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Can this integral be calculated analytically?

$ \int_{-\pi}^\pi \frac{\mathrm dx}{2\pi} \frac{(e^{i y t}+e^{i x t})(e^{ix}+e^{-i(y-z+x)})}{\cos (y-z+x)- \cos x} \left(\frac{1}{1+e^{2\beta(a-b\cos x)}} - \frac{1}{1+e^{2\beta(a-b\cos (y-z+x))}}\right) $

$a$, $\beta$ and $b$ are constants while $y$ and $z$ also need to be integrated over (after the whole thing is multiplied by extra functions of $y$ and $z$). If not, can even this one be done?

$ \int_{-\pi}^\pi \frac{\mathrm dx}{2\pi} \frac{1}{1+e^{2\beta(a-b\cos x)}} $

I have tried it in Mathematica, but it doesn't return a solution. I've also tried various methods by hand but haven't come up with anything.

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    @mixedmath thanks. I can't seem to do it numerically because when integrating the whole thing (including extra functions of y and z) over x, y and z, mathematica can't figure out where the singularities are.2011-08-14

2 Answers 2

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I would consider

$ I(a, b) = \int_{-\pi}^\pi \frac{\mathrm dx}{2\pi} \frac{1}{1+e^{2(a-b\cos x)}} $

By symmetry $x \to -x$, $ I(a, b) = \int_{0}^\pi \frac{\mathrm dx}{\pi} \frac{1}{1+e^{(a-b\cos x)}} $. Now change variables $\cos x = y$, which results in

$ I(a,b) = \int_{-1}^1 \frac{\mathrm{d} y}{ 2 \pi } \frac{1-\tanh(a-b y) }{\sqrt{1-y^2}} = \frac{1}{2} - \int_{-1}^1 \frac{\mathrm{d} y}{ 2 \pi } \frac{\tanh(a-b y)}{\sqrt{1-y^2}} $

Now, because $\sqrt{1-y^2}$ is symmetric in $y$ this further simplifies to

$ I(a,b) = \frac{1}{2} - \sinh(2a) \int_{-1}^1 \frac{\mathrm{d} y}{ 2 \pi } \frac{1}{\sqrt{1-y^2}} \frac{1}{\cosh(2 a) + \cosh(2 b y)} $

Now notice that $ \int_{-1}^1 \frac{\mathrm{d} y}{ \pi } \frac{\cosh(c y)}{\sqrt{1-y^2}} = I_0(c)$. Hence a possible strategy for approximating your integral is to use the following expansion $ \frac{1}{\cosh(2 a) + \cosh(2 b y)} = \frac{1}{\cosh(2a)} \sum_{k>=0} \left( \frac{\cosh(2 b y)}{\cosh (2a)} \right)^k$ and reduce powers of $(\cosh(2 by))^k$ into a sum over multiple arguments (see this page).

I doubt the integral would admit closed form expression.

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Do you have a closed form for $ \frac{1}{2\pi}\int_{-\pi}^{\pi} \frac{1}{1 + \operatorname{e} ^{\operatorname{cos} (x)}} d x $ If not, there is no use asking for something more general.

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    I don't, but sometimes an integral that looks more complicated is actually simpler, and I wanted to make sure I wasn't missing something like that.2011-08-14