A definite integral of the exponential of cos

During some calculs, I came across the following definite integral, $$\int_0^{2\pi} \frac{\sin^2 \theta}{1+b\cos\theta} \exp(a\cos\theta) d\theta$$ with $a$ and $b$ constants. I tried to look it up in the Gradshteyn and Ryzhik, for example in Section 3.93: Trigonometric and exponential functions of trigonometric functions, but find nothing helpful. I also tried the Poisson Integral via complex analyse, apparently the exponential function is a little particular, if it's a $\log$ function instead of $\exp$ that's done, but with exponential I have not yet found the solution.

Thanks in advance if anyone has any idea :)

• I'd look at the residue theorem. Let $z = \exp(i\theta)$ and transform it into an integral over the unit circle. The integrand is holomorphic except for isolated singularities. – Daniel Fischer Jan 12 '18 at 14:30
• I hope you have constraints on $a$ and $b$, since I don't think it converges for all values... Moreover you can reduce the number of parameters and still keep it general – Shashi Jan 12 '18 at 14:35
• @DanielFischer The essential singularity at $z=0$ makes that approach, I believe, quite a mess. – Mark Viola Jan 12 '18 at 16:09
• @Mark Yes, it's probably not going to be pretty with $z + 1/z$ in the argument of the exponential. But it may be the least ugly of all ways. I didn't see a nicer way. – Daniel Fischer Jan 12 '18 at 16:14
• A related integral can be expanded using incomplete gamma function: $$\int_{0}^{2\pi} \frac{e^{a\cos\theta}}{1+b\cos\theta} \,d\theta = 2\pi \sum_{n=0}^{\infty} \frac{e^{-a/b}\Gamma(2n+1,-a/b)}{(n!)^2} \left(\frac{b}{2}\right)^{2n}.$$ – Sangchul Lee Jan 12 '18 at 16:30

Through the Poisson kernel you may derive the Fourier cosine series of $\frac{\sin^2\theta}{1+b\cos\theta}$; the coefficients of the Fourier cosine series of $\exp(a\cos\theta)$ are given by $I_k(a)$, where $I_k$ is a modified Bessel function of the first kind and $I_k(a)\leq \frac{\alpha^k}{2^k k!}I_0(a)$. By the orthogonality relations it follows that such integral can be converted into a fast-convergent series, with the minor drawback that the involved coefficients do not have a nice closed form, but they are pretty simple to compute numerically through continued fractions.