Asymptotic expansion of an integral I came up with a simpler example which illustrates the technical difficulty I have encountered in my work.
Consider an integral depending on parameter $\epsilon$:
\begin{equation}
\int\limits^{\infty}_{1 + \frac{2 \epsilon^{2}}{R^{2} - \epsilon^{2}}}
\frac{1}{t^{2}-1} \frac{1}{\sqrt{R^{2} - \epsilon^{2}\left(t+1\right)/\left(t-1\right)\,}}\,{\rm d}t.
\end{equation}
I am interested in the behaviour of this integral $\left(~\mbox{evaluated at}\ t = \infty~\right)$ as $\epsilon \to 0$. 
This integral can be performed analytically, and it is straightforward to see that the end result diverges logarithmically as $\epsilon \to 0$. My question is: how to extract this logarithmic divergence without performing full integration ?. Naive expansion of the integrand in Taylor series in $\epsilon$ never leads to a logarithmic divergence.
In my real problem the integral cannot be evaluated analytically, but I am interested only in the logarithmic divergence. Is there a systematic way to extract it?
Thank you in advance!
Yegor
 A: First make the change of variables $x = \epsilon^2\frac{t+1}{t-1}$ to get
$$
I(\epsilon) = \int_{\frac{R^2+\epsilon^2}{R^2-\epsilon^2}}^{\infty} \frac{dt}{(t^2-1)\sqrt{R^2-\epsilon^2 \frac{t+1}{t-1}}} = \frac{1}{2} \int_{\epsilon^2}^{R^2} \frac{dx}{x\sqrt{R^2-x}}.
$$
Fix $0 < \delta < R^2$ and split the integral into the two pieces
$$
I(\epsilon) = I_1(\epsilon) + I_2 = \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{dx}{x\sqrt{R^2-x}} + \frac{1}{2} \int_{\delta}^{R^2} \frac{dx}{x\sqrt{R^2-x}},
$$
each of which is finite for $\epsilon > 0$.  We see here that the singularity comes from $I_1(\epsilon)$, whose largest contribution comes from a neighborhood of $x = \epsilon^2 \approx 0$.  Informally,
$$
I_1(\epsilon) = \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{dx}{x\sqrt{R^2-x}} \approx \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{dx}{x\sqrt{R^2}} \approx - \frac{\log(\epsilon^2)}{2\sqrt{R^2}}.
$$
To make this precise, let's write
$$
\frac{1}{\sqrt{R^2 - x}} = \frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}} + \frac{1}{\sqrt{R^2}},
$$
so that
$$
I_1(\epsilon) = \frac{1}{2\sqrt{R^2}} \int_{\epsilon^2}^{\delta} \frac{dx}{x} + \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{1}{x} \left(\frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}}\right)dx.
$$
The integrand in the integral on the right is bounded by a constant for all $x \in [0,\delta]$ (note that $\lim_{x\to 0}$ exists), and the integral on the left is
$$
\frac{\log(\delta/\epsilon^2)}{2\sqrt{R^2}} = -\frac{\log \epsilon}{|R|} + O(1).
$$
Putting all this together, we conclude that

$$
\int_{\frac{R^2+\epsilon^2}{R^2-\epsilon^2}}^{\infty} \frac{dt}{(t^2-1)\sqrt{R^2-\epsilon^2 \frac{t+1}{t-1}}} = -\frac{\log \epsilon}{|R|} + O(1)
$$
  as $\epsilon \to 0$.

In fact we know what the constant in the $O(1)$ term is; it's just the finite terms we threw out along the way.  Indeed,
$$
I(\epsilon) = -\frac{\log \epsilon}{|R|} + C + o(1),
$$
where
$$
C = \frac{\log \delta}{2|R|} + \frac{1}{2} \int_{0}^{\delta} \frac{1}{x} \left(\frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}}\right)dx + \frac{1}{2} \int_{\delta}^{R^2} \frac{dx}{x\sqrt{R^2-x}}.
$$
This constant depends on $R$ but not on $\delta$, so we may take the limit as $\delta \to R^2$ to obtain
$$
\begin{align}
C &= \frac{\log |R|}{|R|} + \frac{1}{2} \int_{0}^{R^2} \frac{1}{x} \left(\frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}}\right)dx \\
&= \frac{\log(2|R|)}{|R|}.
\end{align}
$$
Thus

$$
\int_{\frac{R^2+\epsilon^2}{R^2-\epsilon^2}}^{\infty} \frac{dt}{(t^2-1)\sqrt{R^2-\epsilon^2 \frac{t+1}{t-1}}} = -\frac{\log \epsilon}{|R|} + \frac{\log(2|R|)}{|R|} + o(1)
$$
  as $\epsilon \to 0$.

A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
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 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
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 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
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 \newcommand{\ol}[1]{\overline{#1}}$
$\ds{%
{\cal J}
\equiv\!\!\!\!\!
\int\limits^{\infty}_{1 + \frac{2 \epsilon^{2}}{R^{2} - \epsilon^{2}}}\!\!\!\!\!
{\dd t \over \pars{t^{2}-1}\sqrt{R^{2} - \epsilon^{2}\left(t+1\right)/\left(t-1\right)\,}}
=
{1 \over \verts{R}}
\int\limits^{\infty}_{1 + \mu^{2} \over 1 - \mu^{2}}\!\!\!
{\dd t \over \pars{t^{2}-1}\sqrt{1 - \mu^{2}\left(t+1\right)/\left(t-1\right)\,}}
}$
where $\ds{\mu \equiv \epsilon/R}$.
\begin{align}
{\cal J}
&=
\int_{1 + \mu^{2} \over 1 - \mu^{2}}^{\infty}
{\dd t \over \root{t - 1}\pars{t + 1}\root{\pars{1 - \mu^{2}}t - 1 - \mu^{2}}}
\\[3mm]&=
\int^{1 - \mu^{2} \over 1 + \mu^{2}}_{0}
{\dd t/t^{2}
 \over
 \root{1/t - 1}\pars{1/t + 1}\root{\pars{1 - \mu^{2}}/t - 1 - \mu^{2}}}
\\[3mm]&=
\int^{1 - \mu^{2} \over 1 + \mu^{2}}_{0}
{\dd t
 \over
 \root{1 - t}\pars{t + 1}\root{\pars{1 - \mu^{2}} - \pars{1 + \mu^{2}}t}}
\\[3mm]&=
-\,{1 \over 2}\int^{1 - \mu^{2} \over 1 + \mu^{2}}_{0}{\dd t \over t - 1}
+
\overbrace{\int^{1 - \mu^{2} \over 1 + \mu^{2}}_{0}\bracks{%
{1
 \over
 \root{1 - t}\pars{t + 1}\root{\pars{1 - \mu^{2}} - \pars{1 + \mu^{2}}t}}
- {1 \over 2\pars{1 - t}}}\,\dd t}^{\ds{\equiv\ {\cal K}\pars{\mu}}}
\\[3mm]&=
-\,{1 \over 2}\,\ln\pars{1 - {1 - \mu^{2} \over 1 + \mu^{2}}} + {\cal K}\pars{\mu}
=
-\ln\pars{\verts{\epsilon \over R}}
+
{1 \over 2}\ln\pars{1 + \bracks{\epsilon \over R}^{2}}
+
{\cal K}\pars{\epsilon \over R}
\end{align}

$$\bbox[15px,border:1px dotted navy]{\ds{
{\cal J} \sim -\ln\pars{\verts{\epsilon \over R}}
\quad\mbox{when}\quad \epsilon \gtrsim 0}}
$$

The 'next contribution' is finite and it's given by
$$
{\cal K}\pars{0} = {1 \over 2}\int_{0}^{1}{\dd t \over t + 1}
=
{1 \over 2}\,\ln\pars{2}
$$
