Approximating $I= \int\limits_0^1 \frac{1}{R^{2}-x^{2}} \exp (-\frac{a}{\sqrt{1-x^{2}}})\ dx$ I am trying to approximate the integral of the form:
$$I= \int\limits_0^1  \dfrac{1}{R^{2}-x^{2}} \exp \left(-\frac{a}{\sqrt{1-x^{2}}}\right)\ dx $$
in the limits, $$a \ll 1 \ \ ; \ \ a \gg 1 \  \  \text{and} \  \ a \longrightarrow \infty$$  
I have found in a paper that in the limit, $$a \ll 1$$ the dominant contributions to the integral $\,I\,$ result from values $$x\leqslant \sqrt{1-a^{2}}$$ How this ?
and hence, we can approximate the existing exponential function by one. And by using the Cauchy principal value, we immediately find for the approximation  $R^{2} < 1$:
$$I\simeq \dfrac{1}{R}\ \textrm{argcoth} \left(\dfrac{\sqrt{1-a^{2}}}{R}\right) $$
as a physicist, I did not understand this. Please, how to calculate it within these limits?
 A: The inequality $x\leqslant \sqrt{1-a^{2}}$ appears from another one:
$$
\frac a{\sqrt{1-x^{2}}}\ge1.
$$
So the exponential factor decimates the values of  $\dfrac{1}{R^{2}-x^{2}}$ in $c=2.7$ and more times only on a little segment $[1-a^2,1]$. Say, if $a=10^{-3}$ then the segment's length is $10^{-6}$. The same reasoning holds for any fixed value of $c>1$. So neglecting this segment one gets
$$
I \approx \int\limits_0^{1-a^2}  \dfrac{1}{R^{2}-x^{2}} \exp \Big(-\frac{a}{\sqrt{1-x^{2}}}\Big)\ dx\approx \int\limits_0^{1-a^2}  \dfrac{1}{R^{2}-x^{2}} \ dx\approx
$$
$$
\int\limits_0^{1}  \dfrac{1}{R^{2}-x^{2}} \ dx
=\dfrac{1}{R}\ \textrm{argcoth} \Big(\dfrac{\sqrt{1-a^{2}}}{R}\Big).
$$
Denote $f(x)=\dfrac{1}{R^{2}-x^{2}} \exp \Big(-\frac{a}{\sqrt{1-x^{2}}}\Big)$.
If $a\to\infty$ then the main contribution will be from points next to $x=0$. So
$$
I=\int_0^1 f(x)\,dx=\int_0^2 (f(x)-f(0))\,dx+f(0)=
\int_0^1 g(x)\,dx+\frac{e^{-a}}{R^2},
$$
where $g(x)=O(x^2)$, $x\to0$. So it looks like
$$
I=\left(\frac{1}{R^2}+o(1)\right)e^{-a},\quad a\to+\infty.
$$
