Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The $\exp\cos\,x$ portion makes me think that a closed form isn't likely. – J. M. 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
up vote 4 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.