a)
wegen f(t) = f(-t) fallen die Sinusanteile weg
\( F(ω) = \frac{1}{\sqrt{2*π}} \int\limits_{-\infty}^{+\infty} e^{-a*|t|}*cos(ωt)dt \)
\( F(ω) = \frac{2}{\sqrt{2*π}}\int\limits_{0}^{\infty} e^{-a*t}*cos(ωt)dt \)
\( F(ω) = \sqrt{ \frac{2}{π}} \int\limits_{0}^{\infty} e^{-a*t}*cos(ωt)dt\)
Die Stammfunktion lautet \( G(t) = \frac{e^{-a*t}(ω*sin(ωt)+a*cos(ωt))}{ω^2+a^2} +C \)
\( G(0) = \frac{a}{ω^2+a^2} + C \) und \( G(+\infty) = 0 + C \) daraus folgt
\( F(ω) = \sqrt{ \frac{2}{π}} * - \frac{a}{ω^2+a^2} \)
b)
\( F(ω) = \frac{1}{\sqrt{2*π}} \int\limits_{0}^{+\infty} t* e^{-t}* cos(ωt) dt - \frac{i}{\sqrt{2*π}} \int\limits_{0}^{+\infty} t* e^{-t}* sin(ωt) dt \)
\( F(ω) = \frac{1}{\sqrt{2*π}} \frac{-ω^2+1}{ω^4+2ω^2+1} - \frac{i}{\sqrt{2*π}}\frac{2ω}{ω^4+2ω^2+1} \)
\( F(ω) = \frac{1}{\sqrt{2*π}} \frac{-ω^2 -2iω + 1}{ω^4+2ω^2+1} = \frac{1}{\sqrt{2*π}} \frac{1}{(1+iω)^2} \)
oder
\( F(ω) = \frac{1}{\sqrt{2*π}} \int\limits_{0}^{+\infty} t* e^{-t}* e^{-i*ωt} dt = \frac{1}{\sqrt{2*π}} \frac{1}{(1+iw)^2} \)
((Hinweis: Integralbestimmung über Apps)
