Formelsammlung Mathematik: Integraltransformationen

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1.1

$F(s)={\mathcal {F}}[f(t)](s)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }f(t)e^{-ist}\,dt$

ohne Beweis (Fourier-Transformation)

 

1.2

$f(t)={\mathcal {F}}^{-1}[F(s)](t)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }F(s)e^{ist}\,ds$

ohne Beweis (Rücktransformation)

 

1.3

${\mathcal {F}}[(f*g)(t)](s)={\sqrt {2\pi }}\cdot {\mathcal {F}}[f(t)](s)\cdot {\mathcal {F}}[g(t)](s)$

Beweis (Faltung)

${\mathcal {F}}[(f*g)(t)](s)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f(u)\,g(t-u)\,du\,\,e^{-ist}\,dt$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f(u)\,e^{-isu}\,g(t-u)\,e^{-is(t-u)}\,dt\,du$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f(u)\,e^{-isu}\,g(v)\,e^{-isv}\,dv\,du$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }f(u)\,e^{-isu}\,du\cdot \int _{-\infty }^{\infty }g(v)\,e^{-isv}\,dv={\sqrt {2\pi }}\cdot {\mathcal {F}}[f(t)](s)\cdot {\mathcal {F}}[g(t)](s)$

 

2.1

$F(s)={\mathcal {L}}[f(t)](s)=\int _{0}^{\infty }f(t)e^{-st}\,dt$

ohne Beweis (Laplace-Transformation)

 

2.2

$f(t)={\mathcal {L}}^{-1}[F(s)](t)={\frac {1}{2\pi i}}\int _{a-i\infty }^{a+i\infty }F(s)e^{st}\,ds$

Beweis (Rücktransformation, Bromwich-Integral)

Im Folgenden sei $f(t)=0\,$ für $t<0\,$.

$f(t)e^{-at}={\mathcal {F}}^{-1}\left[{\mathcal {F}}[f(\tau )\,e^{-a\tau }](\omega )\right](t)$

$={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f(\tau )\,e^{-a\tau }\,e^{-i\omega \tau }\,d\tau \,e^{i\omega t}\,d\omega$

$\Rightarrow f(t)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{\infty }f(\tau )\,e^{-(a+i\omega )\tau }\,d\tau \,e^{(a+i\omega )t}d\omega$

Substituiert man $s=a+i\omega \,\,(ds=i\,d\omega )$, so ist

$f(t)={\frac {1}{2\pi i}}\int _{a-i\infty }^{a+i\infty }\underbrace {\int _{0}^{\infty }f(\tau )e^{-s\tau }\,d\tau } _{F(s)}\,e^{st}\,ds$.

 

2.3

${\mathcal {L}}[(f*g)(t)](s)=F(s)\cdot G(s)$

Beweis (Faltung)

$F(s)\cdot G(s)=\int _{0}^{\infty }f(\sigma )\,e^{-s\sigma }\,d\sigma \cdot \int _{0}^{\infty }g(\tau )\,e^{-s\tau }\,d\tau =\int _{0}^{\infty }\int _{0}^{\infty }f(\sigma )\,e^{-s(\sigma +\tau )}\,d\sigma \,g(\tau )\,d\tau$

Nach Substitution $t=\sigma +\tau \,(dt=d\sigma )$ ist das $\int _{0}^{\infty }\int _{\tau }^{\infty }f(t-\tau )\,e^{-st}\,dt\,g(\tau )\,d\tau$.

Und nach Vertauschung der Integrationsreihenfolge ist das

$\int _{0}^{\infty }\int _{0}^{t}f(t-\tau )\,g(\tau )\,d\tau \,e^{-st}\,dt=\int _{0}^{\infty }(f*g)(t)\,e^{-st}\,dt={\mathcal {L}}[(f*g)(t)](s)$.

 

3.1

$F(s)={\mathcal {M}}[f(t)](s)=\int _{0}^{\infty }f(t)\,t^{s-1}\,dt$

ohne Beweis (Mellin-Transformation)

 

3.2

$f(t)={\mathcal {M}}^{-1}[F(s)](t)={\frac {1}{2\pi i}}\int _{a-i\infty }^{a+i\infty }F(s)\,t^{-s}\,ds$

Beweis (Rücktransformation)

$f(e^{t})\,e^{at}={\mathcal {F}}\left[{\mathcal {F}}^{-1}[f(e^{\tau })\,e^{a\tau }](\omega )\right](t)$

$={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f(e^{\tau })\,e^{a\tau }\,e^{i\omega \tau }\,d\tau \,e^{-i\omega t}\,d\omega$

$\Rightarrow f(e^{t})={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f(e^{\tau })\,e^{(a+i\omega )\tau }\,d\tau \,e^{-(a+i\omega )t}\,d\omega$

Substituiert man $s=a+i\omega \,\,(ds=i\,d\omega )$, so ist

$f(e^{t})={\frac {1}{2\pi i}}\int _{a-i\infty }^{a+i\infty }\int _{-\infty }^{\infty }f(e^{\tau })\,e^{s\tau }\,d\tau \,e^{-st}\,ds$.

Substituiert man $u=e^{\tau }\,$, so ist

$f(e^{t})={\frac {1}{2\pi i}}\int _{a-i\infty }^{a+i\infty }\underbrace {\int _{0}^{\infty }f(u)\,u^{s-1}\,du} _{F(s)}\,e^{-st}\,ds$.

Ersetze $e^{t}\,$ durch $t\,$, um die gewünschte Formel zu erhalten.

 

3.3

${\mathcal {M}}\left[(f*g)(t)\right](s)=F(s)\cdot G(s)$

Beweis (Faltung)

${\mathcal {M}}\left[(f*g)(t)\right](s)=\int _{0}^{\infty }\int _{0}^{\infty }f\left({\frac {t}{u}}\right)g(u)\,{\frac {du}{u}}\cdot t^{s-1}\,dt$

$=\int _{0}^{\infty }\int _{0}^{\infty }f\left({\frac {t}{u}}\right)t^{s-1}\,dt\cdot g(u)\,{\frac {du}{u}}=\int _{0}^{\infty }\int _{0}^{\infty }f(t)\,(ut)^{s-1}\,u\,dt\cdot g(u)\,{\frac {du}{u}}$

$=\int _{0}^{\infty }\int _{0}^{\infty }f(t)\,t^{s-1}\,dt\cdot g(u)\,u^{s-1}\,du=F(s)\cdot G(s)$