Zurück zu Bestimmte Integrale
Verwende die Formel Jν(z)=12πi∫Cez2(t−1t)dttν+1{\displaystyle J_{\nu }(z)={\frac {1}{2\pi i}}\int _{C}e^{{\frac {z}{2}}\left(t-{\frac {1}{t}}\right)}\,{\frac {dt}{t^{\nu +1}}}} für Re(z)>0{\displaystyle {\text{Re}}(z)>0\,} . C{\displaystyle C\,} sei hierbei die Kurve, die geradlinig von −∞−iε{\displaystyle -\infty -i\varepsilon } nach −1−iε{\displaystyle -1-i\varepsilon } läuft, von dort im Kreisbogen Kε{\displaystyle K_{\varepsilon }} mit Radius 1+ε2{\displaystyle {\sqrt {1+\varepsilon ^{2}}}} nach −1+iε{\displaystyle -1+i\varepsilon } läuft, und von dort geradlinig nach −∞+iε{\displaystyle -\infty +i\varepsilon } läuft. Setzt man f(t)=ez2(t−1t)1tν+1{\displaystyle f(t)=e^{{\frac {z}{2}}\left(t-{\frac {1}{t}}\right)}\,{\frac {1}{t^{\nu +1}}}} , so ist Jν(z)=12πi∫−∞−iε−1−iεf(t)dt+12πi∫Kεf(t)dt+12πi∫−1+iε−∞+iεf(t)dt{\displaystyle J_{\nu }(z)={\frac {1}{2\pi i}}\int _{-\infty -i\varepsilon }^{-1-i\varepsilon }f(t)\,dt+{\frac {1}{2\pi i}}\int _{K_{\varepsilon }}f(t)\,dt+{\frac {1}{2\pi i}}\int _{-1+i\varepsilon }^{-\infty +i\varepsilon }f(t)\,dt} =12πi∫1∞f(−t−iε)dt−12πi∫1∞f(−t+iε)dt+12πi∫Kεf(t)dt{\displaystyle ={\frac {1}{2\pi i}}\int _{1}^{\infty }f(-t-i\varepsilon )\,dt-{\frac {1}{2\pi i}}\int _{1}^{\infty }f(-t+i\varepsilon )\,dt+{\frac {1}{2\pi i}}\int _{K_{\varepsilon }}f(t)\,dt} . limε→0+f(−t−iε)−f(−t+iε)=ez2(−t+1t)(−eiπνtν+1−−e−iπνtν+1)=−2iez2(−t+1t)⋅sinνπtν+1{\displaystyle \lim _{\varepsilon \to 0+}f(-t-i\varepsilon )-f(-t+i\varepsilon )=e^{{\frac {z}{2}}\left(-t+{\frac {1}{t}}\right)}\left({\frac {-e^{i\pi \nu }}{t^{\nu +1}}}-{\frac {-e^{-i\pi \nu }}{t^{\nu +1}}}\right)=-2i\,e^{{\frac {z}{2}}\left(-t+{\frac {1}{t}}\right)}\cdot {\frac {\sin \nu \pi }{t^{\nu +1}}}} ⇒12πi∫1∞f(−t−iε)dt−12πi∫1∞f(−t+iε)dt=−sinνππ∫1∞ez2(−t+1t)dttν+1{\displaystyle \Rightarrow {\frac {1}{2\pi i}}\int _{1}^{\infty }f(-t-i\varepsilon )\,dt-{\frac {1}{2\pi i}}\int _{1}^{\infty }f(-t+i\varepsilon )\,dt=-{\frac {\sin \nu \pi }{\pi }}\int _{1}^{\infty }e^{{\frac {z}{2}}\left(-t+{\frac {1}{t}}\right)}\,{\frac {dt}{t^{\nu +1}}}} , was nach Substitution t↦et{\displaystyle t\mapsto e^{t}\,} gleich −sinνππ∫0∞e−zsinht−νtdt{\displaystyle -{\frac {\sin \nu \pi }{\pi }}\int _{0}^{\infty }e^{-z\sinh t-\nu t}\,dt} ist. Und 12πi∫Kεf(t)dt⟶ε→0+∫−ππf(eit)ieitdt=12π∫−ππez2(eit−e−it)dteiνttν+1{\displaystyle {\frac {1}{2\pi i}}\int _{K_{\varepsilon }}f(t)\,dt\,\,{\stackrel {\varepsilon \to 0+}{\longrightarrow }}\,\int _{-\pi }^{\pi }f(e^{it})\,i\,e^{it}\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }e^{{\frac {z}{2}}\left(e^{it}-e^{-it}\right)}\,{\frac {dt}{e^{i\nu t}\,t^{\nu +1}}}} =12π∫−ππeizsint−iνtdt=1π∫0πcos(zsint−νt)dt{\displaystyle ={\frac {1}{2\pi }}\int _{-\pi }^{\pi }e^{iz\sin t-i\nu t}\,dt={\frac {1}{\pi }}\int _{0}^{\pi }\cos(z\sin t-\nu t)\,dt} . Also ist 1π∫0πcos(zsinx−νx)dx−sinνππ∫0∞e−zsinhx−νxdx{\displaystyle {\frac {1}{\pi }}\int _{0}^{\pi }\cos(z\sin x-\nu x)dx-{\frac {\sin \nu \pi }{\pi }}\int _{0}^{\infty }e^{-z\sinh x-\nu x}\,dx} .