# Interesting Logarithmic Integral: $\int_{0}^{1} \frac{\ln^2 x \ln^2(1+x)}{x} \;dx$ [closed]

Other than numerical approximation, how can we calculate the closed form of this integral?

$$\int_{0}^{1} \frac{\ln^2 x \ln^2(1+x)}{x} \;dx$$

• Here is the value to 30 decimal places for those that are interested: 0.195745588199876253442970994564. – Cameron Williams Aug 3 '16 at 20:14
• There's no way ? :/ I have decomposed into a series then stuck on them. – Aditya Narayan Sharma Aug 3 '16 at 20:15
• I believe a closed form exists ... Have you tried expanding $\ln^2(1+x)$ in Taylor and then get into Euler sums? – Tolaso Aug 3 '16 at 20:18
• I think it can be done via a parametric differentiation technique but.. at the end of the day, it'll still be a solution in terms of special functions (i.e. an integral or series). – Cameron Williams Aug 3 '16 at 20:18
• $$I=2 \zeta(2) \,\zeta(3)- \frac{29}{8} \zeta(5).$$ Source: this post. – nospoon Aug 3 '16 at 20:27

OK as I suggested in a comment here is an approach.Let us begin with the Taylor series of $\ln^2(1+x)$. It is known to be:

$$\ln^2 (1+x) = 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} x^n$$

Then we have successively:

\begin{align*} \int_{0}^{1} \frac{\ln^2 x \ln^2 (1+x)}{x} \, {\rm d}x &= 2 \int_{0}^{1} \frac{\ln^2 x}{x} \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} x^n \, {\rm d}x \\ &=2 \int_{0}^{1} \ln^2 x \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} x^{n-1} \, {\rm d}x\\ &= 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} \int_{0}^{1} x^{n-1} \ln^2 x \, {\rm d}x\\ &= 2 \sum_{k=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n^4}\\ &=2 \sum_{k=2}^{\infty} \frac{(-1)^n \left [ \mathcal{H}_n - \frac{1}{n} \right ]}{n^4} \\ &= 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^4} - 2 \sum_{n=2}^{\infty} \frac{(-1)^n}{n^5} \\ &= 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^4} - 2 - \frac{15 \zeta(5)}{16} \end{align*}

Well for the Euler sum first of all we have the generating function:

\begin{align} \sum^\infty_{n=1}\frac{H_n}{n^3}z^n =&2{\rm Li}_4(z)+{\rm Li}_4\left(\tfrac{z}{z-1}\right)-{\rm Li}_4(1-z)-{\rm Li}_3(z)\ln(1-z)-\frac{1}{2}{\rm Li}_2^2\left(\tfrac{z}{z-1}\right)\\ &+\frac{1}{2}{\rm Li}_2(z)\ln^2(1-z)+\frac{1}{2}{\rm Li}_2^2(z)+\frac{1}{6}\ln^4(1-z)-\frac{1}{6}\ln{z}\ln^3(1-z)\\ &+\frac{\pi^2}{12}\ln^2(1-z)+\zeta(3)\ln(1-z)+\frac{\pi^4}{90} \end{align}

Integrating once and plugging $z=-1$ we get the value of the sum. I am not presenting the full calculatios but it suffices to say that:

$$\sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^4} = \frac{\pi^2 \zeta(3)}{4} -\frac{43 \zeta(5)}{32}+1$$

Thus:

$$\int_{0}^{1} \frac{\ln^2 x \ln^2 (1+x)}{x} \, {\rm d}x = \frac{\pi^2 \zeta(3)}{3} - \frac{29 \zeta(5)}{8}$$

• (+1) Yes exactly, Thanks ! I was stuck at the generating function of third order there. Now I've got it – Aditya Narayan Sharma Aug 4 '16 at 5:39
• @AdityaNarayanSharma As for your alternating Euler sum, here is the result: $$\sum^\infty_{n=1}\frac{(-1)^n \mathcal{H}_n}{n^3}=2{\rm Li}_4 \left(\frac{1}{2}\right)-\frac{11\pi^4}{360}+\frac{7}{4}\zeta(3)\ln{2}-\frac{\pi^2}{12}\ln^2{2}+\frac{1}{12}\ln^4{2}$$ – Tolaso Aug 4 '16 at 6:08