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

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 $\epsilon \to 0$.

