Definite Integral of Modified Bessel function, exp and power 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 (I'm a physicist i dont know what that is and how to deal with it later on in the problem), 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.) $z=0$
 A: For $\int_0^Axe^{-ax^2}I_0(x)~dx$ ,
$\int_0^Axe^{-ax^2}I_0(x)~dx$
$=\int_0^A\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}e^{-ax^2}}{4^n(n!)^2}~dx$
$=\int_0^A\sum\limits_{n=0}^\infty\dfrac{x^{2n}e^{-ax^2}}{2^{2n+1}(n!)^2}~d(x^2)$
$=\int_0^{A^2}\sum\limits_{n=0}^\infty\dfrac{x^ne^{-ax}}{2^{2n+1}(n!)^2}~dx$
$=-\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{x^ke^{-ax}}{2^{2n+1}a^{n-k+1}n!k!}\right]_0^{A^2}$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$=\sum\limits_{n=0}^\infty\dfrac{1}{2^{2n+1}a^{n+1}n!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{A^{2k}e^{-aA^2}}{2^{2n+1}a^{n-k+1}n!k!}$
$=\dfrac{e^\frac{1}{4a}}{2a}-\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{A^{2k}e^{-aA^2}}{2^{2n+1}a^{n-k+1}n!k!}$
$=\dfrac{e^\frac{1}{4a}}{2a}-\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{A^{2k}e^{-aA^2}}{2^{2n+2k+1}a^{n+1}(n+k)!k!}$
$=\dfrac{e^\frac{1}{4a}}{2a}-\dfrac{e^{-aA^2}}{2a}\Phi_3\left(1,1;\dfrac{1}{4a},\dfrac{A^2}{4}\right)$ (according to http://en.wikipedia.org/wiki/Humbert_series)
