(α)I(\alpha ) is continuous on RR, if:

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(1)

I(α)=∫01sin⁡2αx2dxI(\alpha ) = \int_{0}^{1} \sin^{2} \alpha x^{2} d x;

(2)

I(α)=∫−12x21+x2+α2x4dxI(\alpha ) = \int_{-1}^{2} \frac{x^{2}}{1+x^{2}+\alpha^{2} x^{4}} d x.

Problem 3.13.2

Find the limit:

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(1)

lim⁡α→0∫011+α2x4dx\lim_{\alpha \rightarrow 0} \int_{0}^{1} \sqrt{1+\alpha^{2} x^{4}} d x;

(2)

lim⁡α→0∫−11x2+α2dx\lim_{\alpha \rightarrow 0} \int_{-1}^{1} \sqrt{x^{2}+\alpha^{2}} d x;

(3)

lim⁡α→1∫24xdx1+x2+α6\lim_{\alpha \rightarrow 1} \int_{2}^{4} \frac{x d x}{1+x^{2}+\alpha^{6}};

(4)

lim⁡α→1∫01x2eαx3dx\lim_{\alpha \rightarrow 1} \int_{0}^{1} x^{2} e^{\alpha x^{3}} d x;

(5)

lim⁡α→0∫0πxcos⁡(1+α)xdx\lim_{\alpha \rightarrow 0} \int_{0}^{\pi } x \cos (1+\alpha ) x d x. on RR.

Problem 3.13.3

Prove that the function I(α)=∫01sign⁡(x−α)dxI(\alpha ) = \int_{0}^{1} \operatorname {sign}(x-\alpha ) d x is continuous RR.

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Problem 3.13.4

Let the function f(x)f(x) be continuous and take positive values on the segment [0;1][0 ; 1]. Prove that the function

I(α)=∫01αx2+α2f(x)dx I(\alpha ) = \int _{0}^{1} \frac{\alpha }{x^{2}+\alpha ^{2}} f(x) d x

is discontinuous at α=0\alpha = 0.

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Problem 3.13.5

Determine whether the equality holds

lim⁡α→0∫01f(x;α)dx=∫01lim⁡α→0f(x;α)dx \lim _{\alpha \rightarrow 0} \int _{0}^{1} f(x ; \alpha ) d x = \int _{0}^{1} \lim _{\alpha \rightarrow 0} f(x ; \alpha ) d x

if:

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(1)

f(x;α)=xα2e−x2/a2f(x ; \alpha ) = \frac{x}{\alpha^{2}} e^{-x^{2} / a^{2}};

(2)

f(x;α)=2xa2(α2+x2)2f(x ; \alpha ) = \frac{2 x a^{2}}{\left(\alpha^{2}+x^{2}\right)^{2}}.

Problem 3.13.6

Let the function f(x)f(x) be continuous on the segment [a;b][a ; b] and a<a0<x<ba < a_{0} < x < b. Prove that

lim⁡α→01α∫α0x[f(t+α)−f(t)]dt=f(x)−f(a0) \lim _{\alpha \rightarrow 0} \frac{1}{\alpha } \int _{\alpha _{0}}^{x}[f(t+\alpha )-f(t)] d t = f(x)-f\left(a_{0}\right)
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Problem 3.13.7

Let {φn(x)}\left\{ \varphi_{n}(x)\right\} be a sequence of functions, Riemann integrable on the segment [a;b][a ; b] and taking nonnegative values on this segment, converging uniformly to zero on the set E={x:0<δ⩽∣x∣⩽1}E = \left\{ x: 0 < \delta \leqslant \left|x\right| \leqslant 1\right\} for every δ>0\delta > 0, and moreover

lim⁡n→∞∫−11φn(x)dx=1 \lim _{n \rightarrow \infty } \int _{-1}^{1} \varphi _{n}(x) d x = 1

Prove that for any function f(x)f(x) continuous on the segment [−1;1][-1 ; 1], the equality holds

lim⁡n→∞∫−11f(x)φn(x)dx=f(0) \lim _{n \rightarrow \infty } \int _{-1}^{1} f(x) \varphi _{n}(x) d x = f(0)
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Problem 3.13.8

Determine whether the integrals are equal

∫01(∫01f(x;α)dα)dx and ∫01(∫01f(x;α)dx)dα \int _{0}^{1}\left(\int _{0}^{1} f(x ; \alpha ) d \alpha \right) d x \quad \text{ and } \quad \int _{0}^{1}\left(\int _{0}^{1} f(x ; \alpha ) d x\right) d \alpha

if:

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(1)

f(x;α)=α2−x2(α2+x2)2f(x ; \alpha ) = \frac{\alpha^{2}-x^{2}}{\left(\alpha^{2}+x^{2}\right)^{2}};

(2)

f(x;α)=α−x(α+x)3f(x ; \alpha ) = \frac{\alpha -x}{(\alpha +x)^{3}};

(3)

f(x;α)=(x5α4−2x3α3)e−x2/αf(x ; \alpha ) = \left(\frac{x^{5}}{\alpha^{4}}-\frac{2 x^{3}}{\alpha^{3}}\right) e^{-x^{2} / \alpha }.

Problem 3.13.9

Let the function f(x;α)f(x ; \alpha ), for each α∈[α1;α2]\alpha \in \left[\alpha_{1} ; \alpha_{2}\right], be integrable in xx on the segment [a;b][a ; b], and suppose on this segment there exists a function φ(x)\varphi (x) such that lim⁡α→α0f(x;α)=φ(x)\lim_{\alpha \rightarrow \alpha_{0}} f(x ; \alpha ) = \varphi (x), where α0∈[α1;α2]\alpha_{0} \in \left[\alpha_{1} ; \alpha_{2}\right], uniformly with respect to x∈[a;b]x \in [a ; b]. Prove that:

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(1)

lim⁡α→α0∫abf(x;α)dx=∫abφ(x)dx\lim_{\alpha \rightarrow \alpha_{0}} \int_{a}^{b} f(x ; \alpha ) d x = \int_{a}^{b} \varphi (x) d x;

(2)

lim⁡α→α0∫abf(x;α)g(x)dx=∫abφ(x)g(x)dx\lim_{\alpha \rightarrow \alpha_{0}} \int_{a}^{b} f(x ; \alpha ) g(x) d x = \int_{a}^{b} \varphi (x) g(x) d x, where g(x)g(x) is a function integrable on the segment [a;b][a ; b].

Problem 3.13.10

Using the formula arctg⁡xx=∫01dα1+x2α2\frac{\operatorname {arctg} x}{x} = \int_{0}^{1} \frac{d \alpha }{1+x^{2} \alpha^{2}}, calculate the integral

∫01arctg⁡xx1−x2dx \int _{0}^{1} \frac{\operatorname {arctg} x}{x \sqrt{1-x^{2}}} d x
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Problem 3.13.11

Using the formula

1sin⁡xln⁡a+bsin⁡xa−bsin⁡x=2ab∫01dta2−b2t2sin⁡2x \frac{1}{\sin x} \ln \frac{a+b \sin x}{a-b \sin x} = 2 a b \int _{0}^{1} \frac{d t}{a^{2}-b^{2} t^{2} \sin ^{2} x}

where a>b>0a > b > 0, calculate the integral

∫0π/2ln⁡a+bsin⁡xa−bsin⁡xdxsin⁡x \int _{0}^{\pi / 2} \ln \frac{a+b \sin x}{a-b \sin x} \frac{d x}{\sin x}
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Problem 3.13.12

Let a>0,b>0a > 0, b > 0. Calculate the integral:

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(1)

∫01sin⁡(ln⁡1x)xb−xaln⁡xdx\int_{0}^{1} \sin \left(\ln \frac{1}{x}\right) \frac{x^{b}-x^{a}}{\ln x} d x;

(2)

∫01cos⁡(ln⁡1x)xb−xaln⁡xdx\int_{0}^{1} \cos \left(\ln \frac{1}{x}\right) \frac{x^{b}-x^{a}}{\ln x} d x.

Problem 3.13.13

Find I′(α)I^{\prime }(\alpha ), if:

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(1)

I(α)=∫01sin⁡(αx)dx;I(\alpha ) = \int_{0}^{1} \sin (\alpha x) d x ;

(2)

I(α)=∫13cos⁡(αx3)xdxI(\alpha ) = \int_{1}^{3} \frac{\cos \left(\alpha x^{3}\right)}{x} d x;

(3)

I(α)=∫12eαx2dxxI(\alpha ) = \int_{1}^{2} e^{\alpha x^{2}} \frac{d x}{x};

(4)

I(α)=∫23ch⁡(α4x2)dxxI(\alpha ) = \int_{2}^{3} \operatorname {ch}\left(\alpha^{4} x^{2}\right) \frac{d x}{x}.

Problem 3.13.14

Find Φ′(α)\Phi^{\prime }(\alpha ), if:

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(1)

Φ(α)=∫0αln⁡(1+αx)xdx\Phi (\alpha ) = \int_{0}^{\alpha } \frac{\ln (1+\alpha x)}{x} d x;

(2)

Φ(α)=∫α2αsin⁡αxxdx\Phi (\alpha ) = \int_{\alpha }^{2 \alpha } \frac{\sin \alpha x}{x} d x;

(3)

Φ(α)=∫sin⁡αcos⁡αeα1−x2dx\Phi (\alpha ) = \int_{\sin \alpha }^{\cos \alpha } e^{\alpha \sqrt{1-x^{2}}} d x;

(4)

Φ(α)=∫3αα2eαx2dx\Phi (\alpha ) = \int_{3 \alpha }^{\alpha^{2}} e^{\alpha x^{2}} d x;

(5)

Φ(α)=∫cos⁡αsin⁡αeα4x2dx\Phi (\alpha ) = \int_{\cos \alpha }^{\sin \alpha } e^{\alpha^{4} x^{2}} d x;

(6)

Φ(α)=∫e−αeαln⁡(1+α2x2)dxx\Phi (\alpha ) = \int_{e^{-\alpha }}^{e^{\alpha }} \ln \left(1+\alpha^{2} x^{2}\right) \frac{d x}{x};

(7)

Φ(α)=∫αe−ααeαln⁡(1+α2x2)dx\Phi (\alpha ) = \int_{\alpha e^{-\alpha }}^{\alpha e^{\alpha }} \ln \left(1+\alpha^{2} x^{2}\right) d x;

(8)

Φ(α)=∫ch⁡αsh⁡αln⁡(1+x2+α2)dx\Phi (\alpha ) = \int_{\operatorname {ch} \alpha }^{\operatorname {sh} \alpha } \ln \left(1+x^{2}+\alpha^{2}\right) d x.

Problem 3.13.15

Can the derivative of the function be calculated using Leibniz's rule

I(α)=∫01ln⁡(x2+α2)dx at α=0? I(\alpha ) = \int _{0}^{1} \ln \left(x^{2}+\alpha ^{2}\right) d x \quad \text{ at } \quad \alpha = 0?
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Problem 3.13.16

Let the function ff be continuous on RR. Prove that the function F(x)=12a∫−aaf(x+t)dtF(x) = \frac{1}{2 a} \int_{-a}^{a} f(x+t) d t, where a>0a > 0, has a continuous derivative on RR, and find F′(x)F^{\prime }(x).

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Problem 3.13.17

By differentiating the integral ∫0bdxx2+α2\int_{0}^{b} \frac{d x}{x^{2}+\alpha^{2}} with respect to the parameter α\alpha, where α>0\alpha > 0, calculate the integral ∫0bdx0(x2+α2)2\int_{0}^{b} \frac{d x^{0}}{\left(x^{2}+\alpha^{2}\right)^{2}}.

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Problem 3.13.18

Using differentiation with respect to the parameter α\alpha, calculate the integral I(α)I(\alpha ), if:

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(1)

I(α)=∫0π/2ln⁡(α2−sin⁡2φ)dφ,α>1I(\alpha ) = \int_{0}^{\pi / 2} \ln \left(\alpha^{2}-\sin^{2} \varphi \right) d \varphi , \quad \alpha > 1

(2)

I(α)=∫0πln⁡(1−2αcos⁡x+α2)dx,∣α∣<1I(\alpha ) = \int_{0}^{\pi } \ln \left(1-2 \alpha \cos x+\alpha^{2}\right) d x,\left|\alpha \right| < 1

(3)

I(α)=∫0πln⁡1+αcos⁡x1−αcos⁡xdxcos⁡x,∣α∣<1I(\alpha ) = \int_{0}^{\pi } \ln \frac{1+\alpha \cos x}{1-\alpha \cos x} \frac{d x}{\cos x},\left|\alpha \right| < 1

(4)

I(α)=∫0π/2−arctg⁡(αtg⁡x)⋅dxI(\alpha ) = \int_{0}^{\pi / 2}-\operatorname {arctg}(\alpha \operatorname {tg} x) \cdot d x.

Problem 3.13.19

Let the function f(x;α)f(x ; \alpha ) be continuous in the rectangle

K={(x;α):a⩽x⩽b,α1⩽α⩽α2} K = \left\{ (x ; \alpha ): a \leqslant x \leqslant b, \alpha _{1} \leqslant \alpha \leqslant \alpha _{2}\right\}

and the function g(x)g(x) is integrable on the segment [a;b][a ; b]. Prove that:

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(1)

the function F(α)=∫abf(x;α)g(x)dxF(\alpha ) = \int_{a}^{b} f(x ; \alpha ) g(x) d x is continuous on the segment [α1;α2];\left[\alpha_{1} ; \alpha_{2}\right] ;

(2)

∫α1α2F(α)dα=∫ab(∫α1α2f(x;α)g(x)dα)dx\int_{\alpha_{1}}^{\alpha_{2}} F(\alpha ) d \alpha = \int_{a}^{b}\left(\int_{\alpha_{1}}^{\alpha_{2}} f(x ; \alpha ) g(x) d \alpha \right) d x

(3)

the function F(α)F(\alpha ) is continuously differentiable on the segment [α1;α2]\left[\alpha_{1} ; \alpha_{2}\right], and moreover

F′(α)=∫ab∂f(x;α)∂αg(x)dx F^{\prime }(\alpha ) = \int _{a}^{b} \frac{\partial f(x ; \alpha )}{\partial \alpha } g(x) d x

under the additional condition that the function ∂f(x;α)∂α\frac{\partial f(x ; \alpha )}{\partial \alpha } is continuous in the rectangle KK.

Problem 3.13.20

Let F(α)=∫0α(x+α)f(x)dxF(\alpha ) = \int_{0}^{\alpha }(x+\alpha ) f(x) d x, where f(x)f(x) is a function differentiable on RR. Find F′′(a)F^{\prime \prime }(a).

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Problem 3.13.21

Let F(a)=∫abf(x)∣x−α∣dxF(a) = \int_{a}^{b} f(x)\left|x-\alpha \right| d x, where f(x)f(x) is a function continuous on the segment [a;b][a ; b]. Find F′′(α)F^{\prime \prime }(\alpha ).

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Problem 3.13.22

Let F(α)=1h2∫0h(∫0hh(ξ+η+α)dη)dξF(\alpha ) = \frac{1}{h^{2}} \int_{0}^{h}\left(\int_{0}^{h} h(\xi +\eta +\alpha ) d \eta \right) d \xi, where h>0,fh > 0, f a function continuous on RR. Find F′′(α)F^{\prime \prime }(\alpha ).

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Problem 3.13.23

Let the function ff be continuous on the segment [a;b],x0∈(a;b),x∈(a;b),k≠0[a ; b], x_{0} \in (a ; b), x \in (a ; b), k \neq 0. Prove that the function

y(x)=1k∫x0xf(t)sin⁡k(x−t)dt y(x) = \frac{1}{k} \int _{x_{0}}^{x} f(t) \sin k(x-t) d t

satisfies the differential equation y′′+k2y=f(x)y^{\prime \prime }+k^{2} y = f(x).

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Problem 3.13.24

Let the function ff be continuous on the segment [a;b],x0∈(a;b),x∈(a;b)[a ; b], x_{0} \in (a ; b), x \in (a ; b). Prove that the function

F(x)=1(n−1)!∫x0x(x−t)n−1f(t)dt, where n∈N F(x) = \frac{1}{(n-1) !} \int _{x_{0}}^{x}(x-t)^{n-1} f(t) d t, \quad \text{ where } \quad n \in N

satisfies the conditions

F(x0)=F′(x0)=…=F(n−1)(x0)=0,F(n)(x)=f(x) F\left(x_{0}\right) = F^{\prime }\left(x_{0}\right) = \ldots = F^{(n-1)}\left(x_{0}\right) = 0, \quad F^{(n)}(x) = f(x)
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Problem 3.13.25

Prove that the function

u(r)=∫0πenrcos⁡θdθ u(r) = \int _{0}^{\pi } e^{n r \cos \theta } d \theta

for any n∈Zn \in Z satisfies the equation

d2udr2+1rdudr−n2u=0 \frac{d^{2} u}{d r^{2}}+\frac{1}{r} \frac{d u}{d r}-n^{2} u = 0
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Problem 3.13.26

Prove that the function u(x)u(x) satisfies Bessel's equation

x2u′′+xu′+(x2−n2)u=0 x^{2} u^{\prime \prime }+x u^{\prime }+\left(x^{2}-n^{2}\right) u = 0

if:

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(1)

u(x)=1π∫0πcos⁡(nφ−xsin⁡φ)dφ,n∈Nu(x) = \frac{1}{\pi } \int_{0}^{\pi } \cos (n \varphi -x \sin \varphi ) d \varphi , \quad n \in N

(2)

u(x)=xn∫0πcos⁡(xcos⁡φ)sin⁡2nφdφ,n∈Nu(x) = x^{n} \int_{0}^{\pi } \cos (x \cos \varphi ) \sin^{2 n} \varphi d \varphi , \quad n \in N.

Problem 3.13.27

Consider the complete elliptic integrals

E(k)=∫0π/21−k2sin⁡2φdφ,K(k)≡F(k)=∫0π/2dφ1−k2sin⁡2φ, E(k) = \int _{0}^{\pi / 2} \sqrt{1-k^{2} \sin ^{2} \varphi } d \varphi , \quad K(k) \equiv F(k) = \int _{0}^{\pi / 2} \frac{d \varphi }{\sqrt{1-k^{2} \sin ^{2} \varphi }},

where 0<k<10 < k < 1. Prove that:

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(1)

E′(k)=E(k)−K(k)k,K′(k)=E(k)k(1−k2)−K(k)kE^{\prime }(k) = \dfrac {E(k)-K(k)}{k}, \quad K^{\prime }(k) = \dfrac {E(k)}{k\left(1-k^{2}\right)}-\dfrac {K(k)}{k};

(2)

E′′(k)+1kE′(k)+E(k)1−k2=0E^{\prime \prime }(k)+\frac{1}{k} E^{\prime }(k)+\frac{E(k)}{1-k^{2}} = 0;

(3)

∫0ktK(t)dt=E(k)−(1−k2)K(k)\int_{0}^{k} t K(t) d t = E(k)-\left(1-k^{2}\right) K(k);

(4)

∫0ktE(t)dt=13((1+k2)E(k)−(1−k2)K(k))\int_{0}^{k} t E(t) d t = \frac{1}{3}\left(\left(1+k^{2}\right) E(k)-\left(1-k^{2}\right) K(k)\right).

Problem 3.13.28

Let φ(x)=sin⁡xx\varphi (x) = \frac{\sin x}{x} for x≠0,φ(0)=1x \neq 0, \varphi (0) = 1. Show that:

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(1)

xn+1φ(n)(x)=∫0xtncos⁡(t+nπ2)dt,n∈N,x∈Rx^{n+1} \varphi^{(n)}(x) = \int_{0}^{x} t^{n} \cos \left(t+\frac{n \pi }{2}\right) d t, n \in N, x \in R;

(2)

∣φ(n)(x)∣⩽1n+1,n∈N,x∈R\left|\varphi^{(n)}(x)\right| \leqslant \frac{1}{n+1}, n \in N, x \in R.

Problem 3.13.29

Let the function φ(x)\varphi (x) and its derivative φ′(x)\varphi^{\prime }(x) be continuous on the segment [0;a][0 ; a], and let

F(t)=∫0tφ(x)dxt−x F(t) = \int _{0}^{t} \frac{\varphi (x) d x}{\sqrt{t-x}}

Prove that for t∈(0;a)t \in (0 ; a) the equality holds

F′(t)=φ(0)t+∫0tφ′(x)t−xdx F^{\prime }(t) = \frac{\varphi (0)}{\sqrt{t}}+\int _{0}^{t} \frac{\varphi ^{\prime }(x)}{\sqrt{t-x}} d x
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Problem 3.13.30

Let K(x;y)={x(1−y), if x⩽y,y(1−x), if x>y,K(x ; y) = \begin{cases} x(1-y), & \text{ if } x \leqslant y, \\ y(1-x), & \text{ if } x > y, \end{cases} and let φ(y)\varphi (y) be a function continuous on the segment [0;1][0 ; 1]. Show that the function

u(x)=∫01K(x;y)φ(y)dy u(x) = \int _{0}^{1} K(x ; y) \varphi (y) d y

satisfies on the segment [0;1][0 ; 1] the equation u′′(x)=−φ(x)u^{\prime \prime }(x) = -\varphi (x).

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Problem 3.13.31

Find a function φ(x)\varphi (x), twice differentiable on RR, satisfying the equation:

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(1)

φ(x)=x+∫0x(y−x)φ(y)dy\varphi (x) = x+\int_{0}^{x}(y-x) \varphi (y) d y

(2)

φ(x)=1+λ∫0x(x−y)φ(y)dy,λ>0\varphi (x) = 1+\lambda \int_{0}^{x}(x-y) \varphi (y) d y, \quad \lambda > 0;

(3)

φ(x)=λ∫0x(x−y)φ(y)dy+x2,λ>0\varphi (x) = \lambda \int_{0}^{x}(x-y) \varphi (y) d y+x^{2}, \quad \lambda > 0.

Problem 3.13.32

Find Fxy′′(x;y)F_{x y}^{\prime \prime }(x ; y), if F(x;y)=∫x/yxy(x−yt)f(t)dtF(x ; y) = \int_{x / y}^{x y}(x-y t) f(t) d t, where f(t)f(t) is a differentiable function, y≠0y \neq 0.

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Problem 3.13.33

Let ff be a twice differentiable function and FF a differentiable function. Prove that the function

u(x;t)=12(f(x−at)+f(x+at))+12a∫x−atx+atF(ξ)dξ u(x ; t) = \frac{1}{2}(f(x-a t)+f(x+a t))+\frac{1}{2 a} \int _{x-a t}^{x+a t} F(\xi ) d \xi

satisfies the wave equation

∂2u∂t2=a2∂2u∂x2,a≠0 \frac{\partial ^{2} u}{\partial t^{2}} = a^{2} \frac{\partial ^{2} u}{\partial x^{2}}, \quad a \neq 0

and the following initial conditions:

u(x;0)=f(x),ut′(x;0)=F(x) u(x ; 0) = f(x), \quad u_{t}^{\prime }(x ; 0) = F(x)
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Problem 3.13.34

Show that if the function ff is continuous on the segment [0;a][0 ; a], ξ∈[0;a]\xi \in [0 ; a] and (x−ξ)2+y2+z2≠0(x-\xi )^{2}+y^{2}+z^{2} \neq 0, then the function

u(x;y;z)=∫0af(ξ)dξ(x−ξ)2+y2+z2 u(x ; y ; z) = \int _{0}^{a} \frac{f(\xi ) d \xi }{\sqrt{(x-\xi )^{2}+y^{2}+z^{2}}}

satisfies Laplace's equation ∂2u∂x2+∂2u∂y2+∂2u∂z2=0\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}} = 0.

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