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?


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

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.

  • $\begingroup$ This is my problem. I need a way to remove the singularity or define the integration differently so I can evaluate the expression. $\endgroup$ – Grady F. Mathews Iv Dec 3 '14 at 18:28
  • $\begingroup$ @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". $\endgroup$ – user_of_math Dec 3 '14 at 18:31

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