Essentially, I am trying to work out an integral of the form:
$\int_{0}^{\pi }{x\cdot a\cdot \cos \left( x \right)e^{-\left[ a\cdot \sin \left( x \right) \right]^{2}}\mbox{erfc}\left[ -a\cdot \cos \left( x \right) \right]}dx$
or, similarly:
$\int_{0}^{\pi }{x\cdot a\cdot \cos \left( x \right)e^{\left[ a\cdot \cos \left( x \right) \right]^{2}}\mbox{erfc}\left[ -a\cdot \cos \left( x \right) \right]}dx$
It seems like the integral is going to be pretty easy to work out by parts and with a substitution, but it ended up being a little tougher than I expected. I had almost given up and resorted to numerical methods after trying every trick I know, when I came across this in a table of integrals (http://nvlpubs.nist.gov/nistpubs/jres/73B/jresv73Bn1p1_A1b.pdf):
Where K is a modified Bessel function of the bth order. I can't find a derivation of this result so I don't know how to see if I can extend this to my integral. Any ideas?
(If you are interested in the specifics, I derived a probability density function: $f(x)=\frac{1}{\pi }e^{-\frac{v^{2}}{2\sigma ^{2}}}+\frac{v\cdot \cos \left( x \right)}{\sigma \sqrt{2\pi }}e^{-\left( \frac{v\cdot \sin \left( x \right)}{\sigma \sqrt{2}} \right)^{2}}\mbox{erfc}\left( -\frac{v\cdot \cos \left( x \right)}{\sigma \sqrt{2}} \right), \ \ 0\leq x\leq \pi$
and I'm hoping to find the expected value of x, which is given by: $\mbox{E}\left( X \right)=\int_{-\infty }^{\infty }{xf\left( x \right)dx}=\frac{\pi }{2}e^{-\frac{v^{2}}{2\sigma ^{2}}}+\int_{0}^{\pi }{x\frac{v\cdot \cos \left( x \right)}{\sigma \sqrt{2\pi }}e^{-\left[ \frac{v\cdot \sin \left( x \right)}{\sigma \sqrt{2}} \right]^{2}}\mbox{erfc}\left[ -\frac{v\cdot \cos \left( x \right)}{\sigma \sqrt{2}} \right]}dx$)