Can the following be integrated?

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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The $\exp\cos\,x$ portion makes me think that a closed form isn't likely. –  Guess who it is. Aug 14 '11 at 0:38
I have tried on a few programs and with a few variants, but I think it must be done numerically. –  mixedmath Aug 14 '11 at 0:47
@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. –  Jane Aug 14 '11 at 11:09

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. –  Jane Aug 14 '11 at 11:03