Zurück zu Bestimmte Integrale
Setzt man f ( z ) = 2 α ( α 2 + z 2 ) cosh π z {\displaystyle f(z)={\frac {2\alpha }{(\alpha ^{2}+z^{2})\cosh \pi z}}} , so ist 2 π i res ( f , i α ) = 2 π cos α π = π cot [ ( 1 4 + α 2 ) π ] − π cot [ ( 3 4 + α 2 ) π ] {\displaystyle 2\pi i\,\;{\text{res}}(f,i\alpha )={\frac {2\pi }{\cos \alpha \pi }}=\pi \cot \left[\left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\pi \right]-\pi \cot \left[\left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)\pi \right]} . Und 2 π i ∑ k = 0 ∞ res ( f , i 2 k + 1 2 ) = ∑ k = 0 ∞ ( − 1 ) k [ 4 2 k + 1 + 2 α − 4 2 k + 1 − 2 α ] {\displaystyle 2\pi i\sum _{k=0}^{\infty }{\text{res}}\left(f,i\,{\frac {2k+1}{2}}\right)=\sum _{k=0}^{\infty }(-1)^{k}\left[{\frac {4}{2k+1+2\alpha }}-{\frac {4}{2k+1-2\alpha }}\right]} = [ ψ ( 3 4 + α 2 ) − ψ ( 1 4 + α 2 ) ] − [ ψ ( 3 4 − α 2 ) − ψ ( 1 4 − α 2 ) ] {\displaystyle =\left[\psi \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right]-\left[\psi \left({\frac {3}{4}}-{\frac {\alpha }{2}}\right)-\psi \left({\frac {1}{4}}-{\frac {\alpha }{2}}\right)\right]} , wobei ψ ( 3 4 − α 2 ) = ψ ( 1 − ( 1 4 + α 2 ) ) = ψ ( 1 4 + α 2 ) + π cot [ ( 1 4 + α 2 ) π ] {\displaystyle \psi \left({\frac {3}{4}}-{\frac {\alpha }{2}}\right)=\psi \left(1-\left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right)=\psi \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)+\pi \cot \left[\left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\pi \right]} und ψ ( 1 4 − α 2 ) = ψ ( 1 − ( 3 4 + α 2 ) ) = ψ ( 3 4 + α 2 ) + π cot [ ( 3 4 + α 2 ) π ] {\displaystyle \psi \left({\frac {1}{4}}-{\frac {\alpha }{2}}\right)=\psi \left(1-\left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)\right)=\psi \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)+\pi \cot \left[\left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)\pi \right]} ist. ∫ − ∞ ∞ f ( x ) d x = lim N → ∞ ∮ γ N f d z = 2 π i ∑ Im > 0 res f = 2 [ ψ ( 3 4 + α 2 ) − ψ ( 1 4 + α 2 ) ] {\displaystyle \int _{-\infty }^{\infty }f(x)\,dx=\lim _{N\to \infty }\oint _{\gamma _{N}}f\,dz=2\pi i\,\sum _{{\text{Im}}>0}{\text{res}}f=2\left[\psi \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right]} . Integriere nun beide Seiten nach α {\displaystyle \alpha \,} : ∫ − ∞ ∞ log ( α 2 + x 2 ) cosh π x d x ⏟ =: U ( α ) = 4 [ log Γ ( 3 4 + α 2 ) − log Γ ( 1 4 + α 2 ) ] ⏟ =: V ( α ) + C {\displaystyle \underbrace {\int _{-\infty }^{\infty }{\frac {\log(\alpha ^{2}+x^{2})}{\cosh \pi x}}\,dx} _{=:U(\alpha )}=\underbrace {4\left[\log \Gamma \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)-\log \Gamma \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right]} _{=:V(\alpha )}+C} Wegen U ( α ) − log ( α 2 ) = U ( α ) − ∫ − ∞ ∞ log ( α 2 ) cosh π x d x = ∫ − ∞ ∞ log ( 1 + x 2 α 2 ) cosh π x d x → 0 {\displaystyle U(\alpha )-\log(\alpha ^{2})=U(\alpha )-\int _{-\infty }^{\infty }{\frac {\log(\alpha ^{2})}{\cosh \pi x}}\,dx=\int _{-\infty }^{\infty }{\frac {\log \left(1+{\frac {x^{2}}{\alpha ^{2}}}\right)}{\cosh \pi x}}\,dx\to 0} für α → ∞ {\displaystyle \alpha \to \infty \,} und V ( α ) − log ( α 2 ) → − 4 log 2 {\displaystyle V(\alpha )-\log(\alpha ^{2})\to -4\log {\sqrt {2}}} muss C = 4 log 2 {\displaystyle C=4\log {\sqrt {2}}} sein. Daher lässt sich die rechte Seite auch schreiben als 4 log ( 2 Γ ( 3 4 + α 2 ) Γ ( 1 4 + α 2 ) ) {\displaystyle 4\log \left({\sqrt {2}}\,\;{\frac {\Gamma \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)}}\right)} .
In der Formel ∫ 0 1 log log ( 1 x ) 1 + 2 cos α π ⋅ x + x 2 d x = π 2 sin α π ( α log 2 π + log Γ ( 1 2 + α 2 ) Γ ( 1 2 − α 2 ) ) 0 < α < 1 {\displaystyle \int _{0}^{1}{\frac {\log \log \left({\frac {1}{x}}\right)}{1+2\cos \alpha \pi \cdot x+x^{2}}}\,dx={\frac {\pi }{2\sin \alpha \pi }}\left(\alpha \log 2\pi +\log {\frac {\Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{2}}-{\frac {\alpha }{2}}\right)}}\right)\qquad 0<\alpha <1} substituiere x ↦ e − x {\displaystyle x\mapsto e^{-x}} .
, so ist
![{\displaystyle 2\pi i\,\;{\text{res}}(f,i\alpha )={\frac {2\pi }{\cos \alpha \pi }}=\pi \cot \left[\left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\pi \right]-\pi \cot \left[\left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)\pi \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c15e031e89b5bd57679aef84cbbe819e9df52615)
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![{\displaystyle 2\pi i\sum _{k=0}^{\infty }{\text{res}}\left(f,i\,{\frac {2k+1}{2}}\right)=\sum _{k=0}^{\infty }(-1)^{k}\left[{\frac {4}{2k+1+2\alpha }}-{\frac {4}{2k+1-2\alpha }}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3336f3a2b706df27f2b46e654e451fc4207e0530)
![{\displaystyle =\left[\psi \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right]-\left[\psi \left({\frac {3}{4}}-{\frac {\alpha }{2}}\right)-\psi \left({\frac {1}{4}}-{\frac {\alpha }{2}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f58b5dd534c5e2ad45b8e4f3fc3d7cac03580347)
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![{\displaystyle \psi \left({\frac {3}{4}}-{\frac {\alpha }{2}}\right)=\psi \left(1-\left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right)=\psi \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)+\pi \cot \left[\left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\pi \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/908197a31593da9c9da4db318f2361976a033789)
![{\displaystyle \psi \left({\frac {1}{4}}-{\frac {\alpha }{2}}\right)=\psi \left(1-\left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)\right)=\psi \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)+\pi \cot \left[\left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)\pi \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2580f4aa67b1cbddab3d33a461e98703d2cd3b84)
ist.
![{\displaystyle \int _{-\infty }^{\infty }f(x)\,dx=\lim _{N\to \infty }\oint _{\gamma _{N}}f\,dz=2\pi i\,\sum _{{\text{Im}}>0}{\text{res}}f=2\left[\psi \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c0127dcebdd6c367c1d9ad81a274406300dea44)
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![{\displaystyle \underbrace {\int _{-\infty }^{\infty }{\frac {\log(\alpha ^{2}+x^{2})}{\cosh \pi x}}\,dx} _{=:U(\alpha )}=\underbrace {4\left[\log \Gamma \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)-\log \Gamma \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right]} _{=:V(\alpha )}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8da9804719b4f1dca8890d155b4a18da801222f9)
für 
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