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Suppose we have the following function: $$\Phi(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{1}{2}\frac{x^2}{\sigma^2})$$ and suppose the variable $x$ is depending on an angle $\alpha$: $$x=sin(\alpha)$$ Is it possible to give an analytic expression of the integral $\Phi(\alpha)$? Thanks.

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up vote 2 down vote accepted

It is possible with the following consideration: $$ x^2=\sin^2(\alpha)=\frac{1}{2}(1-\cos(2\alpha)) $$ and so the function becomes $$ \Phi(\alpha)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{4\sigma^2}(1-\cos(2\alpha))}. $$ Then you note that $$ e^{z\cos\theta}=I_0(z)+2\sum_{n=1}^\infty I_n(z)\cos(n\theta) $$ being in your case $I_n(z)$ modified Bessel funtions, $z=-\frac{1}{4\sigma^2}$ and $\theta=2\alpha$. Finally, the integral can be evaluated provided it is done on a finite interval.

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