everyone, I have an integral of exponential function and modified Bessel function to calculate, as follows
$Q=\int\limits_0^\infty {z\exp \left( { – a{z^2}} \right){I_0}\left( {bz} \right)} dz$,
where ${I_0}\left( \right)$ is the 0th order modified Bessel function of first kind; $a > 0$, $b > 0$ are constant.
Thanks!
Best Answer
In general, $~\displaystyle\int_0^\infty z^{2k-1}~e^{-az^2}~I_0(bz)~dz~=~\frac{\Gamma(k)}{2~a^k}\cdot L_{-k}\bigg(\frac{b^2}{4a}\bigg),~$ where $a,b,k>0$ and L is a Laguerre polynomial. For $k=1\iff2k-1=1$, this becomes $\dfrac{\exp\bigg(\dfrac{b^2}{4a}\bigg)}{2~a}~,~$ and for $k=\dfrac12$ $\iff2k-1=0$, we have $\sqrt{\dfrac\pi a}\cdot\dfrac{\exp\bigg(\dfrac{b^2}{8a}\bigg)}2\cdot I_0\bigg(\dfrac{b^2}{8a}\bigg)$. Using these last two results, we can evaluate the integral for all natural values of $2k-1$, by differentiating with regard to a under the integral sign.