Evaluating the integral $\int_0^{2\pi}e^{-\sqrt{a-b\cos t}}\mathrm dt$ In the course of my research, I have stumbled upon the following integral, which I do not know how to compute:
$$I(a,b)=\int_0^{2\pi}e^{-\sqrt{a-b\cos t}}\mathrm dt,\qquad a>b>0.$$
Help would be greatly appreciated.

My thoughts on the problem:


*

*Using the substitution $z=e^{it}$ yields a contour integral of a function analytic in some neighborhood of the circle $|z|=1$, so techniques from complex analysis might help.

*For $a=b$, the function from the previous point has a singularity at $z=1$, but we have the following explicit result (found with some help from Mathematica): $$I(a,a)=\int_0^{2\pi}e^{-\sqrt{a-a\cos t}}\mathrm dt=2\pi(I_0(\sqrt{2a})-\mathbf{L}_0(\sqrt{2a})),$$ where $I_0$ and $\mathbf{L}_0$ are modified Bessel and Struve functions, respectively.

*Alternatively, using the substitution $s=\sqrt{a-b\cos t}$ and exploiting the fact that the original integrand is symmetric about $t=\pi$, we obtain the following expression for the integral: $$I(a,b)=\int_{\sqrt{a-b}}^{\sqrt{a+b}}\frac{4s e^{-s}\mathrm ds}{\sqrt{b^2-(a-s^2)^2}},$$ however I also do not see how to proceed from here.
 A: I will attempt to gain some results using a form similar to elliptic integrals.
First, we use the symmetry to change the limits:
$$I(a,b)=2 \int_0^{\pi}e^{-\sqrt{a-b\cos t}}\mathrm dt$$
Then we use the following substitutions:

$$u=\sin \frac{t}{2},\qquad q=\sqrt{a+b},\qquad p=\frac{2b}{a+b}$$

We get another form of the integral:

$$I(a,b)=4 \int_0^1 \frac{\exp (-q \sqrt{1-pu^2})}{\sqrt{1-u^2}} \mathrm du$$

While it looks even more complicated, it has some uses. For example, in the simple case considered by the OP, $a=b$ and thus $p=1$, we can expand the exponential function to get:
$$I(a,a)=4 \sum_{n=0}^\infty (-1)^n \frac{q^n}{n!} \int_0^1 (1-u^2)^{\frac{n-1}{2}} \mathrm du=2 \sum_{n=0}^\infty (-1)^n \frac{\sqrt{\pi }~ \Gamma \left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n}{2}+1\right)} \frac{q^n}{n!}$$
This series can be shown to give the closed form from the OP, since $q=\sqrt{2a}$.

In general, let's expand the exponential function and get the series:
$$I(a,b)=4 \sum_{n=0}^\infty (-1)^n \frac{q^n}{n!} I_n$$
Where:
$$I_n=\int_0^1 (1-u^2)^{-1/2} (1-p u^2)^{n/2} \mathrm du$$
Numerically, this doesn't make sense, as the convergence of the series is extremely slow.
However, we can find the closed form for the integrals, substituting $u^2=v$:
$$I_n=\frac{1}{2} \int_0^1 v^{-1/2} (1-v)^{-1/2} (1-p v)^{n/2} \mathrm dv$$
But this is just Euler integral for Gauss hypergeometric function, so:
$$I_n=\frac{\pi}{2} {_2 F_1} \left(-\frac{n}{2}, \frac{1}{2};1; p\right)$$
So we get a series representation:

$$I(a,b)=2 \pi \sum_{n=0}^\infty (-1)^n ~{_2 F_1} \left(-\frac{n}{2}, \frac{1}{2};1; p\right)~ \frac{q^n}{n!}$$

For rational $p$ hypergeometric functions usually have a closed form. For even $n$ and any $p$ they can be represented as polynomials, which follows from the properties of hypergeometric series.

Using the hypergeometric series and changing the order of summation allows us to obtain another series for the integral, in terms of so called Bessel polynomials.
I don't have the time to provide the details of the calculation (I used some help from Mathematica to guess the result), but we have a very nice series:

$$I(a,b)= \sqrt{8 \pi q}~ \sum_{k=0}^\infty \frac{(2 k)! ~ K_{k-\frac{1}{2}}(q)}{ (k!)^3} \frac{p^k q^k}{8^k}$$

Where $K_{\alpha}(q)$ is the modified Bessel function, and in this case it gives polynomials multiplied by exponential function (see the Wikipedia link).
Numerical experiments confirm these results.
