tayo4arsenal
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Can anyone help with
[FONT=MathJax_Size1]∫ exp(-x2)[/FONT]
[FONT=MathJax_Size1]/ Q(x)dx with interval [-infinity to Infinity]. Thanks[/FONT]
Please help with the Integral of exponential divide by a Q-function
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pka
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What is a \(\displaystyle Q-\) function?
tayo4arsenal
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It is the Gaussian Q-function. It is also equivalent to 1/2*erfc(sqrt(2)x) which is the complementary error function
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 \(\displaystyle \int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}\)
\(\displaystyle \text{Q function} = \displaystyle \frac{1}{\sqrt{2\pi}}\int_{z}^{\infty}e^{\frac{-x^{2}}{2}}dx\)
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 \(\displaystyle \int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}\)
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