Definite Integral of Modified Bessel function representation I am trying to express the following integral of the Modified Bessel function either in closed form or even using other special functions. Any ideas ?
$$
\int_{0}^{b}x\exp\left(-\,{x^{2} + z^{2} \over 2\sigma^2}\right)
{\rm I}_{0}\left(\vphantom{\large A}xz \over \sigma^{2}\right)\,{\rm d}x
$$
Note that the integration to infinity is given to be equal to one (the inside function is a probability function), i.e., :
$$
\int_{0}^{\infty}x\exp\left(-\,{x^{2} + z^{2} \over 2\sigma^2}\right)
{\rm I}_{0}\left(\vphantom{\large A}xz \over \sigma^{2}\right)\,{\rm d}x=1
$$
Thank you for your time. 
With respect
 A: As  gammatester told the initial integral is related to Marcum Q-function
$$Q_M(\alpha,\beta)=\frac{1}{\alpha^{M-1}}\int_\beta^\infty x^Me^{-\frac{x^2+
\alpha^2}{2}}\mathrm{I}_{M-1}(\alpha x)dx$$
 in the following manner:
$$\begin{eqnarray}
\int_{0}^{b}xe^{-\,{x^{2} + z^{2} \over 2\sigma^2}}
{\rm I}_{0}\left(\vphantom{\large A}xz \over \sigma^{2}\right)\,{\rm d}x&=&\int_{0}^{\infty}xe^{-\,{x^{2} + z^{2} \over 2\sigma^2}}
{\rm I}_{0}\left(\vphantom{\large A}xz \over \sigma^{2}\right)\,{\rm d}x-\int_{b}^{\infty}xe^{-\,{x^{2} + z^{2} \over 2\sigma^2}}
{\rm I}_{0}\left(\vphantom{\large A}xz \over \sigma^{2}\right)\,{\rm d}x=\\
&=&1-\sigma^2\int_{b}^{\infty}\frac{x}{\sigma}\exp\left(-\,{\left(\frac{x}{\sigma}\right)^{2} + \left(\frac{z}{\sigma}\right)^{2} \over 2}\right)
{\rm I}_{0}\left(\vphantom{\large A}\frac{x}{\sigma}\cdot\frac{z}{\sigma}\right)\,{\rm d}\left(\frac{x}{\sigma}\right)=\\
&\mbox{}&\mbox{changing the variable} \qquad \frac{x}{\sigma}=y \\
&=&1- \sigma^2Q_1\left(\frac{z}{\sigma},\frac{b}{\sigma}\right).
\end{eqnarray}
$$
