I'm a physics student trying to solve a scattering problem on the surface of a topological insulator and am currently stuck with integrals of this kind:
$$\int_0^\pi d\phi \int_\phi^\pi d\phi'\,e^{V(\cos\phi'- \cos\phi)}\sin^2\frac{\phi}{2}\cos^2\frac{\phi'}{2}.$$
Please note the variable limits of integration over $\phi'$. If they were definite, I would get Bessel functions in the result, but unfortunately it's not the case.
I could make use of the fact that $V\gg1$ and expand the trigonometric functions in the exponent, thus substituting them by polynomials ($\cos\phi\approx 1-\phi^2/2$), but it doesn't make the problem integrable.
I'd appreciate any suggestions on how one could integrate/approximate this while still keeping some angle-dependence (i.e., $\phi$- and $\phi'$-dependence) in the exponent.
Edit:
Here's my attempt at a saddle point approximation. Since the exponential function is a rapidly varying function, and since $-V(\cos\phi-\cos\phi')\leq 0$ in the region of integration, only the area around where ($\cos\phi-\cos\phi'$) has a minimum will contribute significantly to the integral.
$\cos\phi-\cos\phi'=0$ on the line $\phi=\phi'$, and $\cos\phi-\cos\phi'>0$ in the rest of the region of integration. I therefore expand $\cos\phi'$ around $\phi'=\phi$:
$$\cos\phi'\approx \cos\phi - (\phi'-\phi)\sin\phi-\frac{1}{2}(\phi'-\phi)^2\cos\phi.$$
This gives me
$$\cos\phi-\cos\phi'\approx (\phi'-\phi)\sin\phi+\frac{1}{2}(\phi'-\phi)^2\cos\phi.$$
Notice that the linear term does not disappear, because the line $\phi=\phi'$ doesn't correspond to a minimum, it simply gives the smallest value for ($\cos\phi-\cos\phi'$) in the region of integration.
The integral to lowest order now looks like this:
$$\int_0^\pi d\phi \int_\phi^\pi d\phi'\,e^{-V\sin\phi\,(\phi'- \phi)}\sin^2\frac{\phi}{2}\cos^2\frac{\phi'}{2}.$$
Edit #2:
The inner integral is now doable, and with the help of Mathematica I obtain
$$I=\int_0^\pi d\phi\,\frac{\tan\frac{\phi}{2}\left ( V^2(1+\cos\phi)-V-\csc^2\phi\left ( e^{V\sin\phi(\phi-\pi)} -1 \right ) \right )}{4V(V^2+\csc^2\phi)}.$$
Edit #3:
We can split this up into two parts $I_1+I_2=I$, where
$$I_1=\int_0^\pi d\phi\,\frac{\tan\frac{\phi}{2}\left ( V^2(1+\cos\phi)-V \right )}{4V(V^2+\csc^2\phi)}$$
and
$$I_2=\int_0^\pi d\phi\,\frac{\tan\frac{\phi}{2}\csc^2\phi\left ( 1-e^{V\sin\phi(\phi-\pi)} \right ) }{4V(V^2+\csc^2\phi)}.$$
Here's how the integrands look as a function of $\phi$ (for the case $V=10$):
Around $\phi=\pi$ the contributions from $I_1$ and $I_2$ nearly cancel out, and since $I_1$ is solvable while $I_2$ isn't, I guess a convenient way to proceed would be to approximate $I$ with $I_1$, only with a changed upper integration limit. Instead of integrating $I_1$ from $0$ to $\pi$, we could only integrate up to the value of $\phi$ at which the integrand of $I_1$ crosses zero. That is, $$I\approx\int_0^{\arccos(1/V-1)} d\phi\,\frac{\tan\frac{\phi}{2}\left ( V^2(1+\cos\phi)-V \right )}{4V(V^2+\csc^2\phi)}.$$ The result looks pretty horrible though: $$\frac{\sqrt{V^2+1} (4 V+\log (2 V)-2)-4 (V+1) \tanh ^{-1}\left(\frac{V}{\sqrt{V^2+1}}\right)}{8 V^2 \sqrt{V^2+1}}+$$ $$\frac{2 (V+1) \tanh ^{-1}\left(\frac{V+i \tan \left(\frac{1}{2} \cos ^{-1}\left(\frac{1}{V}-1\right)\right)}{\sqrt{V^2+1}}\right)+2 (V+1) \tanh ^{-1}\left(\frac{V-i \tan \left(\frac{1}{2} \cos ^{-1}\left(\frac{1}{V}-1\right)\right)}{\sqrt{V^2+1}}\right)}{8 V^2 \sqrt{V^2+1}}.$$