TITLE: Closed-form of $\int_0^1 x^n \operatorname{li}(x^m)\,dx$
QUESTION [9 upvotes]: 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?

REPLY [5 votes]: You could also use the integral representation of the logarithmic integral and then switch the order of integration. You'll end up with the same Frullani integral.
$$ \begin{align} \int_{0}^{1} x^{n} \, \text{li}(x^{m}) \, dx &= \frac{1}{m} \int_{0}^{1} (u^{1/m})^{n} \, \text{li}(u) \, u^{1/m-1} \, du \\ &= \frac{1}{m} \int_{0}^{1} u^{n/m+1/m-1} \text{li} (u) \ du \\  &= \frac{1}{m} \int_{0}^{1} u^{n/m+1/m-1} \int_{0}^{u} \frac{1}{\log t} \, dt \, du \\ &= \frac{1}{m} \int_{0}^{1} \frac{1}{\log t} \int_{t}^{1} u^{n/m+1/m-1} \, du \, dt \\ &= \frac{1}{n+1} \int_{0}^{1} \frac{1}{\log t} \left(1-t^{n/m+1/m} \right) \, dt \\ &= \frac{1}{n+1} \int_{\infty}^{0} \frac{1}{w} \left(1-e^{-w(n/m+1/m)} \right) \, e^{-w} \, dw \\ &= -\frac{1}{n+1} \int_{0}^{\infty} \frac{e^{-w}-e^{-w(n/m+1/m+1)}}{w} \, dw \\ &= - \frac{1}{n+1} \log \left(\frac{n}{m} + \frac{1}{m}+1 \right) \\ &= - \frac{1}{n+1} \log \left(\frac{m+n+1}{m} \right) \end{align}$$