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.
 A: 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.
A: 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.
