# 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?

• Have you tried to get Mathematica to simplify the expression resulting from subtracting the values of the antiderivative at 0 and 1? Surely it should be able to simplify it for you if it can give you the antiderivative. – wltrup Aug 5 '15 at 0:42

## 3 Answers

$$\int_{0}^{1}x^n\text{li}(x^m)\,dx = \frac{1}{m}\int_{0}^{1}z^{(n+1)/m-1}\text{li}(z)\,dz$$ 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.

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}

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.