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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.

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1 Answer 1

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$\int_0^xrI_0(r)e^{-b^2r^2}~dr$

$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{r^{2n+1}e^{-b^2r^2}}{4^n(n!)^2}~dr$

$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{r^{2n}e^{-b^2r^2}}{2^{2n+1}(n!)^2}~d(r^2)$

$=\int_0^{x^2}\sum\limits_{n=0}^\infty\dfrac{t^ne^{-b^2t}}{2^{2n+1}(n!)^2}~dt$

$=-\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{t^ke^{-b^2t}}{2^{2n+1}b^{2n-2k+2}n!k!}\right]_0^{x^2}$ (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)

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