# Compute $\int_0^{\infty} Q_1(y,b) \frac{y}{\sigma^2} \exp{(-y^2/(2\sigma^2))} \, dy$

We know that the first order Marcum Q-function can be represented as $$Q_1(y, b)=\int_{b}^{\infty} x \exp{(-(x^2+y^2)/2)} I_0(y x) \, dx ,$$ where $I_0(\cdot)$ is the modified Bessel function of the first kind. In my case, $y$ and $b$ are positive.

I am trying to compute the following integral: $\int_0^{\infty} Q_1(y,b) \frac{y}{\sigma^2} \exp{(-y^2/(2\sigma^2))} \, dy$.
Note that $f(y) = \frac{y}{\sigma^2} \exp{(-y^2/(2\sigma^2))}$ is the PDF of the Rayleigh distribution.

I have no clue how to derive it. Any idea ?

• the result seems to be $\exp(-b^2/(2+2\sigma^2))$. Change the order of integrations and do the $y$-integral first. – Pierpaolo Vivo Feb 2 '16 at 19:12
• @PierpaoloVivo Could you please detail your comment in an answer so that I can accept it. – din Feb 2 '16 at 20:48
• cannot right now, but will do – Pierpaolo Vivo Feb 2 '16 at 21:11

The integral can be written as $$\frac{1}{\sigma^2}\int_b^\infty dx\ x\ e^{-x^2/2}\int_0^\infty dy\ y e^{-y^2 \left(\frac{1}{2}+\frac{1}{2\sigma^2}\right)}I_0(xy)\ .$$ Now we can use the following formula 6.633.4 of Gradshteyn-Ryzhik $$\int_0^\infty dy\ y\ e^{-\alpha y^2}I_\nu (\beta y)J_\nu(\gamma y)=\frac{1}{2\alpha}\exp\left(\frac{\beta^2-\gamma^2}{4\alpha}\right)J_\nu\left(\frac{\beta\gamma}{2\alpha}\right)\qquad \mathrm{Re}\ \alpha>0,\quad\mathrm{Re}\ \nu\ , >-1$$ to perform the $y$-integral (noting that $J_0(0)=1$), resulting in $$\frac{1}{\sigma^2}\frac{\sigma^2}{1+\sigma^2}\int_b^\infty\ dx\ x\ e^{-x^2/2}e^{\frac{\sigma^2 x^2}{2+2\sigma^2}}\ ,$$ which is elementary, and gives eventually $$\frac{1}{\sigma^2}\frac{\sigma^2}{1+\sigma^2}\left(\sigma ^2+1\right) e^{-\frac{b^2}{2 \sigma ^2+2}}=\exp(-b^2/(2\sigma^2+2))\ .$$