# Limit case of integral with exp and modified Bessel function

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1. Sep 8, 2016

### leialee

1. The problem statement, all variables and given/known data

How to integrate this?

$\int_{0}^{A} x e^{-a x^2}~ I_0(x) dx$

where $I_0$ is modified Bessel function of first kind?

I'm trying per partes and looking trough tables of integrals for 2 days now, and I would really really appreciate some help.

This is a part of a problem, whis would be this:

$T(r,z,t)=C\int_{0}^{B} u^{-\frac{3}{2}} du \int_{0}^{A} dr_0 r_0 e^{-\frac{r^2+r_0^2+z^2}{u}}~2\pi I_0(\frac{2rr_0}{u})$

$A=constant$

$B=4Dt$

$u=4D(t-t_0)$

and if the first integral isnt solvable with something relatively not- fancy as hmm lets say Marcum Q-function (cuz what even is that) how do i go about checking out limits for this second integral? Does anyone have any ideas?

1.) $t\to \infty$

2.) $r\to0$

3.) $z=0$

2. Sep 8, 2016

### Ssnow

I found that $I_{0}(x)=\sum_{k=0}^{+\infty}\frac{x^{2k}}{4^{k}(k!)^2}$, on http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html, so you have that:

$\int_{0}^{A}\sum_{k=0}^{+\infty}\frac{x^{2k+1}}{4^{k}(k!)^2}e^{-ax^{2}}dx$

setting $x^{2}=t$ you have

$\int_{0}^{A^2}\frac{1}{2}\sum_{k=0}^{+\infty}\frac{t^{k}}{4^{k}(k!)^2}e^{-at}dt$

if you can put outside the sum you can do it per partes ... , (alternatively you can use a math software in order to examine the integral...)

3. Sep 10, 2016

### rude man

I'd vote for wolfram alpha! But if you don't have subscription rights, leave it as an indefinite integral. They seem to do any level of complexity indefinite integrals for free but balk at even simple definite ones.