Aufgaben:Problem 9

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Ihr seid ja immer noch am basteln mit dem differenzieren unter dem integral oder? Ich hab hier folgendes gefunden (weil ich recht wenig mmp gemacht habe dieses semester, bin ich nicht sicher, ob das das gleiche ist, was wir hatten):


I am trying to take the derivative with respect to \(a\) of some function \(I(a)=\int_{0}^{\infty}f(a,x)dx\). I would like to make sure that I am using the Leiniz Integral Rule correctly. Various web sources indicate a set of conditions that must hold for \(f(x,a)\) and \(\frac{\partial f(x,a)}{\partial a}\) when integration is done over infinite region. From reading this source (see Theorem 10.3 on page 13) the conditions that \(f(x,a)\) and \(\frac{\partial f(x,a)}{\partial a}\) must obey are:

1. \(f(x,a)\) and \(\frac{\partial f(x,a)}{\partial a}\) are continuous over \(x\in[0,\infty)\) and around \(a\) that we are interested in.

2. There exists an integrable function (over \(x\)) \(g(x)\) such that \(|\frac{\partial f(x,a)}{\partial a}|\leq g(x)\).

3. There exists an integrable function (over \(x\)) \(h(x)\) such that \(|f(x,a)|\leq h(x)\).

Integrable here means \(\int_{-\infty}^{\infty}g(x)dx<\infty\).


Mit \(f(t,x)=\cos(xt) / (1+x^2) \) sind die Bedingungen doch gegeben, oder?

1: trivial


3: \(|\cos(xt)| \leq 1 \; \Rightarrow |f(x,t)| \leq 1/(1+x^2)\) und das ist ja integrierbar
\(\; \int_0^\infty 1/(1+x^2)\: dx = \frac{\pi}{2}\)


2: \(\frac{\partial f(x,t)}{\partial t} = - t\sin(xt) / (1+x^2) - 2x\cos(xt) / (1+x^2)^2\)
\( \begin{align}\left|\frac{\partial f(x,t)}{\partial t}\right| &\leq |t\sin(xt) / (1+x^2)| + |2x\cos(xt) / (1+x^2)^2| \\ &\leq |t / (1+x^2)| + |2x / (1+x^2)^2| \end{align}\)
und die terme sind beide wieder integrierbar

Nik (talk) 17:45, 30 December 2014 (CET)


Reply Beni:


Ne, das stimmt so nicht. Wenn du meine Herleitung anschaust, denn verwende ich den Mittelwertsatz und das bedeutet, dass cos(xt) noch einmal nach t abgleitet wird. Damit wäre die dominierende Funktion dann x/(1+x^2), welche auf (0,unendlich) nicht integrierbar ist. Ich habe die Aufgabe in einem Forum gepostet und darüber bin ich auf den Lösungsweg gekommen. Siehe Link:

http://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=202739

Ich meinte der Lösungsweg über das partiell Integrieren hält sich vom Aufwand her in Grenzen. Beim zweiten Lösungsweg bin ich mir sehr unsicher. Daher bevorzuge ich die aktuelle Lösung.



Ok, dann halt nicht ;)

Neuer Versuch:

Proposition: $$\frac{d}{dt} \int_0^\infty \frac{\cos(xt)}{1+x^2} dx = \int_0^\infty \frac{d}{dt} \left(\frac{\cos(xt)}{1+x^2}\right) dx$$

Proof:

First, we put one of the few things that we learned in Analysis I & II to the test: Taylor (yay!)

We expand \(\cos(a+b) - \cos(b)\) around \(a=0\): $$\begin{align} \cos(a+b)-\cos(b) &= (\cos(a+b)-\cos(b))|_{a=0} + a\left.\left(\frac{d}{da}\cos(a+b)\right)\right|_{a=0} + \mathcal{O}(a^2)\\ &= -a\sin(b) + \mathcal{O}(a^2) \end{align}$$

Ready to differentiate hell outta that integral?

$$\begin{align} \frac{d}{dt} \int_0^\infty \frac{\cos(xt)}{1+x^2} dx &= \lim_{h\rightarrow 0} \frac{\int_0^\infty \frac{\cos(x(t+h)) - \cos(xt)}{1+x^2} dx}{h} \\ &= \lim_{h\rightarrow 0} \int_0^\infty \frac{\cos(x(t+h)) - \cos(xt)}{h(1+x^2)} dx \\ &= \lim_{h\rightarrow 0} \int_0^\infty \frac{-xh\sin(xt) + \mathcal{O}(h^2)}{h(1+x^2)} dx \\ &= \lim_{h\rightarrow 0} \int_0^\infty \frac{-x\sin(xt)}{(1+x^2)} + \mathcal{O}(h) \ dx \end{align}$$

Now, let's have a close look at that integrand: $$\left|\frac{-x\sin(xt)}{(1+x^2)}\right| \leq \left|\frac{x}{(1+x^2)}\right| \leq \sup_{x \in \mathbb{R}} \left|\frac{x}{(1+x^2)}\right| = \frac{1}{2}$$

As \(h\) goes to \(0\) anyways, we can ignore the \(\mathcal{O}(h)\) (or account for it with a infinitesimally small constant) and since we now have a bounded integrand, we can move the limes inside the integral.

Nik (talk) 16:42, 13 January 2015 (CET)


Ok, on second thought, that may or may not be correct. It was never \(\mathcal{O}(h^2)\) but actually \(\mathcal{O}(x^2h^2)\) and as we cannot integrate \(\frac{x^2}{1+x^2}\), this fucks everything up, doesn't it?

Mea Culpa!
Nik (talk) 16:47, 13 January 2015 (CET)


A. found this in another group:

For \(t>0\): $$\begin{align} \int_0^\infty \frac{x\sin(xt)}{1+x^2}dx &\overset{\text{integrand even}}{=} \frac{1}{2}\int_\mathbb{R}\frac{x\sin(xt)}{1+x^2}dx \\ &= \frac{2\pi}{4} \frac{1}{\sqrt{2\pi}} \int_\mathbb{R} x \left(\frac{1}{\sqrt{2\pi}}\frac{2}{1+x^2}\right)\sin(xt)\:dx \\ &= \frac{\pi}{2} \frac{1}{\sqrt{2\pi}} \int_\mathbb{R}(-i)\widehat{\frac{d}{dt}f}(x) \frac{1}{2i}\left(e^{ixt}-e^{ix(-t)}\right)dx \\ &= \frac{\pi}{4}\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\widehat{\frac{d}{dt}f}(x)\left(e^{ix(-t)}-e^{ixt}\right) dx\\ &= \frac{\pi}{4}\left(\frac{df}{dt}(-t)-\frac{df}{dt}(t)\right) \end{align}$$ Where we have used that (\(\rightarrow\) look here, p.95) $$\widehat{\frac{d}{dt}f}(x) = ix\hat f(k)$$ and since $$\frac{df}{dt}(t) = \begin{cases}-e^{-t} & \text{for } t>0 \\ e^t & \text{for } t<0\end{cases},$$ we finally find that $$\int_0^\infty \frac{x\sin(xt)}{1+x^2}dx = \frac{\pi}{2}e^{-t}\;.$$

this would be nice, it is similar to something i tried, but the inverse fouriertrafo is not allowed i think because, the fouriertrafo of the derivative is not in L1. check the script fourier-schwartzadded page 97

"Carl (talk) 17:33, 14 January 2015 (CET)"


I've checked on Wolfram Alpha and the integral of the derivative over \( \mathbb{R} \) seems to converge. - A.


Okay, but the Fouriertransform of the derivative is not in L1. We only showed $$f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \hat f(x) e^{ixt} dt $$ for \( f(x), \hat f(x)\in L^1(\mathbb{R}) \). In this case you are probably allowed to use the inverse fouriertransform even though \(\hat f\) is not in L1, but you would have to come up with a prove where you don‘t have to use the L1 property of \(\hat f\). -Carl


Do I understand it correctly that what you're saying essentially amounts to "We do have the Fourier transform of a function (i.e. of \(\frac{d}{dt}e^{-|t|}\)), but we're not allowed to transform it back to the time domain"? To be honest, I can't find any flaws in your argumentation, but this sounds pretty weird to me :/

Nik (talk) 20:40, 14 January 2015 (CET)


yes, it was a surprise to me too, but it seems to be true that you can‘t always recover the original function from the fourier transform. It would probably take a strange function such that the fouriertransform exist but one can not use the inverse transform, I can't think of one...

see: http://en.wikipedia.org/wiki/Fourier_inversion_theorem under "integrable functions in one dimension" -> "piecewise smooth; one dimension" we could use this theorem, if only it where in the script :)

"Carl (talk) 21:17, 14 January 2015 (CET)"


Too bad, then :(
Thanks for pointing that out, though!
-Nik


Okay, I try one too for the second one in 10. b):

Consider the function: \( \mathcal{F}^{-1} \left( k \cdot \mathcal{F} \left( e^{- |x|} \right) \right) \) which is well-defined since \( k \cdot \mathcal{F} \left( e^{- |x|} \right) \) is an odd function and thus in \( L^1 \).

Now we had the property

$$ \mathcal{F} \left( \frac{d}{dx} f \right) = i \cdot k \mathcal{F} \left( f \right) \Rightarrow -i \frac{d}{dx} f = \mathcal{F}^{-1} \left( k \cdot \mathcal{F} \left( f \right) \right) $$

by just applying the inverse Fourier-operator on both sides.

We then get:

$$ -i \cdot \frac{d}{dt} e^{- |t|} = \frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} \sqrt{\frac{2}{\pi}} \frac{x}{1 + x^2} e^{ixt} \, dx = \frac{1}{\sqrt{2 \pi}} \cdot \sqrt{\frac{2}{\pi}} \left( \underbrace{\int_{\mathbb{R}} \frac{x \cdot \cos (xt)}{1 + x^2} \, dx}_{\text{odd function, so zero}} + i \cdot \int_{\mathbb{R}} \frac{x \cdot \sin (xt)}{1 + x^2} \, dx \right) = \frac{i}{\pi} \cdot \int_{\mathbb{R}} \frac{x \cdot \sin (xt)}{1 + x^2} \, dx $$

And since \( t > 0 \) we can simply calculate:

$$ \frac{2}{\pi} \int_{0}^{\infty} \frac{x \cdot \sin (xt)}{1 + x^2} \, dx = e^{-t} $$

which makes us avoid the long proof for Lebesgue dominated convergence.

Best, A.