I'm trying to calculate
$$\int_0^{+\infty}re^{-\beta r^2}I_0(\beta r\rho)I_0(\beta r a)\,dr,$$
where $I_0$ is the modified Bessel function of first kind and zero order. This integral is equivalent (up to $2\pi$) to
$$\int_0^{+\infty}\int_0^{2\pi}re^{-\beta r(r+a\sin(\theta))}I_0(\beta r\rho) \, dr \, d\theta.$$
All constants here ($\beta$, $\rho$ and $a$) are positive. I see that this converges by plotting the function, but I cannot get any expressions (there's probably a trick I'm not seeing). Any suggestions are welcome
Note: the solution for $a=0$ can be found at
The integral of exponential function and modified Bessel function
Best Answer
Consider $$F\left( a,b,c \right)=\int\limits_{0}^{\infty }{x{{e}^{-c{{x}^{2}}}}{{I}_{0}}\left( ax \right){{I}_{0}}\left( bx \right)dx}$$ Then we have from the addition formula for Bessel functions $${{I}_{0}}\left( w \right)=\sum\limits_{m=0}^{\infty }{{{\left( -1 \right)}^{m}}{{I}_{m}}\left( Z \right){{I}_{m}}\left( z \right)\cos \left( m\phi \right)}$$ where $w=\sqrt{{{Z}^{2}}+{{z}^{2}}-2Zz\cos \left( \phi \right)}$. From this, multiplying by $\cos \left( n\phi \right)$for integer n, integrating over $\left[ 0,\pi \right]$ and using the orthogonality of cosines, we find therefore $$\frac{1}{\pi }\int\limits_{0}^{\pi }{{{I}_{0}}\left( w \right)d\phi }={{I}_{0}}\left( Z \right){{I}_{0}}\left( z \right)$$ So we have then $$F\left( a,b,c \right)=\int\limits_{0}^{\infty }{x{{e}^{-c{{x}^{2}}}}dx}\frac{1}{\pi }\int\limits_{0}^{\pi }{{{I}_{0}}\left( xw \right)d\phi }$$ where $$w=\sqrt{{{a}^{2}}+{{b}^{2}}-2ab\cos \left( \phi \right)}$$ Now consider using the series expansion for the Bessel function $$\begin{align} & \int\limits_{0}^{\infty }{x{{e}^{-c{{x}^{2}}}}{{I}_{0}}\left( xw \right)dx}=\sum\limits_{k=0}^{\infty }{\frac{{{w}^{2k}}}{{{2}^{2k}}k{{!}^{2}}}}\int\limits_{0}^{\infty }{{{x}^{2k+1}}{{e}^{-c{{x}^{2}}}}dx}=\frac{1}{2c}\sum\limits_{k=0}^{\infty }{\frac{{{w}^{2k}}\Gamma \left( k+1 \right)}{{{2}^{2k}}{{c}^{k}}k{{!}^{2}}}} \\ & =\frac{1}{2c}\sum\limits_{k=0}^{\infty }{{{\left( \frac{{{w}^{2}}}{4c} \right)}^{k}}\frac{1}{k!}=}\frac{1}{2c}{{e}^{\frac{{{w}^{2}}}{4c}}} \\ \end{align}$$ Then $$F\left( a,b,c \right)=\frac{1}{\pi }\int\limits_{0}^{\pi }{\frac{1}{2c}{{e}^{\frac{{{w}^{2}}}{4c}}}d\phi }=\frac{1}{2c}{{e}^{\frac{{{a}^{2}}+{{b}^{2}}}{4c}}}\frac{1}{\pi }\int\limits_{0}^{\pi }{{{e}^{-\frac{ab}{2c}\cos \left( \phi \right)}}d\phi }=\frac{1}{2c}{{e}^{\frac{{{a}^{2}}+{{b}^{2}}}{4c}}}{{I}_{0}}\left( \frac{ab}{2c} \right)$$ by the definition of the Bessel function.