Chapter 5

Applications of Residue Theory

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Section 22Expansion of Meromorphic Functions into Series of Simplest Fractions and into Infinite Products[8/0%]
Problem 22.1

Prove the formulas:

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(1)

1ez−ea=e−a[−12+1z−a+∑n=1∞2(z−a)(z−a)2+4n2π2]\frac{1}{e^{z}-e^{a}} = e^{-a}\left[-\frac{1}{2}+\frac{1}{z-a}+\sum_{n = 1}^{\infty } \frac{2(z-a)}{(z-a)^{2}+4 n^{2} \pi^{2}}\right];

(2)

πsin⁡πz=∑n=−∞∞(−1)nz−n\frac{\pi }{\sin \pi z} = \sum_{n = -\infty }^{\infty } \frac{(-1)^{n}}{z-n};

(3)

1ch⁡2z−cos⁡2z=14z2+∑n=1∞n(−1)nsh⁡πn2πz24z4+n4π4\frac{1}{\operatorname {ch} 2 z-\cos 2 z} = \frac{1}{4 z^{2}}+\sum_{n = 1}^{\infty } \frac{n(-1)^{n}}{\operatorname {sh} \pi n} \frac{2 \pi z^{2}}{4 z^{4}+n^{4} \pi^{4}}.

Problem 22.2

Prove the formulas:

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(1)

π2cos⁡2πz=∑n=−∞∞1(z+12−n)2\frac{\pi^{2}}{\cos^{2} \pi z} = \sum_{n = -\infty }^{\infty } \frac{1}{\left(z+\frac{1}{2}-n\right)^{2}};

(2)

1ch⁡z=∑n=1∞(−1)n+1πz2+(n−12)2π2\frac{1}{\operatorname {ch} z} = \sum_{n = 1}^{\infty } \frac{(-1)^{n+1} \pi }{z^{2}+\left(n-\frac{1}{2}\right)^{2} \pi^{2}};

(3)

ez−e−zez+e−z=∑n=1∞2zz2+(n−12)2π2\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}} = \sum_{n = 1}^{\infty } \frac{2 z}{z^{2}+\left(n-\frac{1}{2}\right)^{2} \pi^{2}};

(4)

eazez−1=1z+∑n=1∞2z⋅cos⁡2πna−4πnsin⁡2πnaz2+4n2π2,0<a<1\frac{e^{a z}}{e^{z}-1} = \frac{1}{z}+\sum_{n = 1}^{\infty } \frac{2 z \cdot \cos 2 \pi n a-4 \pi n \sin 2 \pi n a}{z^{2}+4 n^{2} \pi^{2}}, 0 < a < 1.

Problem 22.3

Let F(z)F(z) be an entire function satisfying the inequality

∣F(z+iy)∣⩽Mea∣y∣,−π<a<π, \left|F(z+i y)\right| \leqslant M e^{a\left|y\right|}, \quad -\pi < a < \pi ,

for all real xx and yy. Prove that

πF(z)sin⁡πz=∑n=−∞∞(−1)nF(n)z−n and πF(z)cos⁡πz=∑n=−∞∞(−1)n+1F(n+12)z−n−12. \begin{aligned} \frac{\pi F(z)}{\sin \pi z} & = \sum _{n = -\infty }^{\infty }(-1)^{n} \frac{F(n)}{z-n} \\ \text{ and } \frac{\pi F(z)}{\cos \pi z} & = \sum _{n = -\infty }^{\infty }(-1)^{n+1} \frac{F\left(n+\frac{1}{2}\right)}{z-n-\frac{1}{2}}. \end{aligned}
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Problem 22.4

Prove that

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(1)

sin⁡azsin⁡πz=2π∑n=1∞(−1)nnsin⁡naz2−n2,−π<a<π;\frac{\sin a z}{\sin \pi z} = \frac{2}{\pi } \sum_{n = 1}^{\infty } \frac{(-1)^{n} n \sin n a}{z^{2}-n^{2}},-\pi < a < \pi ;

(2)

ch⁡azsh⁡πz=1πz+1π∑n=1∞(−1)n2zcos⁡naz2+n2,−π<a<π;\frac{\operatorname {ch} a z}{\operatorname {sh} \pi z} = \frac{1}{\pi z}+\frac{1}{\pi } \sum_{n = 1}^{\infty }(-1)^{n} \frac{2 z \cos n a}{z^{2}+n^{2}},-\pi < a < \pi ;

(3)

πsin⁡πz=∑n=−∞∞ei(πn+αn−αz)z−n,−π<α<π\frac{\pi }{\sin \pi z} = \sum_{n = -\infty }^{\infty } \frac{e^{i(\pi n+\alpha n-\alpha z)}}{z-n},-\pi < \alpha < \pi;

(4)

∑n=−∞∞(−1)nsin⁡α(z−n)z−n=0,−π<α<π\sum_{n = -\infty }^{\infty }(-1)^{n} \frac{\sin \alpha (z-n)}{z-n} = 0,-\pi < \alpha < \pi;

(5)

π2sin⁡2πz=πα+∑n=−∞∞cos⁡α(z−n)(z−n)2,0⩽α<π\frac{\pi^{2}}{\sin^{2} \pi z} = \pi \alpha +\sum_{n = -\infty }^{\infty } \frac{\cos \alpha (z-n)}{(z-n)^{2}}, 0 \leqslant \alpha < \pi.

Problem 22.5

Prove the formulas:

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(1)

e(α+β)z−1(eαz−1)(eβz−1)=(1α+1β)1z+∑k=1∞(2αzα2z2+4k2π2+2βzβ2z2+4k2π2)\frac{e^{(\alpha +\beta ) z}-1}{\left(e^{\alpha z}-1\right)\left(e^{\beta z}-1\right)} = \left(\frac{1}{\alpha }+\frac{1}{\beta }\right) \frac{1}{z}+\sum_{k = 1}^{\infty }\left(\frac{2 \alpha z}{\alpha^{2} z^{2}+4 k^{2} \pi^{2}}+\frac{2 \beta z}{\beta^{2} z^{2}+4 k^{2} \pi^{2}}\right), αβ≠0,α≠β;\alpha \beta \neq 0, \alpha \neq \beta ;

(2)

ctg⁡(z−a)+ctg⁡a=z∑n=−∞∞1(a−πn)(z−a+πn),sin⁡a≠0\operatorname {ctg}(z-a)+\operatorname {ctg} a = z \sum_{n = -\infty }^{\infty } \frac{1}{(a-\pi n)(z-a+\pi n)}, \sin a \neq 0;

(3)

sin⁡acos⁡z−cos⁡a=∑n=−∞∞2aa2−(z+2πn)2,sin⁡a≠0\frac{\sin a}{\cos z-\cos a} = \sum_{n = -\infty }^{\infty } \frac{2 a}{a^{2}-(z+2 \pi n)^{2}}, \sin a \neq 0;

(4)

cos⁡asin⁡a−sin⁡z=∑n=−∞∞π−2a(z−a+2πn)(z+a+(2n−1)π),cos⁡a≠0\frac{\cos a}{\sin a-\sin z} = \sum_{n = -\infty }^{\infty } \frac{\pi -2 a}{(z-a+2 \pi n)(z+a+(2 n-1) \pi )}, \cos a \neq 0.

Problem 22.6

Let f(z)f(z) be a meromorphic function with poles a1,a2,…a_{1}, a_{2}, \ldots and zeros b1,b2,…b_{1}, b_{2}, \ldots (each zero is written as many times as its multiplicity, and each pole as many times as its order), where the point z=0z = 0 is neither a zero nor a pole of the function f(z)f(z). Suppose that

f′(z)f(z)=∑n=1∞1z−bn−∑n=1∞1z−an, \frac{f^{\prime }(z)}{f(z)} = \sum _{n = 1}^{\infty } \frac{1}{z-b_{n}}-\sum _{n = 1}^{\infty } \frac{1}{z-a_{n}},

and both series converge uniformly in every bounded part of the plane. Prove that

f(z)=∏n=1∞(1−zbn)∏n=1∞(1−zan)f(0) f(z) = \frac{\prod _{n = 1}^{\infty }\left(1-\frac{z}{b_{n}}\right)}{\prod _{n = 1}^{\infty }\left(1-\frac{z}{a_{n}}\right)} f(0)

and both products also converge uniformly in every bounded part of the plane.

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Problem 22.7

Prove the formulas:

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(1)

sh⁡z=z∏n=1∞(1+z2n2π2)\operatorname {sh} z = z \prod_{n = 1}^{\infty }\left(1+\frac{z^{2}}{n^{2} \pi^{2}}\right);

(2)

cos⁡z=∏n=1∞(1−z2(n−(1/2))2π2)\cos z = \prod_{n = 1}^{\infty }\left(1-\frac{z^{2}}{(n-(1 / 2))^{2} \pi^{2}}\right);

(3)

th⁡z=z∏n=1∞[1+(z/πn)2][1+(zπ(n−(1/2)))2]\operatorname {th} z = z \prod_{n = 1}^{\infty } \frac{\left[1+(z / \pi n)^{2}\right]}{\left[1+\left(\frac{z}{\pi (n-(1 / 2))}\right)^{2}\right]};

(4)

eaz−ebz=(a−b)ze(1/2)(a+b)z∏n=1∞[1+(a−b)2z24n2π2]e^{a z}-e^{b z} = (a-b) z e^{(1 / 2)(a+b) z} \prod_{n = 1}^{\infty }\left[1+\frac{(a-b)^{2} z^{2}}{4 n^{2} \pi^{2}}\right];

(5)

ez−ea=(z−a)e(z+a)/2∏n=1∞[1+(z−a2πn)2]e^{z}-e^{a} = (z-a) e^{(z+a) / 2} \prod_{n = 1}^{\infty }\left[1+\left(\frac{z-a}{2 \pi n}\right)^{2}\right];

(6)

ch⁡z−cos⁡z=z2∏n=1∞(1+z44n4π4)\operatorname {ch} z-\cos z = z^{2} \prod_{n = 1}^{\infty }\left(1+\frac{z^{4}}{4 n^{4} \pi^{4}}\right);

(7)

cos⁡πz−cos⁡πa=−π22(z2−a2)∏n=1∞[1−(z+a2n)2]∏n=1∞[1−(z−a2n)2]\cos \pi z-\cos \pi a = -\frac{\pi^{2}}{2}\left(z^{2}-a^{2}\right) \prod_{n = 1}^{\infty }\left[1-\left(\frac{z+a}{2 n}\right)^{2}\right] \prod_{n = 1}^{\infty }\left[1-\left(\frac{z-a}{2 n}\right)^{2}\right];

(8)

cos⁡πz=π(2z+1)∏n=1∞(1−z(z+1)n2−(1/4))\cos \pi z = \pi (2 z+1) \prod_{n = 1}^{\infty }\left(1-\frac{z(z+1)}{n^{2}-(1 / 4)}\right);

(9)

cos⁡z−cos⁡a1−cos⁡a∏n=−∞∞(1−z2(2πn+a)2)\frac{\cos z-\cos a}{1-\cos a} \prod_{n = -\infty }^{\infty }\left(1-\frac{z^{2}}{(2 \pi n+a)^{2}}\right);

(10)

sin⁡(z−a)+sin⁡a=z(π+2a−z)π+2a∏n≠0{1+z(π+2a−z)2πn(π(2n−1)−2a)}\sin (z-a)+\sin a = \frac{z(\pi +2 a-z)}{\pi +2 a} \prod_{n \neq 0}\left\{ 1+\frac{z(\pi +2 a-z)}{2 \pi n(\pi (2 n-1)-2 a)}\right\};

(11)

sin⁡(z−a)sin⁡a−sin⁡z=−∏n=0∞1−z(z−2a)π2(2n+1)2−a21−z(z+2a)π2(2n+1)2−a2\frac{\sin (z-a)}{\sin a-\sin z} = -\prod_{n = 0}^{\infty } \frac{1-\frac{z(z-2 a)}{\pi^{2}(2 n+1)^{2}-a^{2}}}{1-\frac{z(z+2 a)}{\pi^{2}(2 n+1)^{2}-a^{2}}};

(12)

ez2+e2z−1=2e(1/2)(z2+2z−1)∏n=1∞(1+(z−1)4π2(2n−1)2)e^{z^{2}}+e^{2 z-1} = 2 e^{(1 / 2)\left(z^{2}+2 z-1\right)} \prod_{n = 1}^{\infty }\left(1+\frac{(z-1)^{4}}{\pi^{2}(2 n-1)^{2}}\right);

(13)

e2ch⁡z−e2sh⁡z=2eez−z∏n=1∞(1+e−2zπ2n2)e^{2 \operatorname {ch} z}-e^{2 \operatorname {sh} z} = 2 e^{e^{z}-z} \prod_{n = 1}^{\infty }\left(1+\frac{e^{-2 z}}{\pi^{2} n^{2}}\right).

Problem 22.8

Prove the equalities:

∑n=1∞1n2+a2=π2a(cth⁡aπ−1aπ)∑n=1∞1(n2+a2)2=14a4(π2a2sh⁡2aπ+aπcth⁡aπ−2) \begin{aligned} \sum _{n = 1}^{\infty } \frac{1}{n^{2}+a^{2}} & = \frac{\pi }{2 a}\left(\operatorname {cth} a \pi -\frac{1}{a \pi }\right) \\ \sum _{n = 1}^{\infty } \frac{1}{\left(n^{2}+a^{2}\right)^{2}} & = \frac{1}{4 a^{4}}\left(\frac{\pi ^{2} a^{2}}{\operatorname {sh}^{2} a \pi }+a \pi \operatorname {cth} a \pi -2\right) \end{aligned}

assuming that none of the denominators in these equalities vanishes.

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Section 23Computing Improper Integrals[11/73%]
Problem 23.1

Compute the integrals:

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(1)

∫−∞∞x2dx(x2+1)(x2+9)\int_{-\infty }^{\infty } \frac{x^{2} d x}{\left(x^{2}+1\right)\left(x^{2}+9\right)};

(2)

∫−∞∞x2−x+2x4−10x2−9dx\int_{-\infty }^{\infty } \frac{x^{2}-x+2}{x^{4}-10 x^{2}-9} d x;

(3)

∫−∞∞dxx2−2ix−2\int_{-\infty }^{\infty } \frac{d x}{x^{2}-2 i x-2};

(4)

∫−∞∞x2+1x4+1dx\int_{-\infty }^{\infty } \frac{x^{2}+1}{x^{4}+1} d x;

(5)

∫−∞∞x4+1x6+1dx\int_{-\infty }^{\infty } \frac{x^{4}+1}{x^{6}+1} d x;

(6)

∫−∞∞x2dxx4+6x2+25\int_{-\infty }^{\infty } \frac{x^{2} d x}{x^{4}+6 x^{2}+25};

(7)

∫−∞+∞dx(x2+9)(x2+16)\int_{-\infty }^{+\infty } \frac{d x}{\left(x^{2}+9\right)\left(x^{2}+16\right)};

(8)

∫−∞+∞dx(x2+1)3\int_{-\infty }^{+\infty } \frac{d x}{\left(x^{2}+1\right)^{3}}.

Problem 23.2

Compute the integrals:

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(1)

∫−∞+∞x+x31+x4sin⁡3xdx\int_{-\infty }^{+\infty } \frac{x+x^{3}}{1+x^{4}} \sin 3 x d x;

(2)

∫−∞+∞cos⁡4x1+x6dx\int_{-\infty }^{+\infty } \frac{\cos 4 x}{1+x^{6}} d x;

(3)

∫−∞+∞(x+1)sin⁡2xx2−4x+8dx\int_{-\infty }^{+\infty } \frac{(x+1) \sin 2 x}{x^{2}-4 x+8} d x;

(4)

∫−∞+∞(x−2)cos⁡xx2−6x+10dx\int_{-\infty }^{+\infty } \frac{(x-2) \cos x}{x^{2}-6 x+10} d x;

(5)

∫−∞+∞(x+3)sin⁡x2x2+4x+20dx\int_{-\infty }^{+\infty } \frac{(x+3) \sin \frac{x}{2}}{x^{2}+4 x+20} d x;

(6)

∫−∞+∞(x−4)cos⁡23xx2−6x+90dx\int_{-\infty }^{+\infty } \frac{(x-4) \cos \frac{2}{3} x}{x^{2}-6 x+90} d x;

(7)

∫−∞+∞(2−x)cos⁡(3x−2)x2−2x+2dx\int_{-\infty }^{+\infty } \frac{(2-x) \cos (3 x-2)}{x^{2}-2 x+2} d x;

(8)

∫−∞+∞(x−3)sin⁡(x−1)x2+4x+5dx\int_{-\infty }^{+\infty } \frac{(x-3) \sin (x-1)}{x^{2}+4 x+5} d x;

(9)

∫−∞+∞(2x+3)sin⁡(x+5)x2+4x+8dx\int_{-\infty }^{+\infty } \frac{(2 x+3) \sin (x+5)}{x^{2}+4 x+8} d x;

(10)

∫−∞+∞(x+7)cos⁡(x+2)x2+2x+5dx\int_{-\infty }^{+\infty } \frac{(x+7) \cos (x+2)}{x^{2}+2 x+5} d x;

(11)

∫−∞+∞(x3+7x)sin⁡2xx4+5x2+4dx\int_{-\infty }^{+\infty } \frac{\left(x^{3}+7 x\right) \sin 2 x}{x^{4}+5 x^{2}+4} d x;

(12)

∫−∞+∞4x2cos⁡xx4+8x2+16dx\int_{-\infty }^{+\infty } \frac{4 x^{2} \cos x}{x^{4}+8 x^{2}+16} d x;

(13)

∫−∞+∞cos⁡(3−8x)4x2−7x+5dx\int_{-\infty }^{+\infty } \frac{\cos (3-8 x)}{4 x^{2}-7 x+5} d x;

(14)

∫−∞+∞sin⁡(3−6x)3x2−4x+3dx\int_{-\infty }^{+\infty } \frac{\sin (3-6 x)}{3 x^{2}-4 x+3} d x;

(15)

∫−∞+∞xsin⁡(2−x)x2+2dx\int_{-\infty }^{+\infty } \frac{x \sin (2-x)}{x^{2}+2} d x;

(16)

∫−∞+∞xcos⁡(1−2x)x2+4dx\int_{-\infty }^{+\infty } \frac{x \cos (1-2 x)}{x^{2}+4} d x;

(17)

∫−∞+∞cos⁡(1−2x)x−4+5xdx\int_{-\infty }^{+\infty } \frac{\cos (1-2 x)}{x-4+\frac{5}{x}} d x;

(18)

∫−∞+∞sin⁡(1−3x)x−2+5xdx\int_{-\infty }^{+\infty } \frac{\sin (1-3 x)}{x-2+\frac{5}{x}} d x;

(19)

∫−∞+∞x3cos⁡(1−2x)(2x2+1)(x2+4)dx\int_{-\infty }^{+\infty } \frac{x^{3} \cos (1-2 x)}{\left(2 x^{2}+1\right)\left(x^{2}+4\right)} d x;

(20)

∫−∞+∞x3sin⁡(2−x)(x2+2)2dx\int_{-\infty }^{+\infty } \frac{x^{3} \sin (2-x)}{\left(x^{2}+2\right)^{2}} d x.

Problem 23.3

Let R(z)R(z) be a rational function having poles a1,a2,…,ana_{1}, a_{2}, \ldots , a_{n} in the upper half-plane and poles b1,b2,…,bmb_{1}, b_{2}, \ldots , b_{m} on the real axis (and having no other poles for Im(z)⩾0\mathrm{Im}\left(z\right) \geqslant 0). Prove that if the function R(z)R(z) satisfies the condition R(z)=O(1z)R(z) = O\left(\frac{1}{z}\right) as z→∞z \rightarrow \infty, then the formula holds

 v.p. ∫−∞∞R(x)eixdx=2πi∑k=1nresz=akR(z)eiz+πi∑k=1mresz=bkR(z)eiz \text{ v.p. } \int _{-\infty }^{\infty } R(x) e^{i x} d x = 2 \pi i \sum _{k = 1}^{n} \mathop{\mathrm{res}}_{z = a_{k}} R(z) e^{i z}+\pi i \sum _{k = 1}^{m} \mathop{\mathrm{res}}_{z = b_{k}} R(z) e^{i z}

(provided that the integral on the left-hand side exists).

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Problem 23.4

Compute the integrals:

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(1)

v. p. ∫−∞∞1−eiαxx2dx\int_{-\infty }^{\infty } \frac{1-e^{i \alpha x}}{x^{2}} d x, a) α>0\alpha > 0; b) α<0\alpha < 0;

(2)

∫0+∞1−cos⁡αxx2dx,α>0\int_{0}^{+\infty } \frac{1-\cos \alpha x}{x^{2}} d x, \alpha > 0;

(3)

∫0+∞(sin⁡xx)ndx,n=2,3,4\int_{0}^{+\infty }\left(\frac{\sin x}{x}\right)^{n} d x, n = 2,3,4;

(4)

∫0+∞cos⁡2ax−cos⁡2bxx2dx,a>0,b>0\int_{0}^{+\infty } \frac{\cos 2 a x-\cos 2 b x}{x^{2}} d x, a > 0, b > 0;

(5)

∫0+∞sin⁡axx(x2+b2)dx,a>0,Re(b)>0\int_{0}^{+\infty } \frac{\sin a x}{x\left(x^{2}+b^{2}\right)} d x, a > 0, \mathrm{Re}\left(b\right) > 0;

(6)

∫0+∞x2−b2x2+b2sin⁡axxdx,a>0,Re(b)>0\int_{0}^{+\infty } \frac{x^{2}-b^{2}}{x^{2}+b^{2}} \frac{\sin a x}{x} d x, a > 0, \mathrm{Re}\left(b\right) > 0;

(7)

∫0+∞sin⁡2axx2(x2+b2)dx,a>0,Re(b)>0\int_{0}^{+\infty } \frac{\sin^{2} a x}{x^{2}\left(x^{2}+b^{2}\right)} d x, a > 0, \mathrm{Re}\left(b\right) > 0;

(8)

∫0+∞x−sin⁡xx3(x2+a2)dx\int_{0}^{+\infty } \frac{x-\sin x}{x^{3}\left(x^{2}+a^{2}\right)} d x, Re a>0a > 0.

Problem 23.5

Compute the integrals:

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(1)

∫−22dx(2+x)2(4−x2)4\int_{-2}^{2} \frac{d x}{\sqrt[4]{(2+x)^{2}\left(4-x^{2}\right)}};

(2)

∫02dxx2(2−x)3\int_{0}^{2} \frac{d x}{\sqrt[3]{x^{2}(2-x)}};

(3)

∫01dx(x−3)2x−x2\int_{0}^{1} \frac{d x}{(x-3)^{2} \sqrt{x-x^{2}}};

(4)

∫061x+2x6−xdx\int_{0}^{6} \frac{1}{x+2} \sqrt{\frac{x}{6-x}} d x;

(5)

∫05x(5−x)74x2−5x−6dx\int_{0}^{5} \frac{\sqrt[4]{x(5-x)^{7}}}{x^{2}-5 x-6} d x;

(6)

∫−11x1−x2x2−3x+2dx\int_{-1}^{1} \frac{x \sqrt{1-x^{2}}}{x^{2}-3 x+2} d x;

(7)

∫03dx(x+2)2x3(3−x)4\int_{0}^{3} \frac{d x}{(x+2)^{2} \sqrt[4]{x^{3}(3-x)}};

(8)

∫02x2(2−x)3(x+2)2dx\int_{0}^{2} \frac{\sqrt[3]{x^{2}(2-x)}}{(x+2)^{2}} d x;

(9)

∫03(3−x)x23dx\int_{0}^{3} \sqrt[3]{(3-x) x^{2}} d x;

(10)

∫01x−x220x+39x+2dx\int_{0}^{1} \sqrt{x-x^{2}} \frac{20 x+39}{x+2} d x;

(11)

∫−111+x1−x43x2+13x+20(x+2)2dx\int_{-1}^{1} \sqrt[4]{\frac{1+x}{1-x}} \frac{3 x^{2}+13 x+20}{(x+2)^{2}} d x;

(12)

∫−10−1−1x3x+13x+4dx\int_{-1}^{0} \sqrt{-1-\frac{1}{x}} \frac{3 x+1}{3 x+4} d x.

Problem 23.6

Compute the integrals:

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(1)

∫1+∞x−1ln⁡(x−1)x2+3xdx\int_{1}^{+\infty } \frac{\sqrt{x-1} \ln (x-1)}{x^{2}+3 x} d x;

(2)

∫2+∞ln⁡(x−2)(x2−1)x−2dx\int_{2}^{+\infty } \frac{\ln (x-2)}{\left(x^{2}-1\right) \sqrt{x-2}} d x;

(3)

∫0+∞xln⁡xdxx2+1\int_{0}^{+\infty } \frac{\sqrt{x} \ln x d x}{x^{2}+1};

(4)

∫0+∞ln⁡x(x+1)x34dx\int_{0}^{+\infty } \frac{\ln x}{(x+1) \sqrt[4]{x^{3}}} d x;

(5)

∫0+∞ln⁡xdx(x+1)(2x+1)x\int_{0}^{+\infty } \frac{\ln x d x}{(x+1)(2 x+1) \sqrt{x}};

(6)

∫1+∞dx(x2+1)x−1\int_{1}^{+\infty } \frac{d x}{\left(x^{2}+1\right) \sqrt{x-1}};

(7)

∫0+∞dxx5(x+1)2\int_{0}^{+\infty } \frac{d x}{\sqrt[5]{x}(x+1)^{2}};

(8)

∫0+∞xln⁡x(x+1)(x+2)dx\int_{0}^{+\infty } \frac{\sqrt{x} \ln x}{(x+1)(x+2)} d x.

Problem 23.7

Compute the integrals:

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(1)

∫0+∞ln⁡x(x+1)2dx\int_{0}^{+\infty } \frac{\ln x}{(x+1)^{2}} d x;

(2)

∫0+∞ln⁡xdxx2+2x+2\int_{0}^{+\infty } \frac{\ln x d x}{x^{2}+2 x+2};

(3)

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

(4)

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

(5)

∫0+∞xln⁡x(x2+1)2dx\int_{0}^{+\infty } \frac{x \ln x}{\left(x^{2}+1\right)^{2}} d x;

(6)

∫0+∞x2ln⁡x(x2+1)2dx\int_{0}^{+\infty } \frac{x^{2} \ln x}{\left(x^{2}+1\right)^{2}} d x;

(7)

∫0+∞ln⁡xx2−1dx\int_{0}^{+\infty } \frac{\ln x}{x^{2}-1} d x;

(8)

∫0+∞(ln⁡xx−1)2dx\int_{0}^{+\infty }\left(\frac{\ln x}{x-1}\right)^{2} d x;

(9)

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

(10)

∫0+∞(ln⁡xx−1)3dx\int_{0}^{+\infty }\left(\frac{\ln x}{x-1}\right)^{3} d x;

(11)

∫−∞+∞xsh⁡xdx\int_{-\infty }^{+\infty } \frac{x}{\operatorname {sh} x} d x;

(12)

∫0+∞xdxsh⁡xch⁡x\int_{0}^{+\infty } \frac{x d x}{\operatorname {sh} x \operatorname {ch} x};

(13)

∫0+∞(xsh⁡x)2dx\int_{0}^{+\infty }\left(\frac{x}{\operatorname {sh} x}\right)^{2} d x;

(14)

∫0+∞(xsh⁡x)3dx\int_{0}^{+\infty }\left(\frac{x}{\operatorname {sh} x}\right)^{3} d x;

(15)

∫0+∞x2ch⁡xdx\int_{0}^{+\infty } \frac{x^{2}}{\operatorname {ch} x} d x;

(16)

∫01ln⁡1−xxdx1+x2\int_{0}^{1} \ln \frac{1-x}{x} \frac{d x}{1+x^{2}};

(17)

∫01(ln⁡x1−x)2dxx+1\int_{0}^{1}\left(\ln \frac{x}{1-x}\right)^{2} \frac{d x}{x+1};

(18)

∫−11(ln⁡1+x1−x)2dxx2\int_{-1}^{1}\left(\ln \frac{1+x}{1-x}\right)^{2} \frac{d x}{x^{2}};

(19)

∫0+∞dx(x2+π2)ch⁡x\int_{0}^{+\infty } \frac{d x}{\left(x^{2}+\pi^{2}\right) \operatorname {ch} x};

(20)

∫−∞+∞exdx(x2+π2)ch⁡x\int_{-\infty }^{+\infty } \frac{e^{x} d x}{\left(x^{2}+\pi^{2}\right) \operatorname {ch} x};

(21)

∫0+∞dx(x2+9π2)ch⁡x\int_{0}^{+\infty } \frac{d x}{\left(x^{2}+9 \pi^{2}\right) \operatorname {ch} x};

(22)

∫0+∞dx(x+1)(ln⁡2x+π2)\int_{0}^{+\infty } \frac{d x}{(x+1)\left(\ln^{2} x+\pi^{2}\right)};

(23)

∫−11dx(ln⁡1+x1−x)2+9π2\int_{-1}^{1} \frac{d x}{\left(\ln \frac{1+x}{1-x}\right)^{2}+9 \pi^{2}};

(24)

∫01xdx(ln⁡x1−x)2+π2\int_{0}^{1} \frac{x d x}{\left(\ln \frac{x}{1-x}\right)^{2}+\pi^{2}};

(25)

∫0+∞ln⁡xdxx2+2axcos⁡λ+a2,a>0,a<λ<π\int_{0}^{+\infty } \frac{\ln x d x}{x^{2}+2 a x \cos \lambda +a^{2}}, a > 0, a < \lambda < \pi.

Problem 23.8

Let R(z)R(z) be a rational function, continuous and real for real values of zz, and let a1,…,ana_{1}, \ldots , a_{n} be its poles lying in the upper half-plane. Assuming the convergence of the integrals, prove the formulas:

?
(1)

∫−∞+∞R(x)ln⁡∣x−a∣dx=−2πIm(∑s=1n)resz=asR(z)g(z)\int_{-\infty }^{+\infty } R(x) \ln \left|x-a\right| d x = -2 \pi \mathrm{Im}\left(\sum_{s = 1}^{n}\right) \mathop{\mathrm{res}}_{z = a_{s}} R(z) g(z) (here aa is a real number, and g(z)g(z) is an arbitrary branch of the multivalued function Ln⁡(z−a)\operatorname {Ln}(z-a), regular in the upper half-plane);

(2)

∫−∞+∞R(x)∣x−a∣α−1dx=−2πsin⁡πα2Im(∑s=1n)resz=asR(z)g(z)\int_{-\infty }^{+\infty } R(x)\left|x-a\right|^{\alpha -1} d x = -\frac{2 \pi }{\sin \frac{\pi \alpha }{2}} \mathrm{Im}\left(\sum_{s = 1}^{n}\right) \mathop{\mathrm{res}}_{z = a_{s}} R(z) g(z) (here aa is a real number, α>0\alpha > 0, and g(z)g(z) is the branch of the multivalued function (z−ai)α−1\left(\frac{z-a}{i}\right)^{\alpha -1}, regular in the upper half-plane and positive for positive values of z−ai\frac{z-a}{i});

(3)

∫−∞+∞R(x)ln⁡∣x2−a2∣dx=−2πIm(∑s=1n)resz=asR(z)g(z)\int_{-\infty }^{+\infty } R(x) \ln \left|x^{2}-a^{2}\right| d x = -2 \pi \mathrm{Im}\left(\sum_{s = 1}^{n}\right) \mathop{\mathrm{res}}_{z = a_{s}} R(z) g(z) (here a>0a > 0, and g(z)g(z) is an arbitrary branch of the multivalued function Ln⁡(z2−a2)\operatorname {Ln}\left(z^{2}-a^{2}\right), regular in the upper half-plane);

(4)

∫−∞+∞R(x)ln⁡(x2+a2)dx=−4πIm(∑s=1n)resz=asR(z)g(z)\int_{-\infty }^{+\infty } R(x) \ln \left(x^{2}+a^{2}\right) d x = -4 \pi \mathrm{Im}\left(\sum_{s = 1}^{n}\right) \mathop{\mathrm{res}}_{z = a_{s}} R(z) g(z) (here a>0a > 0, and g(z)g(z) is an arbitrary branch of the multivalued function Ln⁡(z+ai)\operatorname {Ln}(z+a i), regular in the upper half-plane).

Problem 23.9

Compute the integrals:

?
(1)

∫−∞+∞∣x−1∣3x2+1dx\int_{-\infty }^{+\infty } \frac{\sqrt[3]{\left|x-1\right|}}{x^{2}+1} d x;

(2)

∫−∞+∞ln⁡∣x2−1∣x2+1dx\int_{-\infty }^{+\infty } \frac{\ln \left|x^{2}-1\right|}{x^{2}+1} d x;

(3)

∫−∞+∞ln⁡(x4+2x2+2)x2+4dx\int_{-\infty }^{+\infty } \frac{\ln \left(x^{4}+2 x^{2}+2\right)}{x^{2}+4} d x;

(4)

∫−∞+∞x2+1−xdxx2+1\int_{-\infty }^{+\infty } \sqrt{\sqrt{x^{2}+1}-x} \frac{d x}{x^{2}+1};

(5)

∫−∞+∞dx(x2+1)∣x−1∣\int_{-\infty }^{+\infty } \frac{d x}{\left(x^{2}+1\right) \sqrt{\left|x-1\right|}};

(6)

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

(7)

∫0+∞ln⁡(1+x2)x2dx1+x2\int_{0}^{+\infty } \frac{\ln \left(1+x^{2}\right)}{x^{2}} \frac{d x}{1+x^{2}};

(8)

∫0+∞arctg⁡xxdx1+x2\int_{0}^{+\infty } \frac{\operatorname {arctg} x}{x} \frac{d x}{1+x^{2}};

(9)

∫0+∞(arctg⁡x)2dxx2\int_{0}^{+\infty }(\operatorname {arctg} x)^{2} \frac{d x}{x^{2}};

(10)

∫−∞+∞arctg⁡xx2−4x+5ln⁡(x2+1)dx\int_{-\infty }^{+\infty } \frac{\operatorname {arctg} x}{x^{2}-4 x+5} \ln \left(x^{2}+1\right) d x;

(11)

∫0+∞x2+1+x−x2+1−xx(x2+1)dx\int_{0}^{+\infty } \frac{\sqrt{\sqrt{x^{2}+1}+x}-\sqrt{\sqrt{x^{2}+1}-x}}{x\left(x^{2}+1\right)} d x;

(12)

∫0+∞(1+x2+x+1+x2−x−2)dxx2\int_{0}^{+\infty }\left(\sqrt{\sqrt{1+x^{2}}+x}+\sqrt{\sqrt{1+x^{2}}-x}-2\right) \frac{d x}{x^{2}};

(13)

∫0+∞1+x2+x−1+x2−xx1+x2dx\int_{0}^{+\infty } \frac{\sqrt{\sqrt{1+x^{2}}+x}-\sqrt{\sqrt{1+x^{2}}-x}}{x \sqrt{1+x^{2}}} d x;

(14)

∫0+∞(x2+a2)(x2+b2)+x2−abdx(x2+1)2,a>0,b>0\int_{0}^{+\infty } \sqrt{\sqrt{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}+x^{2}-a b} \frac{d x}{\left(x^{2}+1\right)^{2}}, a > 0, b > 0.

Problem 23.10

Prove that

∫0+∞x2mdx1+x2n=π2n1sin⁡2m+12nπ \int _{0}^{+\infty } \frac{x^{2 m} d x}{1+x^{2 n}} = \frac{\pi }{2 n} \frac{1}{\sin \frac{2 m+1}{2 n} \pi }

where mm and n,m<nn, m < n, are nonnegative integers, by considering the integral of the function z2m(1+z2n)−1z^{2 m}\left(1+z^{2 n}\right)^{-1} over the boundary of the angle 0<arg⁡z<πn0 < \arg z < \frac{\pi }{n}.

?
Problem 23.11

Compute the integrals:

?
(1)

∫0+∞xdx(4x2+π2)sh⁡x\int_{0}^{+\infty } \frac{x d x}{\left(4 x^{2}+\pi^{2}\right) \operatorname {sh} x};

(2)

∫−∞+∞exdx(4x2+π2)ch⁡x\int_{-\infty }^{+\infty } \frac{e^{x} d x}{\left(4 x^{2}+\pi^{2}\right) \operatorname {ch} x};

(3)

∫0+∞sh⁡2xdx(4x2+π2)ch⁡2x\int_{0}^{+\infty } \frac{\operatorname {sh}^{2} x d x}{\left(4 x^{2}+\pi^{2}\right) \operatorname {ch}^{2} x};

(4)

∫0+∞xdx(4x2+9π2)sh⁡x\int_{0}^{+\infty } \frac{x d x}{\left(4 x^{2}+9 \pi^{2}\right) \operatorname {sh} x};

(5)

∫0+∞ln⁡xdx(x2+1)(4ln⁡2x+9π2)\int_{0}^{+\infty } \frac{\ln x d x}{\left(x^{2}+1\right)\left(4 \ln^{2} x+9 \pi^{2}\right)};

(6)

∫0+∞xln⁡x(x2−1)(ln⁡2x+π2)dx\int_{0}^{+\infty } \frac{\sqrt{x} \ln x}{\left(x^{2}-1\right)\left(\ln^{2} x+\pi^{2}\right)} d x;

(7)

∫1+∞ln⁡(x+x2−1)x2dx\int_{1}^{+\infty } \frac{\ln \left(x+\sqrt{x^{2}-1}\right)}{x^{2}} d x;

(8)

∫1+∞ln⁡x+x2−1x−x2−1dxx2+1\int_{1}^{+\infty } \ln \frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}} \frac{d x}{x^{2}+1};

(9)

∫0+∞ln⁡xdx(x2+a2)(4ln⁡2x+π2),a>0\int_{0}^{+\infty } \frac{\ln x d x}{\left(x^{2}+a^{2}\right)\left(4 \ln^{2} x+\pi^{2}\right)}, a > 0;

(10)

∫a+∞ln⁡(x+x2−a2)x2+a2dx,a>0\int_{a}^{+\infty } \frac{\ln \left(x+\sqrt{x^{2}-a^{2}}\right)}{x^{2}+a^{2}} d x, a > 0;

(11)

∫a+∞arctg⁡axdxx2−a2,a>0\int_{a}^{+\infty } \operatorname {arctg} \frac{a}{x} \frac{d x}{\sqrt{x^{2}-a^{2}}}, a > 0;

(12)

∫0+∞ln⁡(ex+e2x−e2a)ch⁡xdx\int_{0}^{+\infty } \frac{\ln \left(e^{x}+\sqrt{e^{2 x}-e^{2 a}}\right)}{\operatorname {ch} x} d x.

Section 24Integrals Reducible to the Gamma Function[14/0%]
Problem 24.1

Prove the formulas:

?
(1)

B(z,ζ)=B(ζ,z)\mathrm{B}(z, \zeta ) = \mathrm{B}(\zeta , z);

(2)

B(m,n)=1m+n−1Cnm(m,n=1,2,…,m<n)\quad \mathrm{B}(m, n) = \frac{1}{m+n-1} C_{n}^{m} \quad (m, n = 1,2, \ldots , m < n);

(3)

B(z+1,ζ)=zz+ζB(z,ζ)\mathrm{B}(z+1, \zeta ) = \frac{z}{z+\zeta } \mathrm{B}(z, \zeta );

(4)

Γ(n+z+1)=(n+z)(n+z−1)⋯zΓ(z)(n=1,2,…)\Gamma (n+z+1) = (n+z)(n+z-1) \cdots z \Gamma (z)(n = 1,2, \ldots ).

Problem 24.2

Prove the formulas:

?
(1)

∫0+∞xze−x2dx=12Γ(z+12)(Re(z)>−1);\int_{0}^{+\infty } x^{z} e^{-x^{2}} d x = \frac{1}{2} \Gamma \left(\frac{z+1}{2}\right) \quad (\mathrm{Re}\left(z\right) > -1) ;

(2)

∫0+∞xze−xpdx=1pΓ(z+1p)(p>0,Re(z)>−1)\int_{0}^{+\infty } x^{z} e^{-x^{p}} d x = \frac{1}{p} \Gamma \left(\frac{z+1}{p}\right)(p > 0, \mathrm{Re}\left(z\right) > -1);

(3)

∫0+∞tz−1e−ζtdt=ζ−zΓ(z)(Re(z)>0,∣arg⁡ζ∣<π2)\quad \int_{0}^{+\infty } t^{z-1} e^{-\zeta t} d t = \zeta^{-z} \Gamma (z) \quad \left(\mathrm{Re}\left(z\right) > 0,\left|\arg \zeta \right| < \frac{\pi }{2}\right);

(4)

∫0+∞tz−1cos⁡tdt=Γ(z)cos⁡πz2(0<Re(z)>1)\quad \int_{0}^{+\infty } t^{z-1} \cos t d t = \Gamma (z) \cos \frac{\pi z}{2} \quad (0 < \mathrm{Re}\left(z\right) > 1);

(5)

∫0+∞tz−1sin⁡tdt=\quad \int_{0}^{+\infty } t^{z-1} \sin t d t =

{Γ(z)sin⁡πz2,z≠0(−1<Re(z)<1),π2,z=0; \begin{cases} \Gamma (z) \sin \frac{\pi z}{2}, & z \neq 0(-1 < \mathrm{Re}\left(z\right) < 1), \\ \frac{\pi }{2}, & z = 0 ; \end{cases}
(6)

∫0+∞sin⁡xpxpdx=1p−1Γ(1p)cos⁡π2p(p>12)\int_{0}^{+\infty } \frac{\sin x^{p}}{x^{p}} d x = \frac{1}{p-1} \Gamma \left(\frac{1}{p}\right) \cos \frac{\pi }{2 p} \quad \left(p > \frac{1}{2}\right);

(7)

∫0+∞1−cos⁡xpx2pdx=p(p−1)(2p−1)Γ(1p)cos⁡π2p(p>12)\int_{0}^{+\infty } \frac{1-\cos x^{p}}{x^{2 p}} d x = \frac{p}{(p-1)(2 p-1)} \Gamma \left(\frac{1}{p}\right) \cos \frac{\pi }{2 p} \quad \left(p > \frac{1}{2}\right);

(8)

∫0+∞e−xpcos⁡pλcos⁡(xpsin⁡pλ)dx=1pΓ(1p)cos⁡λ(p>0,−π2p<λ<π2p);\int_{0}^{+\infty } e^{-x^{p} \cos p \lambda } \cos \left(x^{p} \sin p \lambda \right) d x = \frac{1}{p} \Gamma \left(\frac{1}{p}\right) \cos \lambda \left(p > 0,-\frac{\pi }{2 p} < \lambda < \frac{\pi }{2 p}\right) ;

(9)

∫0+∞xz−1e−xpcos⁡λsin⁡(xpsin⁡λ)dx=1pΓ(zp)sin⁡λzp\int_{0}^{+\infty } x^{z-1} e^{-x^{p} \cos \lambda } \sin \left(x^{p} \sin \lambda \right) d x = \frac{1}{p} \Gamma \left(\frac{z}{p}\right) \sin \frac{\lambda z}{p} (Re(z)>0,p>1,−π2<λ<π2)\left(\mathrm{Re}\left(z\right) > 0, p > 1,-\frac{\pi }{2} < \lambda < \frac{\pi }{2}\right).

Problem 24.3

Prove the formulas:

?
(1)

∫01tα−1(1−t2)β−1dt=12 B(α2,β)(Re(α)>0,Re(β)>0)\int_{0}^{1} t^{\alpha -1}\left(1-t^{2}\right)^{\beta -1} d t = \frac{1}{2} \mathrm{~ B}\left(\frac{\alpha }{2}, \beta \right) \quad (\mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\beta \right) > 0);

(2)

∫01tα−1(1−tp)β−1dt=1p B(αp,β)(p>0,Re(α)>0,Re(β)>0)\int_{0}^{1} t^{\alpha -1}\left(1-t^{p}\right)^{\beta -1} d t = \frac{1}{p} \mathrm{~ B}\left(\frac{\alpha }{p}, \beta \right)(p > 0, \mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\beta \right) > 0);

(3)

∫0+∞tα−1dt(1+t)α+β=B(α,β)(Re(α)>0,Re(β)>0)\int_{0}^{+\infty } \frac{t^{\alpha -1} d t}{(1+t)^{\alpha +\beta }} = \mathrm{B}(\alpha , \beta ) \quad (\mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\beta \right) > 0);

(4)

∫0+∞tα−1dt(1+t2)β=12 B(α2,β−α2)(Re(α)>0,Re((β−α2))>0)\int_{0}^{+\infty } \frac{t^{\alpha -1} d t}{\left(1+t^{2}\right)^{\beta }} = \frac{1}{2} \mathrm{~ B}\left(\frac{\alpha }{2}, \beta -\frac{\alpha }{2}\right) \quad \left(\mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\left(\beta -\frac{\alpha }{2}\right)\right) > 0\right);

(5)

∫0+∞tα−1dt(1+tp)β=1p B(αp,β−αp)(p>0,Re(α)>0,Re((β−αp))>0)\int_{0}^{+\infty } \frac{t^{\alpha -1} d t}{\left(1+t^{p}\right)^{\beta }} = \frac{1}{p} \mathrm{~ B}\left(\frac{\alpha }{p}, \beta -\frac{\alpha }{p}\right)\left(p > 0, \mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\left(\beta -\frac{\alpha }{p}\right)\right) > 0\right).

Problem 24.4

Prove the formulas:

?
(1)

∫01tα−1(1−t2)β−1(1+t2)(α/2)+βdt=2−(α/2)−1 B(α2,β)(Re(α)>0,Re(β)>0)\int_{0}^{1} \frac{t^{\alpha -1}\left(1-t^{2}\right)^{\beta -1}}{\left(1+t^{2}\right)^{(\alpha / 2)+\beta }} d t = 2^{-(\alpha / 2)-1} \mathrm{~ B}\left(\frac{\alpha }{2}, \beta \right) \quad (\mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\beta \right) > 0);

(2)

∫01tα−1(1−tp)β−1(1+tp)(α/p)+βdt=1p2−(α/p)B(αp,β)\int_{0}^{1} \frac{t^{\alpha -1}\left(1-t^{p}\right)^{\beta -1}}{\left(1+t^{p}\right)^{(\alpha / p)+\beta }} d t = \frac{1}{p} 2^{-(\alpha / p)} \mathrm{B}\left(\frac{\alpha }{p}, \beta \right) (here p>0,Re(α)>0,Re(β)>0p > 0, \mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\beta \right) > 0);

(3)

∫−11(1+t)α−1(1−t)β−1(t+a)α+βdt=2α+β−1(a+1)−α(a−1)−βB(α,β)\int_{-1}^{1} \frac{(1+t)^{\alpha -1}(1-t)^{\beta -1}}{(t+a)^{\alpha +\beta }} d t = 2^{\alpha +\beta -1}(a+1)^{-\alpha }(a-1)^{-\beta } \mathrm{B}(\alpha , \beta ) (here a>1,Re(α)>0,Re(β)>0a > 1, \mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\beta \right) > 0);

Problem 24.5
?
(1)

Let Re(α)>0,Re(β)>0\mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\beta \right) > 0. Prove that

∫−11(1+x)2α−1(1−x)2β−1(1+x2)α+βdx=2α+β−2 B(α,β) \int _{-1}^{1} \frac{(1+x)^{2 \alpha -1}(1-x)^{2 \beta -1}}{\left(1+x^{2}\right)^{\alpha +\beta }} d x = 2^{\alpha +\beta -2} \mathrm{~ B}(\alpha , \beta )
(2)

For a>0,Re(α)>0,Re(β)>0a > 0, \mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\beta \right) > 0, prove that

∫0π2(sin⁡φ)2α−1(cos⁡φ)2β−1(a+sin⁡2φ)α+βdφ=12(a+1)−αa−βB(α,β). \int _{0}^{\frac{\pi }{2}} \frac{(\sin \varphi )^{2 \alpha -1}(\cos \varphi )^{2 \beta -1}}{\left(a+\sin ^{2} \varphi \right)^{\alpha +\beta }} d \varphi = \frac{1}{2}(a+1)^{-\alpha } a^{-\beta } B(\alpha , \beta ).
(3)

For ∣Re(α)∣<∣Re(β)∣\left|\mathrm{Re}\left(\alpha \right)\right| < \left|\mathrm{Re}\left(\beta \right)\right|, prove that ∫0∞eαx(ch⁡x)−βdx=2β−1 B(β+α2,β−α2)\int_{0}^{\infty } e^{\alpha x}(\operatorname {ch} x)^{-\beta } d x = 2^{\beta -1} \mathrm{~ B}\left(\frac{\beta +\alpha }{2}, \frac{\beta -\alpha }{2}\right).

(4)

For ∣Re(α)∣<∣Re(β)∣\left|\mathrm{Re}\left(\alpha \right)\right| < \left|\mathrm{Re}\left(\beta \right)\right|, prove that

∫0+∞ch⁡αx(ch⁡x)βdx=2β−1 B(β+α2,β−α2). \int _{0}^{+\infty } \frac{\operatorname {ch} \alpha x}{(\operatorname {ch} x)^{\beta }} d x = 2^{\beta -1} \mathrm{~ B}\left(\frac{\beta +\alpha }{2}, \frac{\beta -\alpha }{2}\right).
(5)

For Re(α)>0,Re(α+β)<0\mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\alpha +\beta \right) < 0, prove that

∫0+∞(sh⁡x)2α−1(ch⁡x)2β−1dx=12 B(α,−α−β) \int _{0}^{+\infty }(\operatorname {sh} x)^{2 \alpha -1}(\operatorname {ch} x)^{2 \beta -1} d x = \frac{1}{2} \mathrm{~ B}(\alpha ,-\alpha -\beta )
Problem 24.6

Let Γ\Gamma be an arbitrary rectifiable curve going from the point z=−π2z = -\frac{\pi }{2} to the point z=π2z = \frac{\pi }{2}, remaining in the half-plane {z:Im(z)>0}\left\{ z: \mathrm{Im}\left(z\right) > 0\right\}. By (sin⁡z)α(\sin z)^{\alpha } on the curve Γ\Gamma we mean the branch of this function in the half-plane {z:Im(z)>0}\left\{ z: \mathrm{Im}\left(z\right) > 0\right\} which equals one at the point z=−π2z = -\frac{\pi }{2}. Prove that for all complex values of α\alpha the formula holds

∫Γ(sin⁡z)αdz=1+eπiα2B(α+12,12) \int _{\Gamma }(\sin z)^{\alpha } d z = \frac{1+e^{\pi i \alpha }}{2} B\left(\frac{\alpha +1}{2}, \frac{1}{2}\right)
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Problem 24.7

Let Γ\Gamma be an arbitrary rectifiable curve going from the point z=−iz = -i to the point z=iz = i, remaining in the half-plane {z:Re(z)>0}\left\{ z: \mathrm{Re}\left(z\right) > 0\right\}. By zα−1z^{\alpha -1} and (z2+1)β−1\left(z^{2}+1\right)^{\beta -1} on the curve Γ\Gamma we mean the branches of these functions in the half-plane {z:Re(z)>0}\left\{ z: \mathrm{Re}\left(z\right) > 0\right\} which equal one at the points z=1z = 1 and z=0z = 0 respectively. Prove that for Re(β)>0\mathrm{Re}\left(\beta \right) > 0 and any complex value of α\alpha the formula holds

∫Γzα−1(z2+1)β−1dz=icos⁡πα2 B(α2,β) \int _{\Gamma } z^{\alpha -1}\left(z^{2}+1\right)^{\beta -1} d z = i \cos \frac{\pi \alpha }{2} \mathrm{~ B}\left(\frac{\alpha }{2}, \beta \right)
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Problem 24.8

Prove that

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(1)

for any α\alpha and for Re(β)>0\mathrm{Re}\left(\beta \right) > 0

∫0π2(cos⁡φ)β−1cos⁡αφdφ=2−βsin⁡π(β−α)2 B(α−β+12,β); \int _{0}^{\frac{\pi }{2}}(\cos \varphi )^{\beta -1} \cos \alpha \varphi d \varphi = 2^{-\beta } \sin \frac{\pi (\beta -\alpha )}{2} \mathrm{~ B}\left(\frac{\alpha -\beta +1}{2}, \beta \right) ;
(2)

for any α\alpha and for Re(β)>−12\mathrm{Re}\left(\beta \right) > -\frac{1}{2}

∫0π/2(cos⁡φ)2βcos⁡2αφdφ=2−2β−2cos⁡π(α−β)B(α−β,2β)+12 B(β+12,12); \int _{0}^{\pi / 2}(\cos \varphi )^{2 \beta } \cos ^{2} \alpha \varphi d \varphi = 2^{-2 \beta -2} \cos \pi (\alpha -\beta ) \mathrm{B}(\alpha -\beta , 2 \beta )+\frac{1}{2} \mathrm{~ B}\left(\beta +\frac{1}{2}, \frac{1}{2}\right) ;
(3)

for Re(β)<1\mathrm{Re}\left(\beta \right) < 1 and Re(α−β)>0\mathrm{Re}\left(\alpha -\beta \right) > 0

∫0∞e−αx(sh⁡x)βdx=2β−1 B(α+β2,1−β) \int _{0}^{\infty } \frac{e^{-\alpha x}}{(\operatorname {sh} x)^{\beta }} d x = 2^{\beta -1} \mathrm{~ B}\left(\frac{\alpha +\beta }{2}, 1-\beta \right)
(4)

for Re(β)>−1,Re(α+β)<0,Re(α−β)>0\mathrm{Re}\left(\beta \right) > -1, \mathrm{Re}\left(\alpha +\beta \right) < 0, \mathrm{Re}\left(\alpha -\beta \right) > 0

∫0∞(sh⁡x)βsh⁡αxdx=2−β−2[ B(β+1,−α+β2)−B(β+1,α−β2)] \int _{0}^{\infty }(\operatorname {sh} x)^{\beta } \operatorname {sh} \alpha x d x = 2^{-\beta -2}\left[\mathrm{~ B}\left(\beta +1,-\frac{\alpha +\beta }{2}\right)-\mathrm{B}\left(\beta +1, \frac{\alpha -\beta }{2}\right)\right]
Problem 24.9

Prove that for Re(α)>0,Re(β)>0\mathrm{Re}\left(\alpha \right) > 0, \mathrm{Re}\left(\beta \right) > 0

∫−π/4π/4(cos⁡φ−sin⁡φ)2α−1(cos⁡φ+sin⁡φ)2β−1dφ=2α+β−2 B(α,β) \int _{-\pi / 4}^{\pi / 4}(\cos \varphi -\sin \varphi )^{2 \alpha -1}(\cos \varphi +\sin \varphi )^{2 \beta -1} d \varphi = 2^{\alpha +\beta -2} \mathrm{~ B}(\alpha , \beta )
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Problem 24.10

Prove that for 0<Re(α)<10 < \mathrm{Re}\left(\alpha \right) < 1 the formula holds

∫−π/4π/4(cos⁡φ−sin⁡φcos⁡φ+sin⁡φ)2α−1dφ=π2sin⁡πα \int _{-\pi / 4}^{\pi / 4}\left(\frac{\cos \varphi -\sin \varphi }{\cos \varphi +\sin \varphi }\right)^{2 \alpha -1} d \varphi = \frac{\pi }{2 \sin \pi \alpha }
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Problem 24.11

Prove the formula

12πi∫Γet⋅t−zdt=1Γ(z), \frac{1}{2 \pi i} \int _{\Gamma } e^{t} \cdot t^{-z} d t = \frac{1}{\Gamma (z)},

where Γ\Gamma is the positively oriented boundary of the domain

{t:∣t∣>ρ,∣arg⁡t∣<π−η},(0⩽η⩽π2) \left\{ t:\left|t\right| > \rho , \quad \left|\arg t\right| < \pi -\eta \right\} , \quad \left(0 \leqslant \eta \leqslant \frac{\pi }{2}\right)
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Problem 24.12

Denote by Dσ,ρD_{\sigma , \rho } the half-plane {z:Re(z)<σ}\left\{ z: \mathrm{Re}\left(z\right) < \sigma \right\} from which the disks {z:∣z+n∣<ρ},n=1,2,…\left\{ z:\left|z+n\right| < \rho \right\} , n = 1,2, \ldots have been removed. Prove that for any fixed values of the constants ρ>0,σ\rho > 0, \sigma, and mm the inequality holds

∣Γ(z)∣⩽M(1+∣z∣)−m(z∈Dσ,ρ) \left|\Gamma (z)\right| \leqslant M(1+\left|z\right|)^{-m} \quad \left(z \in D_{\sigma , \rho }\right)

with some constant MM depending on the choice of the numbers ρ,σ,m\rho , \sigma , m.

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Problem 24.13

Prove that

∣Γ(12+iξ)∣=πch⁡πξ,−∞<ξ<∞. \left|\Gamma \left(\frac{1}{2}+i \xi \right)\right| = \sqrt{\frac{\pi }{\operatorname {ch} \pi \xi }}, \quad -\infty < \xi < \infty .
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Problem 24.14

Prove that for any values of ss lying in the angle {s:∣arg⁡s∣<π2}\left\{ s:\left|\arg s\right| < \frac{\pi }{2}\right\} the formula holds

12πi∫12−i∞12+i∞Γ(z)s−zdz=e−s \frac{1}{2 \pi i} \int _{\frac{1}{2}-i \infty }^{\frac{1}{2}+i \infty } \Gamma (z) s^{-z} d z = e^{-s}
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