Evaluate the integral
$$ P(\gamma)=\int_{W(-1/\gamma)}^{W(1/\gamma)}\frac{e^{-u} \,\text{d}u}{\sqrt{1-(\gamma u e^{u})^2}} $$
where $\gamma$ is a real number not equal to $0$ and has whatever properties it needs to have to make the product logs in the limits defined. Note that the limits are exactly where the denominator of the integrand goes to zero. No way there's an elementary anti-derivative so there's probably some sort of contour integration involved but I can't quite figure it out. I can express the integral as an infinite sum of special functions by expanding the square root, but I was wondering if there was a more compact form.
Given that $\gamma\geq e$, we have: $$ I = \int_{-1}^{1}W(t/\gamma)W'(t/\gamma)\frac{dt}{t\sqrt{1-t^2}},$$ and since: $$ W(z)W'(z) = -\sum_{n=1}^{+\infty}\frac{(-1)^n (n+1)^{n-1}}{(n-1)!}z^n,$$ it follows that: $$\begin{eqnarray*} I &=& -\sum_{n=1}^{+\infty}\frac{(-1)^n (n+1)^{n-1}}{\gamma^n (n-1)!}\int_{-1}^{1}\frac{z^{n-1}}{\sqrt{1-z^2}}\\&=&\sum_{m=0}^{+\infty}\frac{ (2m+2)^{2m}}{\gamma^{2m+1} (2m)!}\int_{-1}^{1}\frac{z^{2m}}{\sqrt{1-z^2}}\\&=&\sum_{m=0}^{+\infty}\frac{ (2m+2)^{2m}}{\gamma^{2m+1} (2m)!}\int_{-\pi/2}^{\pi/2}\sin^{2m}(\theta)\,d\theta\\&=&\pi\sum_{m=0}^{+\infty}\frac{ (m+1)^{2m}}{\gamma^{2m+1}(m!)^2},\tag{1}\end{eqnarray*}$$ that is not so terrible to deal with, but I do not think has a "nice" expression - I wonder if $\frac{1}{\pi}I(\gamma)$ can be expressed as the inverse of an elementary function, but I found that $$ I(\gamma)\approx \frac{\pi}{\sqrt{\gamma^2-e^2}}$$ is a very good approximation (especially for $\gamma\geq 5$).