Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,sin,sinh)

0.1

\int _{-\infty }^{\infty }{\frac {\sin \pi x^{2}}{\sinh ^{2}\pi x}}\,dx={\frac {2-{\sqrt {2}}}{2}}

ohne Beweis

2.1

\int _{-\infty }^{\infty }{\frac {\sin {\frac {p}{q}}\,\pi x^{2}}{\sinh ^{2}\pi x}}\,dx={\frac {1}{2}}\,{\text{Im}}\left(2\pi i\,\sum _{k=0}^{4pq}{\text{res}}\left(f,i\,{\frac {k}{2p}}\right)-i\pi \,{\text{res}}(f,0)-i\pi \,{\text{res}}\left(f,i\cdot 2q\right)\right)

mit

f(z)={\frac {e^{i{\frac {p}{q}}\,z^{2}}}{\sinh ^{2}\pi z}}\cdot {\frac {e^{2\pi pz}}{\sinh 2\pi pz}}

ohne Beweis

3.1

\int _{0}^{\infty }{\frac {\sin ax}{x}}\,{\frac {\sinh \beta x}{\sinh \gamma x}}\,dx=\arctan \left(\tan {\frac {\beta \pi }{2\gamma }}\,\,\tanh {\frac {a\pi }{2\gamma }}\right)\qquad |{\text{Re}}(\beta )|+|{\text{Im}}(a)|<{\text{Re}}(\gamma )

Beweis

Betrachte die Formel

$\int _{-\infty }^{\infty }{\frac {\sinh ux}{\sinh \gamma x}}\,dx={\frac {\pi }{\gamma }}\,\tan \left({\frac {u\pi }{2\gamma }}\right)$ für $u=\beta +ia$ mit $\left|{\text{Re}}(\beta )\right|+\left|{\text{Im}}(a)\right|<{\text{Re}}(\gamma )$ .

Wegen $\sinh(\beta +ia)x+\sinh(\beta -ia)x=2\cos ax\,\sinh \beta x$ ist

$\int _{-\infty }^{\infty }\cos ax\,\,{\frac {\sinh \beta x}{\sinh \gamma x}}\,dx={\frac {\pi }{2\gamma }}\tan {\frac {(\beta +ia)\pi }{2\gamma }}+{\frac {\pi }{2\gamma }}\tan {\frac {(\beta -ia)\pi }{2\gamma }}$ .

Integriere nach $a\,$ :

$\int _{-\infty }^{\infty }{\frac {\sin ax}{x}}\,\,{\frac {\sinh \beta x}{\sinh \gamma x}}\,dx=i\,\log \left(\cos {\frac {(\beta +ia)\pi }{2\gamma }}\right)-i\,\log \left(\cos {\frac {(\beta -ia)\pi }{2\gamma }}\right)$

Und das lässt sich schreiben als

$=i\,\log \left(\cos {\frac {\beta \pi }{2\gamma }}\,\cosh {\frac {a\pi }{2\gamma }}-i\,\sin {\frac {\beta \pi }{2\gamma }}\,\sinh {\frac {a\pi }{2\gamma }}\right)-i\,\log \left(\cos {\frac {\beta \pi }{2\gamma }}\,\cosh {\frac {a\pi }{2\gamma }}+i\,\sin {\frac {\beta \pi }{2\gamma }}\,\sinh {\frac {a\pi }{2\gamma }}\right)$

$=i\,\log \left(1-i\,\tan {\frac {\beta \pi }{2\gamma }}\,\tanh {\frac {a\pi }{2\gamma }}\right)-i\,\log \left(1+i\,\tan {\frac {\beta \pi }{2\gamma }}\,\tanh {\frac {a\pi }{2\gamma }}\right)$

$=2\,\arctan \left(\tan {\frac {\beta \pi }{2\gamma }}\,\,\tanh {\frac {a\pi }{2\gamma }}\right)$ , da $\arctan z={\frac {i}{2}}{\Big (}\log(1-iz)-\log(1+iz){\Big )}$ ist.