Please help with the Integral of exponential divide by a Q-function

[tayo4arsenal]

Can anyone help with

[FONT=MathJax_Size1]∫ exp(-x²)[/FONT]​

[FONT=MathJax_Size1]/ Q(x)dx with interval [-infinity to Infinity]. Thanks[/FONT]​

 

[pka]

Can anyone help with

[FONT=MathJax_Size1]∫ exp(-x²)[/FONT]​

[FONT=MathJax_Size1]/ Q(x)dx with interval [-infinity to Infinity]. Thanks[/FONT]​

What is a $\displaystyle Q-$ function?

 

[tayo4arsenal]

It is the Gaussian Q-function. It is also equivalent to 1/2*erfc(sqrt(2)x) which is the complementary error function

 

[galactus]

Oh, the Q function would then be the complement of the Gaussian Normal Distribution from Statistics. I had not heard a name for it before.

$\displaystyle \text{Q function} = \displaystyle \frac{1}{\sqrt{2\pi}}\int_{z}^{\infty}e^{\frac{-x^{2}}{2}}dx$

How exactly is your problem set up?. It appears to be:

$\displaystyle \displaystyle \int_{-\infty}^{\infty}\frac{e^{-x^{2}}}{\frac{1}{\sqrt{2\pi}}\displaystyle \int_{z}^{\infty} e^{\frac{-x^{2}}{2}}dx}dx$.

Looks rather ugly.

But, using the erf, the above would be:

$\displaystyle \displaystyle \frac{-2\sqrt{\pi}}{erf\left(\frac{z}{\sqrt{2}}\right)-1}$

Note that the famous Gauss integral is $\displaystyle \int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$