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I would like to solve the integral $$F_n(\kappa,\theta,\phi)=\int_{-\pi}^{\pi}{\rm e}^{\kappa\cos(x-\theta)}\cos(n\, x-\phi)\,{\rm d}x$$ that appears related to the identity $$I_n(\kappa)=\frac{1}{\pi}\int_{0}^{\pi}{\rm e}^{\kappa\cos(x)}\cos(n\, x)\,{\rm d}x,$$ where $I_n(\kappa)$ is the Modified Bessel Function of the first kind. Any ideas?

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After elementary trig manipulations it seems that the problem could be solved if the integral $$\int_{-\pi}^{\pi}e^{\kappa \cos(x)}\sin(n\, x){\rm d} x$$ was known. – Arrigo Benedetti Sep 13 '12 at 0:52
And this integral is zero so I think the problem is solved... – Arrigo Benedetti Sep 13 '12 at 1:14
Why is the integral zero for any $\kappa$ and $n$? – marty cohen Sep 13 '12 at 1:25

Reduce to known integral

Assume $n$ is a non-negative integer. Then the integrand is periodic with period $2\pi$. Then: $$ F_n= \int_{-\pi}^\pi \exp\left( \kappa \cos(x-\theta) \right) \cos(n x - \phi) \mathrm{d}x = \int_{-\theta-\pi}^{-\theta+\pi} \exp\left( \kappa \cos(x) \right) \cos(n x + n \theta- \phi) \mathrm{d}x $$ The latter integral, using periodicity can be reduced to pieces of $(-\pi, \pi)$ interval, which can be rearranged, so that $$ \int_{-\pi}^\pi \exp\left( \kappa \cos(x-\theta) \right) \cos(n x - \phi) \mathrm{d}x = \int_{-\pi}^{+\pi} \exp\left( \kappa \cos(x) \right) \cos(n x + n \theta- \phi) \mathrm{d}x $$ Denote $\varphi = n\theta - \phi$, and use $\cos(n x + \varphi) = \cos(n x) \cos(\varphi) - \sin(n x) \sin(\varphi)$ to write: $$ \begin{eqnarray} F_n &=& \cos(\varphi) \int_{-\pi}^{\pi} \exp\left( \kappa \cos(x) \right) \cos(n x) \mathcal{d} x - \sin(\varphi) \int_{-\pi}^{\pi} \exp\left( \kappa \cos(x) \right) \sin(n x) \mathcal{d} x \\ &=& 2 \pi \cos(\varphi) I_n(\kappa) - \sin(\varphi) \cdot 0 \end{eqnarray} $$ The last integral is zero as integral of an odd function $\sin(n x) \exp(\kappa \cos(x))$ over a symmetric domain. Thus: $$ F_n(\kappa, \theta, \theta) = 2 \pi \cos(n \theta - \phi) I_n(\kappa) $$

Differentiate under integral sign

Alternatively, we could establish that $F_n$ satisfies ODE in $\kappa$. $$ \begin{eqnarray} \kappa^2 \partial_\kappa^2 F_n + \kappa \partial_\kappa F_n &=& \int_{-\pi}^\pi \exp(\kappa \cos(x-\theta)) \left(\kappa^2 \underbrace{\cos^2(x-\theta)}_{1-\sin^2(x-\theta)}+ \kappa \cos(x-\theta) \right) \cos(n x - \phi) \mathrm{d} x \\ &=& \kappa^2 F_n - \int_{-\pi}^\pi \left(\frac{\mathrm{d}^2}{\mathrm{d}x^2} \exp(\kappa \cos(x-\theta)) \right) \cos(n x-\phi) \mathrm{d} x \\ &\stackrel{\text{by parts}}{=}& (\kappa^2 + n^2) F_n + \text{boundary terms} \end{eqnarray} $$ where the boundary term vanishes, if $n$ is an integer: $$\begin{eqnarray} \text{boundary terms} &=& - \left(\left. \left(\frac{\mathrm{d}}{\mathrm{d}x} \exp(\kappa \cos(x-\theta)) \right) \cos(n x-\phi) \right|_{x=-\pi}^{x=\pi} \right) \\ &\phantom{=}& - \left(\left. n \exp(\kappa \cos(x-\theta) ) \sin(n x-\phi) \right|_{x=-\pi}^{x=\pi} \right) \\ &=& 2 \sin(\pi n) \exp( -\kappa \cos(\theta) ) \left( n \cos(\phi) - \kappa \sin(\theta) \sin(\phi) \right) = 0 \end{eqnarray} $$ Thus $F_n$ satisfies the differential equation of $I_n(\kappa)$: $$ \kappa^2 \frac{\mathrm{d}^2}{\mathrm{d} \kappa^2} F_n + \kappa \frac{\mathrm{d}}{\mathrm{d} \kappa} F_n - (\kappa^2 + n^2) F_n = 0 $$ Because for $\kappa = 0$, $F_n$ is finite: $$ \left.F_n \right|_{\kappa = 0} = \int_{-\pi}^\pi \cos(n x -\phi) \mathrm{d} x = 2 \cos(\phi) \frac{\sin(\pi n)}{n} = 2 \pi \cos(\phi) \delta_{n,0} $$ we conclude that $$ F_n = g_n(\theta, \phi) I_n(\kappa) $$ Thus $g_{0}(\theta, \phi) = \cos(\phi)$. One can similarly establish an ordinary differential equation for $F_n$ as a function of $\theta$: $$ \frac{\mathrm{d}^2}{\mathrm{d} \theta^2} F_n = \int_{-\pi}^\pi \left(\frac{\mathrm{d}^2}{\mathrm{d}x^2} \exp(\kappa \cos(x-\theta)) \right) \cos(n x-\phi) \mathrm{d} x = -n^2 F_n $$ and trivially $$ \frac{\mathrm{d}^2}{\mathrm{d} \phi^2} F_n = - F_n $$ Combining these, with initial conditions, we arrive at the same result: $$ F_n(\kappa, \theta, \phi) = 2\pi \cos\left(n \theta - \phi\right) I_n\left( \kappa \right) $$

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Sahsha, thanks for the detailed explanation. – Arrigo Benedetti Sep 15 '12 at 1:12

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