Some interesting integral identities.
$$I_n=\int_0^\infty \exp\left(-x^n\right)\,dx=\Gamma\left(\frac{n+1}{n}\right)\
J_n=\int_{-\infty}^\infty x^{-2n}\exp\left(-x^{-2n}\right)\,dx=\frac{1}{n}\Gamma\left(\frac{2n-1}{2n}\right)$$
and one that seems related to $J_n$, call it ${J_n}^*$,
$${J_n}^*=\int_0^\infty x^{-n}\exp\left(-x^{-n}\right)\,dx=\frac{1}{n}\Gamma\left(\frac{n-1}{n}\right)$$
where $I_n$ and $J_n$ holds for natural $n\ge1$, and ${J_n}^*$ for natural $n\ge2$.

To be clear, $\large\exp(f(x))=e^{f(x)}$ and $\Gamma$ denotes the gamma function defined by
$$\Gamma(t)=\int_0^\infty x^{t-1}e^{-x}\,dx$$

Question

Some interesting integral identities.
$$I_n=\int_0^\infty \exp\left(-x^n\right)\,dx=\Gamma\left(\frac{n+1}{n}\right)\
J_n=\int_{-\infty}^\infty x^{-2n}\exp\left(-x^{-2n}\right)\,dx=\frac{1}{n}\Gamma\left(\frac{2n-1}{2n}\right)$$
and one that seems related to $J_n$, call it ${J_n}^*$,
$${J_n}^*=\int_0^\infty x^{-n}\exp\left(-x^{-n}\right)\,dx=\frac{1}{n}\Gamma\left(\frac{n-1}{n}\right)$$
where $I_n$ and $J_n$ holds for natural $n\ge1$, and ${J_n}^*$ for natural $n\ge2$.

To be clear, $\large\exp(f(x))=e^{f(x)}$ and $\Gamma$ denotes the gamma function defined by
$$\Gamma(t)=\int_0^\infty x^{t-1}e^{-x}\,dx$$

anonymous · Answer

These results were obtained using Mathematica, but I'm interested in seeing an analytical approach.

I attempted to verify $I_n$ using the usual polar conversion argument for $I_2$, which doesn't seem to work out nicely. I've also tried using the power series for $e^t$ and integrating term by term, but I'm running into a brick wall :/

I'm currently trying out some parameterizations to see if differentiating under the integral sign might work.

zarkon · Answer

$$\Gamma\left(\frac{n+1}{n}\right)=\left(\frac{n+1}{n}-1\right)\Gamma\left(\frac{n+1}{n}-1\right)$$

$$=\frac{1}{n}\Gamma\left(\frac{1}{n}\right)=\frac{1}{n}\int\limits_{0}^{\infty}u^{1/n-1}e^{-u}du$$

let $u=x^n$

$$du=nx^{n-1}dx$$



$$=\frac{1}{n}\int\limits_{0}^{\infty}x^{1-n}e^{-x^n}nx^{n-1}dx$$
$$=\int\limits_{0}^{\infty}e^{-x^n}dx$$

zarkon · Answer

$J_n$ follows from starting with $\frac{1}{n}\Gamma\left(\frac{2n-1}{2n}\right)$

use the definition of the gamma function and then do a simple substitution

finally use the fact that the integrand is an even function to get the final result.

anonymous · Answer

That's quite a neat approach!

zepdrix · Answer

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anonymous · Answer

Aside from working backwards from Zarkon's method, I'm hoping to come up with a forward (left-to-right) proof of the equality. Potentially with some clever DUIS setup.

kainui · Answer

This is interesting I'll be thinking about this for a while I think to see what I can come up with.

kainui · Answer

I guess I'll just write the relationship between J and J star. 
$$\Large 2J_{2n}^* = J_n$$
The 2n part in the subscript is a direct substitution, and the 2 multiplying the outside just comes from the fact that 2n we just plugged in makes it an even function so it's symmetrical, and half of the original J_n.

I'm not sure if that's what Zarkon already discovered and explained since I'm not entirely sure I understand his argument yet. I'm playing around with the integrals more to see if there's a clever little trick.

kainui · Answer

I don't think this is what you're looking for but just starting with the definition we can work it backwards.$$\Gamma(t)=\int_0^\infty x^{t-1}e^{-x}\,dx$$ $$\Gamma(\frac{n+1}{n})=\int_0^\infty x^{\frac{n+1}{n}-1}e^{-x}\,dx=\int_0^\infty x^{\frac{1}{n}}e^{-x}\,dx$$

anonymous · Answer

Right, that's what Zarkon took advantage of in his proof.

kainui · Answer

I see what you're trying to do though I think. You want to get to the answer without knowing the answer already right? Or are you more concerned with finding the method that gets this with DUIS for fun?

anonymous · Answer

Yeah I'm curious to see how one would obtain this result without knowing it beforehand. I remember seeing a tricky approach to evaluating $I_2$ in a pdf about DUIS (probably this one: http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf )

anonymous · Answer

Actually, it's not in there, but I've come across another pdf very similar to it...

anonymous · Answer

Here we are, under section 4: http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf