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?

  • $\begingroup$ 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. $\endgroup$ – wltrup Aug 5 '15 at 0:42

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


$\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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.