I think I found all the pieces you need for an analytic solution, but my hunch is that it is useless. If you want insight, you will likely get more from pushing a bunch of a's and b's into the expression and doing the integrals numerically and observing behavior.
My analytical solution is an infinite series of terms that include Bessel, modified Bessel, modified Struve, and Weber functions, some with imaginary arguments (although of course in the end all the imaginary terms must cancel). Here it goes. Except where noted, all equation numbers refer to "handbook of mathematical functions" edited by Abramowitz and Stegun (you can find it free online at
http://www.math.ucla.edu/~cbm/aands/intro.htm#006 ).
I start by using basic trig to change the integral to
<br />
P = \exp(a/2) \int_{0}^{\pi}\sin\theta\exp\left(\frac{-a}{2}\cos(2 \theta)+ b\sin\theta\right)d\theta<br />
Then I use (9.6.34) which is the generating function for modified Bessel functions of the first kind, I_k(x), to get
<br />
\exp\left(\frac{-a}{2}\cos(2 \theta) \right) = \exp\left(\frac{a}{2}\cos(2 \theta-\pi) \right)<br />
<br />
= I_0(a/2) + 2 \sum_{k=1}^{\infty} I_k(a/2) \cos (2 k \theta - k \pi )<br />
So we can write the integral as
<br />
P = \exp(a/2) \int_{0}^{\pi} d\theta \, \sin\theta \exp\left(b \sin \theta \right)<br />
\left[ I_0(a/2) + 2 \sum_{k=1}^{\infty} I_k(a/2) \cos (2 k \theta - k \pi ) \right]<br />
The first integral in the series was solved by Mathematica Alpha
<br />
\int_{0}^{\pi} d\theta \, \sin\theta \exp\left(b \sin \theta \right) = <br />
\pi \left[L_{-1}(b) + I_1(b) \right]<br />
Where L_{-1}(b) is the modified Struve function of order -1. So it is already getting crazy. So I now write the integral as
<br />
P = \pi \exp(a/2) I_0(a/2) \left[L_{-1}(b) + I_1(b) \right] + <br />
\exp(a/2) \sum_{k=1}^{\infty} I_k(a/2) \left[\alpha_k^{+}(b) + \alpha_k^{-}(b) \right] <br />
where I have used basic trig identities and defined
<br />
\alpha_k^{\pm}(b) = \int_{0}^{\pi} d\theta \, \sin\left( (2k \pm 1) \theta - k \pi \right) <br />
\exp\left(b \sin \theta \right)<br />
Note that the modified bessel functions of the first kind decrease rapidly with increasing order once the order is larger than the argument. So as long as a isn't too large very few terms will be required.
Anyway, we basically just need to evaluate integrals of the form
<br />
Q_\ell (b) = \int_{0}^{\pi} d\theta \, \sin\left( \ell \theta \right) \, \exp\left(b \sin \theta \right)<br />
where \ell is odd. Numerically these can be evaluated. Analytically I only come up with something crazy. I do the following
<br />
Q_\ell (b) = \int_{0}^{\pi} d\theta \, \sin\left( \ell \theta \right) \, \exp\left(b \sin \theta \right)<br />
= \frac{1}{2i} \int_{0}^{\pi} d\theta \, \left[\exp (i (-ib \sin\theta + \ell\theta )) - <br />
\exp (i (-ib \sin\theta - \ell\theta)) \right]<br />
where i=\sqrt{-1}. We then continue ...
<br />
Q_\ell (b) = \frac{1}{2i} \int_{0}^{\pi} d\theta \, <br />
\cos( \ell\theta - i b \sin \theta ) + i \sin( \ell\theta - i b \sin \theta ) <br />
- \cos( -\ell\theta - i b \sin \theta ) - i \sin( -\ell\theta - i b \sin \theta )<br />
Now I use (9.1.21) and (12.3.3) to write these as Bessel functions of first kind and Weber's functions
<br />
Q_\ell (b) = \frac{\pi}{2i} \left[J_\ell(ib) + i E_\ell(ib) - J_{-\ell} (ib) - i E_{-\ell}(ib)\right]<br />
Where E_\ell(x) is Weber's function.
So I think all the pieces are here for an infinite series representation of the integral, if you want to go through the trouble of puting it all together. As I stated before, I suspect this approach is useless, but it was kind of fun to play with. Numerical evaluation of the original integral will lead to some insight. Series and asymptotic expansions of it may also help.
