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Let $$f : (-\infty, \infty) \rightarrow \mathbb{C}$$ be an even ,smooth, and compactly supported, vanishing at zero.

Is there a way to solve integral

$\int_0^\infty f'(x)/x \; d x$

in a suitable interpretation a la integration by parts.

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I'm not sure what you are aiming at. Is it the $\infty$ integral boundary? If $f$ is compactly supported there is $ b\in \mathbb{R}$ such that you may replace $\infty$ by $b$. The singualarity at $0$ may cause trouble, though. – user20266 May 28 '12 at 14:19
Yes, the singularity at zero is the trouble. I would like to see this integral expressed only as integral over plus, perhaps some dirac delta like $f'(0)$ or there like. – May 28 '12 at 19:19
Note that the integral does not converge in general. For example, it diverges if $f(x)$ looks like $\frac{1}{\ln x}$ near $0$, since $\ln(-\ln x)=\int \frac{1}{x \ln x} \, dx$ is singular at $0$. – Micah May 28 '12 at 19:46
Thanks Micah, I have added some information, which I thought. Although in your example $f$ is not smooth at $0$. – May 29 '12 at 10:37
up vote 1 down vote accepted

The question in the title and the question in the text are different. I assume the one in the text is the correct one. In that case, let $\varepsilon > 0$. We get (by partial integration)

$$ \int_{\varepsilon}^\infty \frac{f'(x)}x\,dx = \left[ \frac{f(x)}x \right]_\varepsilon^\infty + \int_{\varepsilon}^\infty \frac{f(x)}{x^2}\,dx.$$

Let $\varepsilon \searrow 0$. Then the first term tends to $-f'(0)$ (the value at the upper bound is $0$ since $f$ is compactly supported). The second term could very well diverge though. Take for example $f(x) = x\psi(x)$ where $\psi$ is a cutoff function that is equal to $1$ on a neighborhood of $0$.

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So I have to impose in that case $f(0) \neq 0$ or double zero, nice. – May 29 '12 at 13:36
Imposing $f(0) \neq 0$ doesn't help. In this case the first term is divergent. – mrf May 29 '12 at 13:37
Right, it was probably to early here;) – May 29 '12 at 15:52

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