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
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.