Deriving property of the Gamma Function

Question

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

Answer 1

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

Answer 2

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

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

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

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

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

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