Zurück zu Bestimmte Integrale
Für k = 1 , 2 {\displaystyle k=1,2\,} sei u k ( t ) = Γ ( α k + i t ) β k α k + i t {\displaystyle u_{k}(t)={\frac {\Gamma (\alpha _{k}+it)}{\beta _{k}^{\alpha _{k}+it}}}} und f k ( z ) = Γ ( z ) ( β k e ω ) z {\displaystyle f_{k}(z)={\frac {\Gamma (z)}{(\beta _{k}\,e^{\omega })^{z}}}} mit ω ∈ R {\displaystyle \omega \in \mathbb {R} } . Berechne die Fouriertransformierte F [ u k ] ( ω ) = ∫ − ∞ ∞ Γ ( α k + i t ) β k α k + i t e − i ω t d t {\displaystyle {\mathcal {F}}[u_{k}](\omega )=\int _{-\infty }^{\infty }{\frac {\Gamma (\alpha _{k}+it)}{\beta _{k}^{\alpha _{k}+it}}}\,e^{-i\omega t}\,dt} = − i e ω α k ∫ − ∞ ∞ Γ ( α k + i t ) ( β k e ω ) α k + i t i d t = − i e ω α k ∫ α k + i R f k d z {\displaystyle =-i\,e^{\omega \alpha _{k}}\int _{-\infty }^{\infty }{\frac {\Gamma (\alpha _{k}+it)}{(\beta _{k}\,e^{\omega })^{\alpha _{k}+it}}}\,i\,dt=-i\,e^{\omega \alpha _{k}}\int _{\alpha _{k}+i\mathbb {R} }f_{k}\,dz} = − i e ω α k lim N → ∞ ∮ γ N f k d z = 2 π e ω α k ∑ n = 0 ∞ res ( f k , − n ) {\displaystyle =-i\,e^{\omega \alpha _{k}}\,\lim _{N\to \infty }\oint _{\gamma _{N}}f_{k}\,dz=2\pi \,e^{\omega \alpha _{k}}\,\sum _{n=0}^{\infty }{\text{res}}(f_{k},-n)} = 2 π e ω α k ∑ n = 0 ∞ ( − 1 ) n n ! ( β k e ω ) n = 2 π e ω α k e β k e ω {\displaystyle =2\pi \,e^{\omega \alpha _{k}}\,\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\,\left(\beta _{k}\,e^{\omega }\right)^{n}=2\pi \,{\frac {e^{\omega \alpha _{k}}}{e^{\beta _{k}\,e^{\omega }}}}} . Also ist ∫ − ∞ ∞ Γ ( α 1 − i t ) β 1 α 1 − i t Γ ( α 2 + i t ) β 2 α 2 + i t d t = ∫ − ∞ ∞ u 1 ( 0 − t ) u 2 ( t ) d t = ( u 1 ∗ u 2 ) ( 0 ) {\displaystyle \int _{-\infty }^{\infty }{\frac {\Gamma (\alpha _{1}-it)}{\beta _{1}^{\alpha _{1}-it}}}\,{\frac {\Gamma (\alpha _{2}+it)}{\beta _{2}^{\alpha _{2}+it}}}\,dt=\int _{-\infty }^{\infty }u_{1}(0-t)\,u_{2}(t)\,dt=(u_{1}*u_{2})(0)} nach der Faltungsformel 1 2 π ∫ − ∞ ∞ F [ u 1 ] ( ω ) ⋅ F [ u 2 ] ( ω ) d ω = 2 π ∫ − ∞ ∞ e ω α 1 e β 1 e ω e ω α 2 e β 2 e ω d ω {\displaystyle {\frac {1}{2\pi }}\int _{-\infty }^{\infty }{\mathcal {F}}[u_{1}](\omega )\cdot {\mathcal {F}}[u_{2}](\omega )\,d\omega =2\pi \int _{-\infty }^{\infty }{\frac {e^{\omega \alpha _{1}}}{e^{\beta _{1}e^{\omega }}}}\,{\frac {e^{\omega \alpha _{2}}}{e^{\beta _{2}e^{\omega }}}}\,d\omega } . Und das ist nach Substitution e ω = t {\displaystyle e^{\omega }=t\,} gleich 2 π ∫ 0 ∞ t α 1 e β 1 t t α 2 e β 2 t d t t = 2 π ∫ 0 ∞ t α 1 + α 2 − 1 e − ( β 1 + β 2 ) t d t = 2 π Γ ( α 1 + α 2 ) ( β 1 + β 2 ) α 1 + α 2 {\displaystyle 2\pi \int _{0}^{\infty }{\frac {t^{\alpha _{1}}}{e^{\beta _{1}t}}}\,{\frac {t^{\alpha _{2}}}{e^{\beta _{2}t}}}\,{\frac {dt}{t}}=2\pi \int _{0}^{\infty }t^{\alpha _{1}+\alpha _{2}-1}\,e^{-(\beta _{1}+\beta _{2})t}\,dt=2\pi {\frac {\Gamma (\alpha _{1}+\alpha _{2})}{(\beta _{1}+\beta _{2})^{\alpha _{1}+\alpha _{2}}}}} .
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=\int _{-\infty }^{\infty }{\frac {\Gamma (\alpha _{k}+it)}{\beta _{k}^{\alpha _{k}+it}}}\,e^{-i\omega t}\,dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f25614151d410f3ecf76c8413248dd7b5119cf4d)
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\cdot {\mathcal {F}}[u_{2}](\omega )\,d\omega =2\pi \int _{-\infty }^{\infty }{\frac {e^{\omega \alpha _{1}}}{e^{\beta _{1}e^{\omega }}}}\,{\frac {e^{\omega \alpha _{2}}}{e^{\beta _{2}e^{\omega }}}}\,d\omega }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e680f5c0a2938cb3174ba6e8fbe206f0f209fe3)
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