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In physics, we have an non-Gaussian Distribution which can be simply written as $f(x)=\exp(-ax^2-bx^3)$, and we may need to calculate the integral of this distribution, simply written as $\int_0^\infty f(x)dx$, so how to calculate this integral?

and then we may calculate the expectation functions like $\langle xx\rangle=\int_0^\infty xxf(x)dx$.

Maybe you can refer to arxiv:astro-ph/0210603https://arxiv.org/abs/astro-ph/0210603, and the question is just how to gain equation (5.13) from Distribution of (5.12).

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    $\begingroup$ Even for a=0, b=1, where your integral is pretty much Gamma(1/3), there is no closed form formula; calling it Gamma(1/3) is just giving a name to the integral we can’t evaluate. $\endgroup$
    – David Loeffler
    Commented Jan 9 at 7:08
  • $\begingroup$ Wolfram Alpha will tell you this integral can be written in terms of a hypergeometric function; I don't quite understand the question you write with reference to Maldacena's paper... $\endgroup$
    – Carlo Beenakker
    Commented Jan 9 at 7:26
  • $\begingroup$ Thanks to David Loeffler, I know that result of Gamma(1/3) and finally know the reason. $\endgroup$
    – ZhengJiang
    Commented Jan 9 at 8:39
  • $\begingroup$ In fact I think this question is from the Maldacena's paper, and I want to understand the method of calculating expectations of the non-Gaussion distributions. Thanks to Carlo Beenakker, but it seems different from Maldacena's "simple results" $\endgroup$
    – ZhengJiang
    Commented Jan 9 at 8:42

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