Gamma Functions and Gaussian-like function forms?

Question

This isn't for a class, but it's simply something interesting I thought of.

Take a Gaussian:
Int_(-infinity, infinity) Exp[-x^2] dx
Now of course this evaluates to Sqrt(pi)

What if we let y = x^2 and took the Gaussian in y?  It goes to 2*Gamma[5/4]...y=x^3?  2*Gamma[7/6].  This pattern follows all the way to at least n = 100, for y = x^n...

Now, this is something I'd like to know a starting point on (I know this should probably be done using induction, and I have the n=1 case done, as Sqrt(pi) = 2*Gamma[3/2]; What I need a starting point on is how to carry out the inductive step)

For any natural number n, prove that
Int_(-infinity, infinity) Exp[-x^(2n)] dx = 2 * Gamma[(2*n+1)/(2*n)]

Further, is there a finite natural number n > 1 such that this result is expressible in simple closed form (i.e. an exact numerical expression, and not in terms of the Gamma function)

Scythian1950 · Accepted Answer

Right off, it should be noted that  Γ(n+1) = n Γ(n), so that  Γ(1+1/2n) = (1/2n) Γ(1/2n).  Unfortunately, Γ(1/2n) has a value in terms of elementary constants only if n = 1.  I will get back to you with the rest of the problem when I get time later on.

The Gamma function can be found by the definite integral:

Γ(n) = ∫ (0 to ∞) t^(x-1) e^(-t) dt

We make the substitution t = x^2n, which also runs from 0 to ∞

dt = 2n x^(2n-1) dx = 2n (t^(1/2n))^(2n-1) dx, so that we have:

Γ(x) =  ∫ (0 to ∞) t^(x-1)  2n (t^(1/2n))^(2n-1) e^(-x^2n) dx

(1/2n)Γ(x) =  ∫ (0 to ∞) t^(x-1/2n) e^(-x^2n) dx

(SEE THE TERM t^(x-1/2n), WHAT CAN WE DO TO MAKE THIS GO AWAY?)

So, if x = 1/2n, we have:

(1/2n)Γ(1/2n) = Γ((2n+1)/2n) =  ∫ (0 to ∞) e^(-x^2n) dx

2 Γ((2n+1)/2n) =  ∫ (-∞ to ∞) e^(-x^2n) dx

which is the expression you seek.

Gamma Functions and Gaussian-like function forms?

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