Gamma Functions and Gaussian-like Functions Part 2...?

Question

As a follow-up to
http://answers.yahoo.com/question/index;_ylt=ArkcqXNFGVCIPgBSRuHAs3jsy6IX;_ylv=3?qid=20080521104305AAmhmRY

If, as scythian said, we allow x=1/2n, it does indeed get rid of the t^(x-1/2n) term, but it also affects the exponential term, and changes the nature of the integral.

For, if we let x=1/2n, then the integral:

(1/2n)Γ(1/2n) = Γ((2n+1)/2n) = ∫ (0 to ∞) e^(-x^2n) dx
=  ∫ (0 to ∞) e^(-(1/2n)^2n) dx

but e^(-(1/2n)^(2n)) is a constant for any fixed n, so the integral above is not defined.

Can anyone tweak this proof?  Thank you.

Scythian1950 · Accepted Answer

Nick S, I'm sorry, it was my fault.  Go back to the original proof, and instead of making the subsitution t = x^2n, make the subsitution t = y^2n.  Then you'll get to the point where we substitute x = 1/2n.  I had been sloppy in not using ANOTHER symbol for what is really supposed to be an independent variable.

The integral that results in the Gamma function is what's known as an integral transform, which is of the form f(x) = ∫ g(x,t)dt.  Thus, with the first substitution, it should have been of the form f(x) = ∫ g(x,y)dy, after which we end up with f(1/2n) = ∫ g(1/2n,y)dy.  Get it now?

Gamma Functions and Gaussian-like Functions Part 2...?

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