# Cauchy Principal Value for log integral

How do I evaluate the expression

$\lim_{\xi\to0}(\int_0^\xi\! \ln(\frac{1}{r})\frac{F}{\xi} \, \mathrm{d}r)$ , where$\ F,\xi$ are real numbers and $\xi\geq0$.

Integration gives the expression

$\lim_{\xi\to0}[\frac{F}{\xi}(r\ln(\frac{1}{r}) + r)]$ where $r$ is evaluated at $\ 0$ and $\xi$.

The first term is singular and be can't be evaluated. Can this integral be defined in a Cauchy Principal Value sense, or evaluated in another way to remove the singularity?

Thanks

• Oops, didn't see you've edited your question. – user_of_math Dec 3 '14 at 18:13

## 1 Answer

The integral itself need not be evaluated as a PV because

$$\lim_{r \rightarrow 0^+} r \ln \frac{1}{r} = 0$$

So you have

$$\lim_{\xi\rightarrow0} \frac{F}{\xi} \left( \xi \ln \frac{1}{\xi} + \xi - 0 \right) \\ = F \lim_{\xi\rightarrow0} \left( \ln \frac{1}{\xi} + 1 \right) \\$$

which does not exist.

• This is my problem. I need a way to remove the singularity or define the integration differently so I can evaluate the expression. – Grady F. Mathews Iv Dec 3 '14 at 18:28
• @GradyF.MathewsIv In some circumstances, people do a Hadamard "finite-part" regularization of integrals that don't quite exist even as a PV. en.wikipedia.org/wiki/Hadamard_regularization. It essentially involves "throwing away an infinity". – user_of_math Dec 3 '14 at 18:31