Contour integral $\int^\pi_{-\pi}(a-\cos\theta)^b\exp(c\cos\theta)d\theta$ assuming $a>1$, $b>0$, $c>0$

Question

Under the condition $a>1$, $b>0$, $c>0$, is there any good function to express the following integral? $$ \int^\pi_{-\pi}\left(a-\cos\theta\right)^b\exp\left(c\cos\theta\right)d\theta $$ I guess some hypergeometric function can do this.

Answer 1

$\int_{-\pi}^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$

$=\int_{-\pi}^0(a-\cos\theta)^be^{c\cos\theta}~d\theta+\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$

$=\int_\pi^0(a-\cos(-\theta))^be^{c\cos(-\theta)}~d(-\theta)+\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$

$=\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta+\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$

$=2\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$

$=2\int_1^{-1}(a-x)^be^{cx}~d(\cos^{-1}x)$

$=2\int_{-1}^1\dfrac{(a-x)^be^{cx}}{\sqrt{1-x^2}}dx$

$=2\int_0^2\dfrac{(a-(x-1))^be^{c(x-1)}}{\sqrt{1-(x-1)^2}}d(x-1)$

$=2e^{-c}\int_0^2x^{-\frac{1}{2}}(2-x)^{-\frac{1}{2}}(a+1-x)^be^{cx}~dx$

$=2e^{-c}\int_0^1(2x)^{-\frac{1}{2}}(2-2x)^{-\frac{1}{2}}(a+1-2x)^be^{2cx}~d(2x)$

$=2(a+1)^be^{-c}\int_0^1x^{-\frac{1}{2}}(1-x)^{-\frac{1}{2}}\left(1-\dfrac{2x}{a+1}\right)^be^{2cx}~dx$

$=2\pi(a+1)^be^{-c}~\Phi_1\left(\dfrac{1}{2},-b,1;\dfrac{2}{a+1},2c\right)$ (according to About the confluent versions of Appell Hypergeometric Function and Lauricella Functions)

Answer 2

For natural values of c, it can be expressed in terms of Bessel functions as $\pi\Big[P\cdot I_0(c)-Q\cdot I_1(c)\Big]$

where P and Q are polynomials of degree b and $b-1$ in a, with factors depending on powers of c,

whose exponents are negative integers, no bigger than b in absolute value. This can be shown by

writing $\cos x$ as $\dfrac{e^{ix}+e^{-ix}}2$ , and then expanding using the binomial theorem. In general, many

integrals containing $f\Big(g(x)\Big)$, with f and g trigonometric, hyperbolic, or exponential functions, can

be expressed in a similar fashion. Hope this helps.