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:


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)