Evaluate $\int_{0}^1 x^{p}(\log x)^q dx$

Question

Evaluate

$$\int_{0}^1 x^{p}(\log x)^q dx$$ for $p \in \mathbb{N}$ and $q \in \mathbb{N}$.

Answer 1

By substituting $x=e^{-u}$, then $u=\frac{v}{p+1}$, we get: $$I(p,q)=\int_{0}^{1}x^p(\log x)^q\,dx = (-1)^{q}\int_{0}^{+\infty}e^{-(p+1)u} u^q\,du=\color{red}{\frac{(-1)^q}{(p+1)^{q+1}}\cdot q!}.$$

Answer 2

Let $$I(p)=\int_0^1x^p\ dx$$ Therefore your integral is $I^{(q)}(p)$ using differentiation under the integral sign. Thus the $q$th derivative of $\frac{1}{p+1}$ is what you are after.

Answer 3

For $\mathbb{R} \ni p>-1$ and $q \in \mathbb{R}$ by setting $x = e^{-t/(p+1)}$ we get

$$\int_0^1 x^{p}(\log x)^q\,dx=\frac{(-1)^q}{(p+1)^{q+1}} \cdot \int_0^\infty e^{-t}t^q \, dt.$$

By definition of the upper incomplete gamma function we get for

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

Using this we get

$$\int_0^1 x^{p}(\log x)^q\,dx = \frac{(-1)^q}{(p+1)^{q+1}} \cdot \Gamma(q+1).$$

Assuming that $p,q \in \mathbb{N}$ we have

$$\int_0^1 x^{p}(\log x)^q\,dx = \frac{(-1)^q}{(p+1)^{q+1}} \cdot q!.$$