How can calculate this integral?

Question

I have the following integral as: $$\int_{0}^{\infty}x\exp\left(-\frac{x^2+a^2}{2}\right)I_0(ax)dx.$$ where $I_0(.)$ is Bessel function of zero degree. Can any one help me calculataing this integral? Thanks!

Answer 1

$$\begin{eqnarray*}\int_0^\infty xe^{-\frac{x^2+a^2}{2}}I_0(ax)\,dx&=&\int_{0}^{+\infty}xe^{-\frac{x^2+a^2}{2}}\sum_{n=0}^{+\infty}\frac{1}{(n!)^2}\left(\frac{ax}{2}\right)^{2n}\\&=&\int_{0}^{+\infty}e^{-\frac{a^2}{2}}\sum_{n=0}^{+\infty}\frac{a^{2n}x^{2n+1}e^{-\frac{x^2}{2}}}{4^n(n!)^2}\\&\color{red}{=}&e^{-\frac{a^2}{2}}\sum_{n=0}^{+\infty}\frac{a^{2n}}{2^n n!}\\&=&e^{-\frac{a^2}{2}}\cdot e^{\frac{a^2}{2}}=1.\end{eqnarray*}$$ $\color{red}{=}$ is a consequence of: $$\int_{0}^{+\infty}x^{2n+1}e^{-x^2/2}\,dx=\frac{1}{2}\int_{0}^{+\infty}y^{n}e^{-y/2}dy=2^n\int_{0}^{+\infty}z^n e^{-z}dz=2^n\cdot n!.$$