Revision as of 09:27, 31 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 and find

$$ \int_{0}^{\infty} \frac{1}{1+x^2} \, dx = \frac{\pi}{2} $$

Claim: $$ \frac{d}{dt} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{d}{dt} \frac{\cos(xt)}{1+x^2} \, dx \, , t \gt 0$$ Beachte: Wir benötigen partielle Integration, da das Integral sonst nicht Lebesgue-Integrable ist und daher das Theorem für dominierende Konvergenz versagt. (Siehe Diskussion).

Proof:

first make some definitions:

\(f(x,t) := \cos(xt) \) and \(g(x) := \frac{1}{1+x^2}\)

\( \partial^{-1}_x f(x) := \int_{0}^{x} f(x') \, dx' \)

assume \(t \gt 0 \) and use partial integration (additional term goes to \( 0 \)) $$ \int_{0}^{\infty} f(x,t)g(x) \, dx = -\int_{0}^{\infty} \partial^{-1}_x f(x,t) \partial_x g(x) \, dx \\ = \frac{2}{t} \int_{0}^{\infty} \frac{x \sin(xt)}{(1+x^2)^2} \, dx $$

now we try to dervate this integral w.r.t \(t\)

define \( G(x,t) := \sin(xt) \), \(t_0 \gt 0 \) and by mean value theorem for \( h \ne 0 \) we find \( \xi(x) \) between \( 0 \) and \( h \) s.t.: $$ \frac{G(x,t_0+h)-G(x,t_0)}{h} = G_t(x,t_0+\xi) \le |x| \\ \Rightarrow \bigg| \frac{G(x,t_0+h)-G(x,t_0)}{h} \frac{x}{(1+x^2)^2} \bigg| \le \frac{x^2}{(1+x^2)^2} \\ d_n(x,t) := \frac{G(x,t+h_n)-G(x,t)}{h_n} \frac{x}{(1+x^2)^2} $$ with sequence \( h_n \ne 0 \) converging to \( 0 \) we have integrable \( d_n(x) \) bounded by an integrable function.

with dominated convergence theorem it follows that we can take the limit inside the integral and get \( \partial_t \) under the integral.

We do again partial integration and use (Leibniz integral rule and exchange of partial dervatives)

$$ -\int_{0}^{\infty} \partial_t (\partial^{-1}_x f(x,t) \partial_x g(x)) \, dx = -\int_{0}^{\infty} \partial^{-1}_x \partial_t f(x,t) \partial_x g(x) \, dx = \int_{0}^{\infty} \partial_t(f(x,t)g(x)) \, dx $$

where the additional term from the partial integration goes to 0. Now we have proven the claim.

Alternativer Beweis (Dies ist nur eine Skizze. Ich bin mir da etwas unsicher(Siehe Diskussion für mehr Infos). Ich halte mich bisher an den obigen Beweis, der vom Aufwand her auch noch ok ist. Ev. kommt jemand anders hier noch auf eine einfachere Lösung, dann ergänze ich das noch)

Consider $$ \begin{align} f_n(t) = \frac{2}{\pi}\int\limits_0^n\frac{\cos{x t}}{1+x^2}\mathrm{d}x \end{align} $$

with \(f_n(t)\) analytic.

further $$ \begin{align} f_n(t)\xrightarrow[n\rightarrow\infty]{glm.}\frac{2}{\pi}\int\limits_0^\infty\frac{\cos{x t}}{1+x^2}\mathrm{d}x=:f(t) \end{align} $$

With $$ \begin{align} \int\limits_0^\infty\frac{1}{1+x^2}\mathrm{d}x<\infty \end{align} $$ it follows $$ \begin{align} f'_n(t)\xrightarrow[n\rightarrow\infty]{glm.}f'(t) \end{align} $$ This implys:\[ \begin{align} -\frac{2}{\pi}\int\limits_0^\infty \frac{x\sin{xt}}{1+x^2}\mathrm{d}x = \frac{\mathrm{d}}{\mathrm{d}t}\frac{2}{\pi}\int\limits_0^\infty \frac{\cos{xt}}{1+x^2}\mathrm{d}x \end{align} \]

And the claim is proven.

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