Revision as of 08:35, 28 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: $$ \frac{d}{dt} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{d}{dt} \frac{\cos(xt)}{1+x^2} \, dx $$

Wenn jemand einen einfacheren Weg weiss, gebt mir bescheid. Ohne partiell Integrieren klappt das Theorem für dominierende Konvergenz nicht.

Ich bin noch an einem einfacheren Beweis mittels Funktionentheorie...

Proof:

first make some definitions:

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

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

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 \( G(x,t) := \sin(xt) \) and by mean value theorem for \( h \ne 0 \) we find \( \xi(x) \) between \( 0 \) and \( h \) s.t.: $$ \frac{G(x,t+h)-G(x,t)}{h} = \partial_t G(x,t+\xi) \le |x| \\ \Rightarrow \bigg| \frac{G(x,t+h)-G(x,t)}{h} \frac{x}{(1+x^2)^2} \bigg| \le \frac{x^2}{(1+x^2)^2} \\ d_n(x,t) := \frac{G(x,t+h_n)-G(x,t)}{h_n} \frac{x}{(1+x^2)^2} $$ 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 (noch nicht ganz ausgereift)

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