The integral $I(a,b)=\int_0^1x^a\log^b(x) \space dx$

Question

I found a formula for the integral $$I(a,b)=\int_0^1x^a\log^b(x) \space dx$$ By just differentiating the integral $I(t)=\int_0^1x^t\space dx $ under the integral sign b times and evaluating the resultant $b$-th derivative at $x=a$. The formula that I came up with states that $$I(a,b)=\frac{(-1)^bb!}{(a+1)^{b+1}} $$ A quick check on some values of $(a,b)$ on desmos like $(1,1), (3,4) ,(1,2)$ quickly verifies that the formula for the integral is true for at least these values of $(a,b)$. My question is: how do I find all pairs $(a,b)$ such that this formula makes sense?

Answer 1

The right-hand side is only defined when $a \neq -1$ and $b \in \{0, 1, 2, \ldots\}$. Moreover, since $|\log x| \geq 1$ on $(0, \frac{1}{e}]$ we have $$\left\vert\int_0^1 x^a \log^b x \,dx\right\vert = \int_0^1 x^a |\log x|^b \,dx \geq \int_0^\frac{1}{e} x^a |\log x|^b \,dx \geq \int_0^\frac{1}{e} x^a \,dx.$$ In particular, our integral diverges if $a \leq -1$.

For the remaining values of $(a, b)$, namely, $a > -1$ and $b \in \{0, 1, 2, \ldots\}$, we can establish that the formula holds in a different way: Substituting $u := -(1 + a) \log x$ gives that the integral is equal to $$\frac{(-1)^b}{(1 + a)^{1 + b}} \int_0^\infty u^b e^{-u} \,du,$$ but applying integration by parts $b$ times gives that this latter integral has value $b!$, establishing the identity.

Remark More generally, for $a, b \in (-1, \infty)$, using the usual integral representation of the gamma function shows that $$\int_0^1 x^a (-\log x)^b \,dx = \frac{\Gamma(1 + b)}{(1 + a)^{1 + b}} .$$

Answer 2

By the Leibniz integral rule, you can differentiate under the integral sign as long as your function and all the derivatives you are taking are continuous. In your case differentiating only adds additional $\log(x)$, which doesn't make your function discontinuous (at $0$ or anywhere else) as long as $a > 0$. Moreover, since you want to take $b$ derivatives, $b$ is necessarily a non-negative integer for your argument to work. Hence you arive at $a \in \mathbb{R}_{>0}$, $b \in \mathbb{N}_0$.

Using the measure theory statement of that theorem, you can in fact show that it suffices to assume $a > -1$.