IS $\int_{-\infty}^\infty e^{-\beta\cdot g(x)}g(x)^{\alpha-1}={\Gamma(\alpha)\over \beta^\alpha}\ \ ?$

Question

Is the following statement true:

Let $g(x)$ be some non-negative continuous function of $x$.We know $$\int_{0}^\infty e^{-\beta x}x^{\alpha-1}dx={\Gamma(\alpha)\over \beta^\alpha}$$

IS $$\int_{0}^\infty e^{-\beta\cdot g(x)}g(x)^{\alpha-1}dx={\Gamma(\alpha)\over \beta^\alpha}\ \ ?$$

Answer 1

You are most likely thinking of the substitution $x=g(u),\ dx=g'(u)du$ where

$g(u)\to\infty,u\to\infty$

$g(u)\to0,u\to0$

$g(u)$ is bijective.

Putting in the substitution, we would get

$$\int_0^\infty e^{-\beta x}x^{\alpha-1}dx=\int_0^\infty g'(u)e^{-\beta g(u)}g(u)^{\alpha-1}du$$