# 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)