Section 24

Integrals Reducible to the Gamma Function

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Problem 24.1

Prove the formulas:

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(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:

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(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:

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(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:

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(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
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(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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