How do we prove that $\int_0^1 (x\ln x)^{-1+n}\,dx = -\left( -\frac{1}{n} \right)^n \Gamma(n)$

Question

I need this solution to prove that

$$\int_0^1 (x\ln x)^{-1+n}\,dx = -\left( -\frac{1}{n} \right)^n \Gamma(n)$$

Thank you!

Answer 1

First, let $t=\ln{x}$ and $dt=\frac{dx}{x}$ or $dx=e^t dt$:

$\int_{-\infty}^0 e^{tn}t^{n-1} \; dt$.

You want the integral to be in the form of $\Gamma(x)=\int_0^{\infty} e^{-t}t^{z-1} \; dt$, so let $u=-tn$ and $\frac{du}{-n}=dt$:

$\int_{\infty}^0 e^{-u} {\left(\frac{-t}{n}\right)}^{n-1} \frac{-du}{n}={\left(\frac{1}{n}\right)}^n{\left(-1\right)}^n\int_0^{\infty} e^{-u} t^{n-1} \; du$.

Notice the integral is equal to $\Gamma(n)$, so the final answer is $\boxed{{\left(\frac{1}{n}\right)}^n{\left(-1\right)}^{n-1} \;\Gamma(n)}$