# Deriving property of the Gamma Function

It's been awhile since I've done much calculus and I've recently been looking at an identity for the gamma function:

$$\frac {\Gamma(s)}{n^s}=\int_0^{\infty}e^{-nx}x^{s-1}dx =\frac 1{n^s} \int_0^{\infty}e^{-x}x^{s-1}dx$$

I understand how to work back from there and show that they are equal using the substitution $$u=nx$$, but I'm having trouble seeing how to derive this from:

$$\frac {\Gamma(s)}{n^s}=\frac 1{n^s} \int_0^{\infty}e^{-x}x^{s-1}dx$$

I see this used fairly often when deriving the integral form of the Zeta function and I'm just wondering where it comes from.

Any help would be appreciated

Let $$f(x)=x^{s-1}e^{-x}, F(y)= \int_0^y f(x)dx$$, $$G(y) = F(ny)$$ then $$G'(x) =n F'(nx)= n f(nx)$$ thus $$\int_0^y n f(nx)dx = G(y)-G(0) = F(ny)=\int_0^{ny}f(x)dx$$

Letting $$y\to \infty$$ you get the change of variable formula $$\Gamma(s)=\int_0^\infty f(x)dx=n\int_0^\infty f(nx)dx= n \int_0^\infty n^{s-1}x^{s-1}e^{-nx}dx$$

$$u=nx, du=ndx, du={1\over n}dx$$, $$x^{s-1}=e^{(s-1)ln(x)}=e^{(s-1)ln({u\over n})}$$.

$$\int_0^{\infty}e^{-nx}x^{s-1}dx=$$

$${1\over n}\int_0^{\infty}e^{-u}e^{ln({u\over n})(s-1)}du=$$

$${1\over n}\int_0^{\infty}e^{-u}e^{(s-1)ln(u)-(s-1)ln(n)}du$$

$${1\over n}\int_0^{\infty}e^{-u}e^{(s-1)ln(u)}e^{-(s-1)ln(n)}du$$

$$={1\over n}\int_0^{\infty}e^{-u}u^{s-1}n^{1-s}du$$