Integral involving Modified Bessel Function of the First Kind Why is this true?
$$ \int_0^\infty e^{-\frac{1}{2}(b^2+x^2)} I_0(bx) x \,dx = 1 $$
Note that $I_0(x)$ is a modified bessel function of the first kind. The difficulty for me lies in a) translating the bessel function into something basic (the argument is not an integer, hence making it complicated to me), and 2) this doesn't seem to be on integral-table.com.
If it helps, the equation above is identical to:
$$ Q(b,0) = 1,$$
where $Q(a,b)$ is a $Q$-Marcum function defined as:
$$ Q(a,b) = \int_b^\infty e^{-\frac{1}{2}(a^2+x^2)} I_0(ax) \,dx$$
Thanks.
 A: Lets start from the definition of the modified Bessel Function. Recall
$$I_{0}(z)=\sum_{n=0}^{\infty}\frac{1}{4^{n}}\frac{z^{2n}}{n!n!}.$$ So our integral is
$$\int_{0}^{\infty}e^{-\frac{1}{2}\left(x^{2}+b^{2}\right)}x\sum_{n=0}^{\infty}\frac{1}{4^{n}}\frac{b^{2n}x^{2n}}{n!n!}dx.$$ 
Switching the order of sumation and integration we get
$$e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx.$$
Now, let $u=\frac{1}{2}x^{2}$ and $du=xdx$, to see that 
$$\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx=2^{n}\int_{0}^{\infty}e^{-u}u^{n}du=2^{n}\Gamma(n+1)=2^{n}n!.$$ 
Hence 
$$e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx=e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\left(2^nn!\right)$$
$$=e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{\left(\frac{b^{2}}{2}\right)^{n}}{n!}=e^{-\frac{1}{2}b^{2}}e^{\frac{1}{2}b^{2}}=1$$
as desired.
Notice that the exact same argument shows 
$$\int_0^\infty e^{\left(\frac{1}{2}b^2-\frac{1}{2}x^2\right)}J_0(bx)xdx=1$$ where $J_0(x)$ is the Bessel Function.
Hope that helps,
A: See also the Boros-Moll article in  http://www.mat.utfsm.cl/scientia/vol12.html , equation (5.8).
