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While computing densities for some distributions, I stumbled on the following family of parametrized integrals:

$$ p (x) := \sqrt{\frac{2}{\pi}} \int_{\mathbb{R}_+} e^{-\frac{x^2}{2 y^2} - y^2} \ d y.$$

They are obtained by playing around with normal distributions and exponential distributions. Using Wolfram Alpha, the following conjecture seems to hold:

$$ p (x) = \frac{1}{\sqrt{2}} e^{- \sqrt{2} x}.$$

However, I have no idea about how I should prove this equality. I don't think a residue computation may work, because the singularities are essential, but my complex analysis classes are a bit far away, and perhaps there is some nifty trick to solve this problem...

Thank you very much.

D. Thomine
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1 Answers1

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Note that

$$y^2+\frac{x^2}{2 y^2} = \left ( y + \frac{x}{\sqrt{2} y}\right)^2-\sqrt{2} x$$

Ron Gordon
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  • I tried this. It did not help me. In addition, you create a bunch of new terms which cancel later, so I don't think it is the best way to proceed. – D. Thomine Sep 20 '13 at 15:22
  • I think the correct expression to use is $$y^2+\frac{x^2}{2 y^2} = \left ( y - \frac{x}{\sqrt{2} y}\right)^2+\sqrt{2} x$$. – achille hui Sep 20 '13 at 15:25
  • @D.Thomine: see the first solution in that link. Sorry I did not provide more details, but I had to work on something else. – Ron Gordon Sep 20 '13 at 15:27
  • @achillehui: either way should work, but you are right, that sub is better. – Ron Gordon Sep 20 '13 at 15:28