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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.

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  • $\begingroup$ Substituting $x=\cos\theta$, the integral becomes $I(a,b,c)=2\int_{-1}^{1}(a-x)^{b}\exp{(c\,x)}\,\frac{\mathrm{d}x}{\sqrt{1-x^2}}$. $\endgroup$ – David H Jun 22 '14 at 4:49
  • $\begingroup$ Thank you for your suggestion. The form you showed seems to have high compatibility with a function called Gegenbauer's Polynomial, and now I'm playing around it. $\endgroup$ – S. Miyabe Jun 23 '14 at 10:56
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$\int_{-\pi}^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$

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

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

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

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

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

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

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

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

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

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

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

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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.

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  • $\begingroup$ Thank you for your suggestion. But unfortunately, neither b nor c can be regarded to be an integer in this problem. $\endgroup$ – S. Miyabe Jun 22 '14 at 3:22
  • $\begingroup$ For $a=1$ we get an expression in terms of regularized confluent hypergeometric functions, which, for $b=\dfrac12$ , can further be simplified to one in terms of error functions. $\endgroup$ – Lucian Jun 22 '14 at 3:40
  • $\begingroup$ Thank you for further examination. Also, another simplification with $c=0$ is expressed by a Gauss hypergeometric function ${}_2F_1$. $\endgroup$ – S. Miyabe Jun 22 '14 at 4:09

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