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$I_{2n} = \int_{-\infty}^{+\infty} e^{-x^{2n}}dx$

We know very well $I_{2}= \sqrt{\pi}$

Could you please help me find general formula of $I_{2n}$?

Thanks for answers

1 Answers 1

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You may rewrite $I_{2n}$ using the integral definition of the Gamma function :
$\Gamma(s)= \int_0^{\infty} t^{s-1} e^{-t} dt$

Let's set $t=x^{2n}$ then : $\Gamma(s)= 2n \int_0^{\infty} x^{2n(s-1)} e^{-x^{2n}} x^{2n-1}dx$

For $s=\frac1{2n}$ the powers of $x$ disappear and we get : $\Gamma\left(\frac1{2n}\right)= 2n \int_0^{\infty} e^{-x^{2n}} dx$ so that (we multiply by $2$ since the integral of $I_{2n}$ starts from $-\infty$) $I_{2n}=2\frac{\Gamma\left(\frac1{2n}\right)}{2n}=2\Gamma\left(\frac{2n+1}{2n}\right).$

No closed form is known for $\Gamma(1/4)$ nor other fractions (except $\Gamma(1/2)$ of course!) as opposed to the derivatives of $(\ln(\Gamma))$ where closed forms exist for every fraction!

Some interesting equalities concerning $\Gamma(1/n)$ may be found at Wolfram Mathworld (equation (63) and following), the reference to Chudnovsky's proof of the transcendence of $\Gamma(1/3)$ and $\Gamma(1/4)$ and iterative algorithms for $\Gamma(k/24)$.