Difference between revisions of "Aufgaben:Problem 11"
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− | + | Claim: (Lasst mich wissen, wenn ihr eine Möglichkeit seht dies abzukürzen) | |
+ | $$ \frac{d}{dt} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{d}{dt} \frac{\cos(xt)}{1+x^2} \, dx $$ | ||
+ | Proof: | ||
+ | $$ | ||
+ | g(x) := \frac{\cos(xt)}{1+x^2}, t>0 \\ | ||
+ | g_t(x) = -\frac{x \sin(xt)}{1+x^2} \text{ and } |g_t(x)| \le \frac{x}{1+x^2}, t > 0, x \in \mathbb{R}_{\ge 0} \\ | ||
+ | $$ | ||
+ | with mean value theorem and \(\epsilon_n \rightarrow 0 \) | ||
+ | $$ | ||
+ | h_n(x,t) = \frac{g(x,t+\epsilon_n) - g(x,t)}{\epsilon_n} = g_t(x,t + r_n) \text{ where } |r_n| \le |\epsilon_n|, t \gt 0 \\ | ||
+ | \Rightarrow |h_n(x,t)| \le \frac{x}{1+x^2}, x \in \mathbb{R}_{\ge 0}, t \gt 0 \\ | ||
+ | $$ | ||
+ | Because \(h_n\) and \(\frac{x}{1+x^2}\) integrable, \(|h_n|\) dominated and \( \lim\limits_{n \rightarrow \infty}{h_n} \rightarrow g_t(x) \) the claim follows by the dominated convergence theorem. | ||
+ | for t>0 use this proof | ||
$$ | $$ | ||
\begin{align} | \begin{align} | ||
\ \frac{d}{dt} e^{-|t|} = \frac{d}{dt} \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx \\ | \ \frac{d}{dt} e^{-|t|} = \frac{d}{dt} \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx \\ | ||
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− | |||
\Leftrightarrow e^{-t} = \frac{2}{\pi} \int_{0}^{\infty} \frac{x \sin(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} \\ | \Rightarrow \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx = \frac{\pi}{2} e^{-t} \\ |
Revision as of 11:15, 24 December 2014
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: (Lasst mich wissen, wenn ihr eine Möglichkeit seht dies abzukürzen) $$ \frac{d}{dt} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{d}{dt} \frac{\cos(xt)}{1+x^2} \, dx $$ Proof: $$ g(x) := \frac{\cos(xt)}{1+x^2}, t>0 \\ g_t(x) = -\frac{x \sin(xt)}{1+x^2} \text{ and } |g_t(x)| \le \frac{x}{1+x^2}, t > 0, x \in \mathbb{R}_{\ge 0} \\ $$ with mean value theorem and \(\epsilon_n \rightarrow 0 \) $$ h_n(x,t) = \frac{g(x,t+\epsilon_n) - g(x,t)}{\epsilon_n} = g_t(x,t + r_n) \text{ where } |r_n| \le |\epsilon_n|, t \gt 0 \\ \Rightarrow |h_n(x,t)| \le \frac{x}{1+x^2}, x \in \mathbb{R}_{\ge 0}, t \gt 0 \\ $$ Because \(h_n\) and \(\frac{x}{1+x^2}\) integrable, \(|h_n|\) dominated and \( \lim\limits_{n \rightarrow \infty}{h_n} \rightarrow g_t(x) \) the claim follows by the dominated convergence theorem.
for t>0 use this proof $$ \begin{align} \ \frac{d}{dt} e^{-|t|} = \frac{d}{dt} \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx \\ \Leftrightarrow e^{-t} = \frac{2}{\pi} \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx \\ \Rightarrow \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx = \frac{\pi}{2} e^{-t} \\ \text{for } \ t>0 \end{align} $$