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Can anyone do

$$ \int_{0}^{\infty}\int_{0}^{a+bx^{2}}e^{-\beta\left(x^{2}+y^{2}\right)}dydx $$

Perhaps this post could help: Integrating a product of exponentials and error functions

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apg
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    sorry to repost this, but I consider it a different question to a previous post. The down vote is unnecessary. – apg Feb 15 '14 at 00:03
  • have you made any progress? E.g., is it possible to pull the $e^{-\beta x^2}$ into the outer integral? – TooTone Feb 15 '14 at 00:04
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    yes, essentially it becomes an exponential times a error function with the inner upper limit for an argument – apg Feb 15 '14 at 00:05
  • This is indeed

    $$\int^\infty_0e^{-\beta x^2}dx\int^{a+bx^2}_0e^{-\beta y^2}dy=\frac{1}{2}\beta^{-\frac{1}{2}}\Gamma(\frac{1}{2})\int^{a+bx^2}_0e^{-y^2 \beta }dy$$ Where $\Gamma$ is the Gamma function. Hope this helps.

     –  Feb 15 '14 at 00:10
  • But this is wrong, since the second part is x-dependent. – Christopher K Feb 15 '14 at 00:23
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    Have you tried polar co-ordinates? – user27182 Feb 15 '14 at 00:29
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    @SanathDevalapurkar I don't think that's right. The limit of integration on the right integral involves $x$ so you have to keep the $dx$ after the $dy$, i.e. the integrals can't be factored in this way. – TooTone Feb 15 '14 at 00:50
  • @TooTone I understand. Thanks for the clarification. –  Feb 15 '14 at 00:54
  • you're going to need to look at the reference in the OP – apg Feb 15 '14 at 01:23
  • can someone expand on converting this to polar coordinates? – apg Feb 15 '14 at 01:37
  • I think polars makes the limits too complex, leading to an even trickier Gaussian. – apg Feb 15 '14 at 02:23

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