Integral Related to Exponential Function and Modified Bessel Function 1 I'm looking for a closed form solution for the following integral:

$$S(x) = \displaystyle \int\limits_0^x r \,I_0(r) \,e^{-b^2r^2} \,dr $$

where $I_0()$ is the $0$-th order modified Bessel function of first kind,  $b&gt;0$, and $x&gt;0$. 

Thanks.
$\int_0^xrI_0(r)e^{-b^2r^2}~dr$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{r^{2n+1}e^{-b^2r^2}}{4^n(n!)^2}~dr$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{r^{2n}e^{-b^2r^2}}{2^{2n+1}(n!)^2}~d(r^2)$
$=\int_0^{x^2}\sum\limits_{n=0}^\infty\dfrac{t^ne^{-b^2t}}{2^{2n+1}(n!)^2}~dt$
$=-\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{t^ke^{-b^2t}}{2^{2n+1}b^{2n-2k+2}n!k!}\right]_0^{x^2}$ (according to https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#Integrals_of_polynomials)
$=\sum\limits_{n=0}^\infty\dfrac{1}{2^{2n+1}b^{2n+2}n!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{x^{2k}e^{-b^2x^2}}{2^{2n+1}b^{2n-2k+2}n!k!}$
$=\dfrac{e^\frac{1}{4b^2}}{2b^2}-\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{x^{2k}e^{-b^2x^2}}{2^{2n+1}b^{2n-2k+2}n!k!}$
$=\dfrac{e^\frac{1}{4b^2}}{2b^2}-\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{x^{2k}e^{-b^2x^2}}{2^{2n+2k+1}b^{2n+2}(n+k)!k!}$
$=\dfrac{e^\frac{1}{4b^2}}{2b^2}-\dfrac{e^{-b^2x^2}}{2b^2}\Phi_3\left(1,1;\dfrac{1}{4b^2},\dfrac{x^2}{4}\right)$ (according to https://en.wikipedia.org/wiki/Humbert_series)
Using the integral representation for $I_0$ and Fubini's theorem, the given integral equals

$$ \frac{1}{\pi}\int_{0}^{\pi}\int_{0}^{x} r \exp\left(r\cos\theta-b^2 r^2\right)\,dr\,d\theta $$
or
$$ \frac{1-I_0(x)e^{-b^2 x^2}}{2b^2}+\frac{1}{4b^3\sqrt{\pi}}\int_{0}^{\pi}\cos(\theta)e^{\frac{\cos^2\theta}{4b^2}}\left[\text{Erf}\left(\frac{\cos\theta}{2b}\right)-\text{Erf}\left(bx-\frac{\cos\theta}{2b}\right)\right]\,d\theta. $$
The second term is not really elementary, but such representation provides the asymptotics for $r\to 0^+$ or $r\to +\infty$ in a simple way. Would you mind sharing the original probabilistic problem?