Revision as of 09:07, 25 December 2014

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:

fix \(t\) and set \(G(x) := \cos(xt) \), for \(h \ne 0 \) consider $$ \int_{0}^{\infty} \frac{G(t+h) - G(t)}{h} \frac{1}{1+x^2} \, dx $$ by mean value theorem we find \( \epsilon \) between \( 0 \) and \( h \) s.t. $$ \frac{G(t+h) - G(t)}{h} = \frac{d}{dt}G(t+\epsilon) = -x \sin(x(t+\epsilon)) $$

use sequence \(h_n \ne 0\) that converges to \( 0 \) and define $$ g_n := \frac{G(t+h_n)-G(t)}{h_n} \frac{1}{1+x^2} \\ |g_n| \le \frac{|x|}{1+x^2} $$ use dominated convergence theorem (hier ist noch ein fehler. gn ist gar nicht integrierbar) $$ \Rightarrow \lim\limits_{n \rightarrow \infty} \int_{0}^{\infty} \frac{G(t+h_n) - G(t)}{h_n} \frac{1}{1+x^2} \, dx = \int_{0}^{\infty} \frac{-x \sin(xt)}{1+x^2} \, dx $$

use the proof for \( t>0 \) $$ \begin{align} \ \frac{d}{dt} e^{-|t|} = \frac{d}{dt} \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx \\ \Leftrightarrow e^{-t} = \frac{2}{\pi} \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx \\ \Rightarrow \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx = \frac{\pi}{2} e^{-t} \\ \end{align} $$