I have been struggling with an approximation to the following integral \begin{equation} \text{p.v.}\int_{-\infty}^{\infty} {e^{-s^2/2v} \over (e^{-2s}- q a)^2} {ds \over \sqrt{2 \pi v}} \end{equation}
where the parameters $q, a, v$ are positive constants and $0<q \leq 1$. My intention is to find the solution to the following equation
\begin{equation} {1 \over a^2} = \text{p.v.}\int_{-\infty}^{\infty} {e^{-s^2/2v} \over (e^{-2s}- q a)^2} {ds \over \sqrt{2 \pi v}} \end{equation}
I tried expanding ${e^{-s^2/2v} \over (e^{-2s}- q a)^2}$ around $s=0$ and $s=-{1 \over 2}\ln{qa}$, but expansion around the 2nd point gives rise to an expression of the form $$ {C_{-2} \over (s + {1 \over 2}\ln qa)^2} + {C_{-1} \over s + {1 \over 2}\ln qa} + C_0 + C_{1} (s + {1 \over 2}\ln qa) + C_{2} (s + {1 \over 2}\ln qa)^2 $$ Integration the above expression around a small interval $(-{1\over 2}\ln qa - \epsilon, -{1\over 2}\ln qa + \epsilon)$ gives a divergent result.
Plotting the integral as a function of $a$ for fixed $q$ and $v$ (preferably in log-log scale), one can see that the function has many spikes and many intersections with $1/a^2$. Could you share an idea of a possible approximation?
If $a\,q\ne0$, the integral does not converge.
Let $s^*=-\ln(a\,q)/2$. The denominator of the integrand vanishes when $s=s^*$. Close to $s^*$ we have $$ (e^{-2s}-a\,q)^2\sim (s-s^*)^2, $$ meaning $$ \lim_{s\to s^*}\frac{(e^{-2s}-a\,q)^2}{(s-s^*)^2}=c\ne0. $$ Finally, $(s-s^*)^{-2}$ is not integrable on any neighborhood of $s^*$.