Explicit expression for $\int_{0}^{2\pi}e^{-\alpha\cos(\theta)(1+\gamma\cos(\theta))}d\theta$

Question

Working on some probability distributions and I'm stuck with getting an analytical expression for

$$\int_{0}^{2\pi}e^{-\alpha\cos(\theta)(1+\gamma\cos(\theta))}d\theta$$

$\alpha$ is positive, $\gamma$ can be either way. Any tips are welcome!

Answer 1

I do not know if a closed form solution exists but, may be, a series solution.

Writing $$e^{\alpha \cos (\theta ) (1+\gamma \cos (\theta ))}=\sum_{n=0}^\infty \frac{\alpha ^n \gamma ^n}{n!}\, e^{\alpha \, \cos (\theta )} \cos ^{2 n}(\theta )$$ $$I_n=\int_0^{2\pi} e^{\alpha \, \cos (\theta )} \cos ^{2 n}(\theta )\,d\theta$$ $$I_n=2 \sqrt{\pi }\,\,\frac{ \Gamma \left(n+\frac{1}{2}\right)}{n!}\,\,\, _1F_2\left(n+\frac{1}{2};\frac{1}{2},n+1;\frac{\alpha ^2}{4}\right)$$ The term $$T_n=\frac{4^n}{\binom{2 n}{n}}\,\alpha^n \, _1F_2\left(n+\frac{1}{2};\frac{1}{2},n+1;\frac{\alpha ^2}{4}\right)$$ is a linear combination of $I_n(\alpha )$ and $I_{n+1}(\alpha )$ weighted by polynomials in $\alpha$

$$\left( \begin{array}{cc} n & T_n \\ 0 & I_0(\alpha ) \\ 1 & I_1(\alpha )+\alpha I_2(\alpha ) \\ 2 & \left(\alpha ^2+3\right) I_2(\alpha ) \\ 3 & 5 \left(\alpha ^2+3\right) I_3(\alpha )+\alpha \left(\alpha ^2+5\right) I_4(\alpha ) \\ 4 & \left(\alpha ^4+42 \alpha ^2+105\right) I_4(\alpha )+4 \alpha ^3 I_5(\alpha ) \\ 5 & \left(13 \alpha ^4+378 \alpha ^2+945\right) I_5(\alpha )+\alpha \left(\alpha ^4+42 \alpha ^2+189\right) I_6(\alpha ) \\ \end{array} \right)$$