Closed form expression for integral involving modified Bessel function?

Question

In my research, I have come across the following integral $\int_0^{\infty} x e^{-x^2}I_1(2xy)e^{-\alpha x} dx$ where $y\geq 0$ and $\alpha\geq 0$ are parameters, and $I_1$ is the modified Bessel function of order 1. Ideally I would like to know if there exists a closed form expression for this integral in terms of $y$ and $\alpha$. I have checked Watson and I can't find it there (other than the case where $\alpha=0$), so I suspect that it is probably not known (at least classically). Any help would be appreciated.

Answer 1

I did not find any closed form for $$\int_0^{\infty} x e^{-x^2}I_1(2xy)e^{-\alpha x} dx$$ but we can do a few things;

$$x=\frac{t}{2 y}\implies I=\frac 1{4y^2}\exp\left(\frac{\alpha ^2}{4} \right)\int_0^\infty \exp\left(-\left(\frac{t}{2 y}+\frac{\alpha }{2}\right)^2 \right)\,t\, I_1(t)\,dt$$

Now using $$t\, I_1(t)=\sum_{n=0}^\infty \frac{2^{-(2 n+1)} }{n! \, (n+1)!}\,t^{2( n+1)}$$ $$\int_0^\infty \exp\left(-\left(\frac{t}{2 y}+\frac{\alpha }{2}\right)^2 \right)\,t^{2(n+1)}\,dt=e^{-\frac{\alpha ^2}{4}} y^{2 n+3}\, \Gamma (2 n+3)\, U\left(n+\frac{3}{2},\frac{1}{2},\frac{\alpha ^2}{4}\right)$$ where appears Tricomi'sconfluent hypergeometric function.

All of this makes $$\color{blue}{I=\frac{\alpha y}{4 \sqrt{\pi }}\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{3}{2}\right)}{\Gamma (n+1)}\,\,y^{2n}\,\,U\left(n+2,\frac{3}{2},\frac{\alpha ^2}{4}\right)}$$

The summation seems to be converging without any problem. Trying for $y=\alpha=1$, the partial sums are $$\left( \begin{array}{cc} p & \sum_{n=0}^p \\ 0 & 0.159231 \\ 1 & 0.241763 \\ 2 & 0.267294 \\ 3 & 0.273057 \\ 4 & 0.274090 \\ 5 & 0.274243 \\ 6 & 0.274263 \\ 7 & 0.274265 \\ 8 & 0.274266 \end{array} \right)$$

If $\alpha$ is small, we can use the expansion of Tricomi'sconfluent hypergeometric function and then

$$A_n=\frac{\alpha y}{4 \sqrt{\pi }} \frac{\Gamma \left(n+\frac{3}{2}\right)}{\Gamma (n+1)}\,\,y^{2n}\,\,U\left(n+2,\frac{3}{2},\frac{\alpha ^2}{4}\right)$$ will give $$A_n=\frac{y^{2 n+1}}{2 \Gamma (n+1) }\Bigg[\frac{1}{\Gamma (n+2)}\sum_{k=0}^\infty \frac{\Gamma \left(n+k+\frac{3}{2}\right)}{(2 k)!}\,\alpha^{2k}-\frac{1}{ (n+1)}\sum_{k=1}^\infty \frac{(n+1)_k}{(2 k-1)!}\,\alpha^{2k-1}\Bigg] $$

Now, consider $$S_k=\sum_{n=0}^\infty \frac{\Gamma \left(n+k+\frac{3}{2}\right)}{2 \Gamma (n+1)\Gamma (n+2) }y^{2 n+1}$$ they all write $$S_k=\sqrt{\pi }\,y\, e^{\frac{y^2}{2}} \Bigg[P_k(y^2) \,I_0\left(\frac{y^2}{2}\right)+Q_{k-1}(y^2) \,I_1\left(\frac{y^2}{2}\right)\Bigg]$$ where $P_i(u)$ and $Q_i(u)$ are polynomials of degree $i$ in $u$.

Similarly

$$T_k=\sum_{n=0}^\infty \frac{(n+1)_k}{(n+1)\Gamma (n+1)} y^{2n+1}= y\,e^{y^2}\,R_{k-1}(y^2)$$

Edit

To avoid loosing this more than interesting comment from @Maxim

If we expand $U$ into series, we get hypergeometric functions of two variables of order two, which are expressible in terms of the Horn functions. In this case we get the Humbert series $\Psi_2$: $$\color{red}{I = \frac {y \sqrt \pi} 4 \Psi_2 {\left( \frac 3 2, 2, \frac 1 2; y^2, \frac {\alpha^2} 4 \right)} - \frac {\alpha y} 2 \Psi_2 {\left( 2, 2, \frac 3 2; y^2, \frac {\alpha^2} 4 \right)}}$$