Currently I am stuck with the following integral: \begin{equation} \int_0^{x_{\rm th}} \frac{2x_1}{\sigma^2} e^{-\frac{x_1^2}{\sigma^2(1-\mu _2^2)}} I_0\left (\frac {2~\mu_2x_{\rm th}x_1}{\sigma^2(1-\mu_2^2)} \right) \,dx_1, \end{equation} where $\sigma,x_{\rm th}>0$, $\mu_2 \in [0,1]$, and $I_0(\cdot)$ is the zero-order modified Bessel function of the first kind.
I have obtained a solution by considering the following series expansion: $I_v(z)=\sum\limits_{k=0}^{\infty}\frac{1}{k!\:\:\Gamma(v+k+1)}\left( \frac{z}{2} \right)^{v+2k}$, from Gradshteyn. But now I am trying to obtain a solution in terms of some well-defined functions and not any series decomposition.
My solution with the series decomposition:
\begin{align} &\int_0^{x_{\rm th}} x_1 \exp \left(-\frac {x_1^2}{\sigma ^{2}(1-\mu^2)} \right) I_{0}\left ({\frac {2\mu x_{\rm th}x_1}{\sigma ^{2}(1-\mu^2)}}\right) \, dx_1 \\[6pt] = {} & \int_0^{x_{\rm th}} x_1 \exp \left(-\frac {x_1^2}{\sigma ^{2}(1-\mu^2)} \right) \sum_{k=0}^{\infty} \frac{1}{(k!)^2}\left(\frac {\mu x_{\rm th}x_1}{\sigma ^{2}(1-\mu^2)} \right)^{2k} \, dx_1 \\[6pt] = {} & \sum_{k=0}^{\infty} \frac{1}{(k!)^2} \left(\frac {\mu x_{\rm th}}{\sigma ^{2}(1-\mu^2)} \right)^{2k} \int_0^{x_{\rm th}} \exp \left(-\frac {x_1^2}{\sigma^{2}(1-\mu^2)} \right) x_1^{2k+1} \, dx_1 \\[6pt] = {} & \frac{1}{2} \sum_{k=0}^{\infty} \frac{\left(\mu x_{\rm th} \right)^{2k}}{(k!)^2 \left( \sigma ^{2}(1-\mu^2) \right)^{k-1}} \left( \Gamma(k+1)-\Gamma \left( k+1, \frac{x_{\rm th}^2}{\sigma^{2}(1-\mu^2)} \right) \right). \end{align}
Any help or lead in this case will be very helpful. Thanks!