Prove $-\int_0^1x^k\ln x \ dx = \frac{1}{(k+1)^2}$ using the gamma function

Question

Prove that

$$-\int_0^1x^k\ln x \ dx = \frac{1}{(k+1)^2} \tag{$k>-1$}$$

using the gamma function

$$\Gamma(z+1) = \int_0^\infty e^{-t}t^{z}\ dt$$

Which substitution should I take here? It appears that I can find some relation from $\ln x$ and $e^{-t}$, but... how? Also, I must take a substitution that is $0$ on $0$ and $1$ on infinity, but which one?

Answer 1

Start by letting: $$x=e^{-u} \iff dx=-e^{-u}~du$$ Therefore, it follows that $x=0 \implies t\to \infty$ and $x=1 \implies t=0.$ $$-\int_0^1x^k\ln x \ dx =-\int_{\infty}^0 e^{-ku}\cdot -u\cdot -e^{-u}~du=\int_0^{\infty} e^{-ku-u}\cdot u~du=\int_0^{\infty} e^{-u(k+1)}\cdot u~du$$ Now substitute: $$u=\frac{t}{k+1} \iff du=\frac{1}{k+1}~dt$$ Note that the integration limits stay the same due to the condition that $k>-1$. You should eventually get the answer you require.