calculus - Closed-form of $int_0^1 x^n operatorname{li}(x^m),dx$

I've conjectured, that for $n\geq0$ and $m\geq1$ integers

$$\int_0^1 x^n \operatorname{li}(x^m)\,dx \stackrel{?}{=} -\frac{1}{n+1}\ln\left(\frac{m+n+1}{m}\right),$$
where $\operatorname{li}$ is the logarithmic integral.

Although there is a known antiderivative of $x^n \operatorname{li}(x^m)$, the simplification of the expression seems not trivial. I think there are other ways to evaluate this definite integral problem.

How could we prove this identity?

Answer

$$\int_{0}^{1}x^n\text{li}(x^m)\,dx = \frac{1}{m}\int_{0}^{1}z^{(n+1)/m-1}\text{li}(z)\,dz$$
but integration by parts gives:

$$\int_{0}^{1}z^{\alpha-1}\text{li}(z)\,dz = \left.\frac{z^\alpha-1}{\alpha}\text{li}(z)\right|_{0}^{1} - \int_{0}^{1}\frac{z^{\alpha}-1}{\alpha\log z}\,dz$$
and the last integral can be computed through the substitution $z=e^{-t}$ and Frullani's theorem:

$$\int_{0}^{1}\frac{z^{\alpha}-1}{\log(z)}\,dz = \log(\alpha+1)$$
hence:
$$\int_{0}^{1}x^n\text{li}(x^m)\,dx = -\frac{1}{n+1}\,\log\left(\frac{n+1}{m}+1\right)$$
as you claimed.