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?
 A: 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.
$$ \begin{align} \int_{0}^{1} x^{n} \, \text{li}(x^{m}) \, dx &= \frac{1}{m} \int_{0}^{1} (u^{1/m})^{n} \, \text{li}(u) \, u^{1/m-1} \, du \\ &= \frac{1}{m} \int_{0}^{1} u^{n/m+1/m-1} \text{li} (u) \ du \\  &= \frac{1}{m} \int_{0}^{1} u^{n/m+1/m-1} \int_{0}^{u} \frac{1}{\log t} \, dt \, du \\ &= \frac{1}{m} \int_{0}^{1} \frac{1}{\log t} \int_{t}^{1} u^{n/m+1/m-1} \, du \, dt \\ &= \frac{1}{n+1} \int_{0}^{1} \frac{1}{\log t} \left(1-t^{n/m+1/m} \right) \, dt \\ &= \frac{1}{n+1} \int_{\infty}^{0} \frac{1}{w} \left(1-e^{-w(n/m+1/m)} \right) \, e^{-w} \, dw \\ &= -\frac{1}{n+1} \int_{0}^{\infty} \frac{e^{-w}-e^{-w(n/m+1/m+1)}}{w} \, dw \\ &= - \frac{1}{n+1} \log \left(\frac{n}{m} + \frac{1}{m}+1 \right) \\ &= - \frac{1}{n+1} \log \left(\frac{m+n+1}{m} \right) \end{align}$$
A: If
$I(n, m)
=\int_0^1 x^n \operatorname{li}(x^m)\,dx
$,
setting
$y = x^m$,
$x = y^{1/m}$
so
$dx = \frac1{m}y^{1/m-1} dy
$.
Therefore
$\begin{array}\\
I(n, m)
&=\int_0^1 y^{n/m} \operatorname{li}(y)\,\frac1{m}y^{1/m-1}dy\\
&=\frac1{m}\int_0^1 y^{n/m+1/m-1} \operatorname{li}(y)\,dy\\
&=\frac1{m}\int_0^1 y^{(n+1-m)/m} \operatorname{li}(y)\,dy\\
&= \frac1{m}I(\frac{n+1-m}{m}, 1)
\end{array}
$
For a different transformation,
set
$y = x^{n+1}
$,
so
$x = y^{1/(n+1)}
$
and
$dx
=\frac1{n+1}y^{1/(n+1)-1} dy
$.
Then
$\begin{array}\\
I(n, m)
&=\int_0^1  y^{n/(n+1)}\operatorname{li}(y^{m/(n+1)})\,\frac1{n+1}y^{1/(n+1)-1} dy\\
&=\frac1{n+1}\int_0^1  \operatorname{li}(y^{m/(n+1)}) dy\\
&= \frac1{n+1}I(0, \frac{m}{n+1})
\end{array}
$
According to Wolfy,
$\int x^r \operatorname{li}(x) dx 
= \frac{\operatorname{li}(x) x^{r+1}-Ei((r+2) log(x))}{r+1}
$
and
$\int \operatorname{li}(x^r) dx 
= x \operatorname{li}(x^r)-Ei((r+1) log(x))
$.
These might be a good start.
A: $$\int_{0}^{1}x^n\text{li}(x^m)\,dx = \frac{1}{m}\int_{0}^{1}z^{(n+1)/m-1}\text{li}(z)\,dz $$
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.
