Proving $\int_{0}^{\infty} u^{x-1}e^{-u} du = 2 \int_{0}^{\infty} u^{2x-1}e^{-u^2} du$

Question

I am trying to solve exercise 1.1 in Neural Networks for Pattern Recognition by Bishop.

They give the gamma function as:

$\int_{0}^{\infty} u^{x-1}e^{-u} du$

I can only solve it using the form of the gamma function:

$2 \int_{0}^{\infty} u^{2x-1}e^{-u^2} du$

Which is equation 3 here

It is very unclear to me how these two forms of the gamma function are equal. Is there a well-known proof?

Answer 1

Just set $u = t^2$:

$$\int_0^\infty u^{x-1}e^{-u}\,du = 2\int_0^\infty t\cdot t^{2x-2}e^{-t^2}\,dt = 2\int_0^\infty t^{2x-1}e^{-t^2}\,dt$$