How to extract the divergent part from the singular integral How to extract the divergent part of the following integral simply as $u \rightarrow \infty$
$$g(u) = \frac{\sqrt{2u}}{\pi} \int^1_{\frac{1}{u}} dz \frac{\sqrt{z-1}}{\sqrt{z^2-u^{-2}}} $$
 A: 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:

quite a fast convergence
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\down}{\downarrow}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$$
{\rm g}\pars{u}
=
{\root{2u} \over \pi}\int_{1/u}^{1}{\root{z - 1} \over \sqrt{z^{2} - u^{-2}}}\,\dd z
=
{\root{2u} \over \pi}\int_{1/u}^{1}\root{z - 1}\varphi'\pars{z}\,\dd z
$$
where $\ds{\varphi\pars{z} \equiv \int_{1/u}^{z}{\dd t \over \root{t^{2} - u^{-2}}}}$

Then,
\begin{align}
{\rm g}\pars{u}
&=
-\,{\root{2} \over 2\pi}\,u^{1/2}\int_{1/u}^{1}\,
{\varphi\pars{z} \over \root{z - 1}}\,\dd z
\end{align}
With the change of variables $\ds{t \equiv u^{-1}\sec\pars{x}}$:
\begin{align}
\varphi\pars{z}
&=\int_{0}^{\arccos\pars{u^{-1}z^{-1}}}\sec\pars{x}\,\dd x
=\left.\ln\pars{\sec\pars{x} + \tan\pars{x}}\right\vert_{0}^{\arccos\pars{u^{-1}z^{-1}}}
\\[3mm]&=\left.\ln\pars{ut + \root{u^{2}t^{2} - 1}}\right\vert_{u^{-1}}^{z}
\\[3mm]&=\ln\pars{uz + \root{u^{2}z^{2} - 1}}
=\ln\pars{u} + \ln\pars{z + \root{z^{2} - u^{-2}}}
\end{align}

The 'leading term' when $u \gg 1$ becomes:
$$
{\rm g}\pars{u}\sim -\,{\root{2} \over 2\pi}\,u^{1/2}\ln\pars{u}
\int_{u^{-1}}^{1}{\dd z \over \root{z - 1}}
=
{\root{2} \over \pi}\,u^{1/2}\ln\pars{u}\root{u^{-1} - 1}
$$
A: The Maple command $$series(sqrt(z-1)/sqrt(z^2-1/u^2), z = 1/u, 2) assuming u>0$$ produces $$  1/2\,\sqrt {-{\frac {-1+u}{u}}}\sqrt {2}\sqrt {u}{\frac {1}{\sqrt {z-{
u}^{-1}}}}+ \left( -1/8\,\sqrt {-{\frac {-1+u}{u}}}\sqrt {2}{u}^{3/2}-
1/4\,\sqrt {-{\frac {-1+u}{u}}}\sqrt {2}{u}^{3/2} \left( -1+u \right) 
^{-1} \right) \sqrt {z-{u}^{-1}}+O \left(  \left( z-{u}^{-1} \right) ^
{3/2} \right) 
.$$
