# 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, $b>0$, and $x>0$.

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)