Proving that $\Gamma (x) = \int_{0}^{1} \left( \ln \left(\frac{1}{u} \right) \right)^{x-1} du$

Question

I want to prove that
$$ \Gamma (x) = \int_{0}^{1} \left( \ln \left(\frac{1}{u} \right) \right)^{x-1} du $$

I start with $$ \int_{0}^{1} \left( \ln \left(\frac{1}{u} \right) \right)^{x-1} du = $$

$$ \int_{0}^{1} \left( -\ln u \right)^{x-1} du $$

Now, if I set $-\ln u = t$, I have $u=e^{-t}$ and $dt=\frac{-du}{u} \Leftrightarrow du = -dt * u. $

I will set these to my function above:

$$ \int_{0}^{1} t^{x-1}(-e^{t})(-dt) = \int_{0}^{1} t^{x-1}e^{t}dt $$

Which is like the $\Gamma = \int_{0}^{+\infty} t^{x-1}e^{t}dt$, except for the ends.

How do I change the 1 into $+\infty$ ?

Answer 1

You need to change integration limits when doing the change of variable!