How to extract the divergent part of the following integral simply as $u \rightarrow \infty$


$$g(u) = \frac{\sqrt{2u}}{\pi} \int^1\_{\frac{1}{u}} dz \frac{\sqrt{z-1}}{\sqrt{z^2-u^{-2}}} $$



How to extract the divergent part from the singular integral
it's an elliptic integral; a series expansion gives


$$g(u)=\frac{\sqrt{2u}}{\pi} \int^1\_{1/u} dz \frac{\sqrt{z-1}}{\sqrt{z^2-u^{-2}}}=i\frac{1}{\pi}(2u)^{1/2}\;[\ln (8u)-2]+{\cal O}(u^{-1/2})$$


so the integral diverges as $\sqrt{u}\ln u$


here is a plot of $-i(\pi/\sqrt{2u})g(u)$, evaluated numerically, and $\ln(8u)-2$ versus $u$, just as a check:


![](https://ilorentz.org/beenakker/MO/compare_integrals.gif)


quite a fast convergence


