Zurück zu Bestimmte Integrale
Betrachte die Formel ∫ − ∞ ∞ sinh u x sinh γ x d x = π γ tan ( u π 2 γ ) {\displaystyle \int _{-\infty }^{\infty }{\frac {\sinh ux}{\sinh \gamma x}}\,dx={\frac {\pi }{\gamma }}\,\tan \left({\frac {u\pi }{2\gamma }}\right)} für u = β + i a {\displaystyle u=\beta +ia} mit | Re ( β ) | + | Im ( a ) | < Re ( γ ) {\displaystyle \left|{\text{Re}}(\beta )\right|+\left|{\text{Im}}(a)\right|<{\text{Re}}(\gamma )} . Wegen sinh ( β + i a ) x + sinh ( β − i a ) x = 2 cos a x sinh β x {\displaystyle \sinh(\beta +ia)x+\sinh(\beta -ia)x=2\cos ax\,\sinh \beta x} ist ∫ − ∞ ∞ cos a x sinh β x sinh γ x d x = π 2 γ tan ( β + i a ) π 2 γ + π 2 γ tan ( β − i a ) π 2 γ {\displaystyle \int _{-\infty }^{\infty }\cos ax\,\,{\frac {\sinh \beta x}{\sinh \gamma x}}\,dx={\frac {\pi }{2\gamma }}\tan {\frac {(\beta +ia)\pi }{2\gamma }}+{\frac {\pi }{2\gamma }}\tan {\frac {(\beta -ia)\pi }{2\gamma }}} . Integriere nach a {\displaystyle a\,} : ∫ − ∞ ∞ sin a x x sinh β x sinh γ x d x = i log ( cos ( β + i a ) π 2 γ ) − i log ( cos ( β − i a ) π 2 γ ) {\displaystyle \int _{-\infty }^{\infty }{\frac {\sin ax}{x}}\,\,{\frac {\sinh \beta x}{\sinh \gamma x}}\,dx=i\,\log \left(\cos {\frac {(\beta +ia)\pi }{2\gamma }}\right)-i\,\log \left(\cos {\frac {(\beta -ia)\pi }{2\gamma }}\right)} Und das lässt sich schreiben als = i log ( cos β π 2 γ cosh a π 2 γ − i sin β π 2 γ sinh a π 2 γ ) − i log ( cos β π 2 γ cosh a π 2 γ + i sin β π 2 γ sinh a π 2 γ ) {\displaystyle =i\,\log \left(\cos {\frac {\beta \pi }{2\gamma }}\,\cosh {\frac {a\pi }{2\gamma }}-i\,\sin {\frac {\beta \pi }{2\gamma }}\,\sinh {\frac {a\pi }{2\gamma }}\right)-i\,\log \left(\cos {\frac {\beta \pi }{2\gamma }}\,\cosh {\frac {a\pi }{2\gamma }}+i\,\sin {\frac {\beta \pi }{2\gamma }}\,\sinh {\frac {a\pi }{2\gamma }}\right)} = i log ( 1 − i tan β π 2 γ tanh a π 2 γ ) − i log ( 1 + i tan β π 2 γ tanh a π 2 γ ) {\displaystyle =i\,\log \left(1-i\,\tan {\frac {\beta \pi }{2\gamma }}\,\tanh {\frac {a\pi }{2\gamma }}\right)-i\,\log \left(1+i\,\tan {\frac {\beta \pi }{2\gamma }}\,\tanh {\frac {a\pi }{2\gamma }}\right)} = 2 arctan ( tan β π 2 γ tanh a π 2 γ ) {\displaystyle =2\,\arctan \left(\tan {\frac {\beta \pi }{2\gamma }}\,\,\tanh {\frac {a\pi }{2\gamma }}\right)} , da arctan z = i 2 ( log ( 1 − i z ) − log ( 1 + i z ) ) {\displaystyle \arctan z={\frac {i}{2}}{\Big (}\log(1-iz)-\log(1+iz){\Big )}} ist.