Aufgaben:Problem 11
Contents
Problem a)
$$f(t)=e^{-|t|} \in L^1(\mathbb{R})$$
Compute the Fourier transform of \(f(t)\)
Solution
$$ \begin{align} \hat f(x) \ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \mathrm e^{-|t|}e^{-ixt}\,\mathrm dt \\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{0} \mathrm e^{t(1-ix)}\,\mathrm dt + \int_{0}^{\infty} \mathrm e^{-t(1+ix)}\,\mathrm dt \\ &= \frac{1}{\sqrt{2\pi}} \left[ \frac{1}{1-ix}e^{t(1-ix)} \bigg \vert_{t=-\infty}^{t=0} - \frac{1}{1+ix} e^{-t(1+ix)} \bigg \vert_{t=0}^{t=\infty} \right] \\ &= \frac{1}{\sqrt{2\pi}} \left[ \frac{2}{1+x^2} \right] \\ &= \sqrt{\frac{2}{\pi}} \frac{1}{1+x^2} \end{align} $$
Problem b)
Using the result from a), compute $$ \int_{0}^{\infty} \frac{1}{1+x^2} \,\mathrm dx $$ and $$ \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2}\,\mathrm dx \ , t>0 $$
Solution
by using inverse fourier transform $$ \begin{align} e^{-|t|} &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \sqrt{\frac{2}{\pi}} \frac{1}{1+x^2} e^{ixt} \, dx \\ &= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{e^{ixt}}{1+x^2} \, dx \\ &= \frac{1}{\pi} \int_{0}^{\infty} \frac{e^{ixt} + e^{-ixt}}{1+x^2} \, dx \\ &= \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx \end{align} $$
thus we can set t=0
$$ \begin{align} 1 = \frac{2}{\pi} \int_{0}^{\infty} \frac{1}{1+x^2} \, dx \\ \Rightarrow \int_{0}^{\infty} \frac{1}{1+x^2} \, dx = \frac{\pi}{2} \end{align} $$
Claim: $$ \frac{d}{dt} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{-x \sin(xt)}{1+x^2} \, dx , t \gt 0 $$ Proof: Muss mittels gleichmässiger Konvergenz bzgl. dem oberen Grenzwert erfolgen... tbd
use the proof for \( t>0 \) $$ \begin{align} \ \frac{d}{dt} e^{-|t|} = \frac{d}{dt} \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx \\ \Leftrightarrow e^{-t} = \frac{2}{\pi} \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx \\ \Rightarrow \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx = \frac{\pi}{2} e^{-t} \\ \end{align} $$
