# integration of product of Q function with exp.

I have the below integration that I'm trying to evaluate. The problem is that the final answer is finite which should be a variable of x. So what I'm doing wrong? \label{eq:dots}\begin{aligned} &\dfrac{\sqrt{2}}{\sqrt{\pi}} \left[ \int_x^\infty e^{ \tfrac{- f^2}{2}} Q\left( f \right) +\int_{ -x}^\infty e^{ \tfrac{-f^2}{2}} Q\left(f\right) \right] df\\ =&\dfrac{\sqrt{2}}{\pi\sqrt{\pi}} \left[ \int_x^\infty e^{ \tfrac{- f^2}{2}} \left( \int_0^{\tfrac{\pi}{2}} e^{\tfrac{-f^2}{2sin^2\phi}} d \phi \right) df+\int_{-x}^\infty e^{ \tfrac{-f^2}{2}} \left( \int_0^{\tfrac{\pi}{2}} e^{\tfrac{-f^2}{2sin^2\phi}} d \phi \right)df \right] \\ =&\dfrac{2}{\pi} \left[ \int_0^{\tfrac{\pi}{2}} \left( \dfrac{1}{\sqrt{2\pi}}\int_x^\infty e^{ \tfrac{- f^2}{2} \left(1+ \tfrac{1}{sin^2\phi}\right)}df \right) d \phi +\int_0^{\tfrac{\pi}{2}} \left( \dfrac{1}{\sqrt{2\pi}}\int_{-x}^\infty e^{ \tfrac{- f^2}{2} \left(1+ \tfrac{1}{sin^2\phi}\right)} df \right)d \phi \right] \\ =&\dfrac{2}{\pi} \left[ \int_0^{\tfrac{\pi}{2}} Q \left(x \sqrt{1+ \tfrac{1}{sin^2\phi}}\right) \tfrac{d\phi}{\sqrt{1+ \tfrac{1}{sin^2\phi}}}+\int_0^{\tfrac{\pi}{2}} Q \left(-x \sqrt{1+ \tfrac{1}{sin^2\phi}}\right) \tfrac{d\phi}{\sqrt{1+ \tfrac{1}{sin^2\phi}}} \right] \\ =&\dfrac{2}{\pi} \left[ \int_0^{\tfrac{\pi}{2}} Q \left(x \sqrt{1+ \tfrac{1}{sin^2\phi}}\right) \tfrac{d\phi}{\sqrt{1+ \tfrac{1}{sin^2\phi}}}+\int_0^{\tfrac{\pi}{2}} 1- Q \left(x \sqrt{1+ \tfrac{1}{sin^2\phi}}\right) \tfrac{d\phi}{\sqrt{1+ \tfrac{1}{sin^2\phi}}} \right] \\=& \dfrac{2}{\pi} \int_0^{\tfrac{\pi}{2}} \tfrac{d\phi}{\sqrt{1+ \tfrac{1}{sin^2\phi}}} \end{aligned} step 1 to step 2, I replaced $Q\left(f\right) = \dfrac{1}{\pi}\int_0^{\tfrac{\pi}{2}} e^{\tfrac{-f^2}{2sin^2\phi}} d \phi$ , step 2 to step 3 I exchanged variables since the first integration is a function of f while the second integartion is a function of $\phi$. step 3 to step 4 I used $Q(y)=\dfrac{1}{\sqrt{2\pi}} \int_y^\infty e^{\tfrac{-u^2}{2}} du$ after subsitution. step 4 to step 5 Q(g)=1-Q(-g).

• Not sure how you get from 2nd to 3rd line. The $e^{-f^2 \over 2}$ term can't be brought inside the integral? Also, how did a factor of $\sqrt2 \over \sqrt \pi$ come out? – Shuri2060 Jul 1 '17 at 22:02
• I didn't check all details, but under your assumptions, $Q(-f)=Q(f),$ so your expression has to be constant, it doesn't depend on $x$. – Professor Vector Jul 2 '17 at 9:08
• Thanks for your notes, I updated the post to explain what i did in each step – Diana Jul 2 '17 at 17:58