Empirically, i have obtained the following value: \begin{align}K&=\int_0^1 \frac{\arctan x\ln^2 x}{1+x^2}\,dx\\ &=\frac{151}{11520}\pi^4-\frac{1}{24}\ln^4 2-\text{Li}_4\left(\frac{1}{2}\right)+\frac{1}{24}\pi^2\ln^2 2-\frac{7}{8}\zeta(3)\ln 2\end{align}

How to prove this?

My attempt:
Observe: \begin{align}K&=\int_0^1 \int_0^1\frac{x\ln^2 x}{(1+x^2)(1+t^2x^2)}\,dt\,dx\\ \end{align} On the other hand,

\begin{align}K&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^2 t}{1+t^2}\,dt\right)\arctan x\right]_0^1-\int_0^1 \int_0^1\frac{x\ln(tx)^2}{(1+x^2)(1+t^2x^2)}\,dt\,dx\\ &=\frac{\pi^4}{64}-K-\int_0^1\int_0^1 \frac{x\ln^2 t}{(1+x^2)(1+t^2x^2)}\,dt\,dx-2\int_0^1\int_0^1 \frac{x\ln t\ln x}{(1+x^2)(1+t^2x^2)}\,dt\,dx\\ \end{align} Moreover, one can prove: \begin{align}\int_0^1 \int_0^1\frac{x\ln^2 t}{(1+x^2)(1+t^2x^2)}\,dt\,dx&=\frac{1}{64}\pi^4-\text{G}^2\end{align}

Unfortunately, $\displaystyle U= \int_0^1\int_0^1 \frac{x\ln t\ln x}{(1+x^2)(1+t^2x^2)}\,dt\,dx$ seems not easier to compute than $K$

Edit: \begin{align}U&=\int_0^1\int_0^1 \frac{x\ln t\ln x}{(1-t^2)(1+x^2)}\,dt\,dx -\int_0^1\int_0^1 \frac{xt^2\ln t\ln x}{(1-t^2)(1+t^2x^2)}\,dt\,dx\\ &=\frac{1}{384}\pi^4-\int_0^1\int_0^1 \frac{xt^2\ln t\ln(tx)}{(1-t^2)(1+t^2x^2)}\,dt\,dx+\int_0^1\int_0^1 \frac{xt^2\ln^2 t}{(1-t^2)(1+t^2x^2)}\,dt\,dx\\ \end{align} The last one is doable and, \begin{align}V&=\int_0^1\int_0^1 \frac{xt^2\ln t\ln(tx)}{(1-t^2)(1+t^2x^2)}\,dt\,dx\\ &=\int_0^1 \frac{\ln t}{1-t^2}\left(\int_0^t \frac{u\ln u}{1+u^2}\,du\right)\,dt\\ &=\frac{1}{4}\int_0^1 \frac{\ln t}{1-t^2}\left(\int_0^{t^2} \frac{\ln u}{1+u}\,du\right)\,dt\\ \end{align}


Since for $t\neq 1$, $\displaystyle \frac{1}{1-t^2}=\frac{1}{2}\times \frac{2t}{1-t^2}+\frac{1}{1+t}$ then,

\begin{align}V&=\frac{1}{4}\int_0^1 \left(\frac{1}{2}\times \frac{2t}{1-t^2}+\frac{1}{1+t}\right)\ln t\left(\int_0^{t^2} \frac{\ln u}{1+u} \,du\right)\,dt\\ &=\frac{1}{4}\int_0^1 \frac{\ln t}{1+t}\left(\int_0^{t^2} \frac{\ln u}{1+u}\,du\right)\,dt+\frac{1}{16}\int_0^1 \frac{\ln t}{1-t}\left(\int_0^t \frac{\ln u}{1+u}\,du\right)\,dt \end{align}

  • $\begingroup$ a related integral is here math.stackexchange.com/questions/3260176/… $\endgroup$ Commented Dec 30, 2019 at 21:57
  • $\begingroup$ The integral is $I_2$ from here. $$\int_0^1 \frac{\arctan x\ln^2 x}{1+x^2}dx=\frac{1}{4} \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^3} + \frac{1}{8} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^3}$$ Both series appears here at $(650)$ and $(663)$ and most likely both were posted on MSE too. $\endgroup$
    – Zacky
    Commented Dec 30, 2019 at 22:03
  • $\begingroup$ I see that "harmonic number" is related enough to be tagged unless you want your integral evaluated without using harmonic series. $\endgroup$ Commented Dec 30, 2019 at 22:43
  • 2
    $\begingroup$ Ali Shather: tag added. Anyway, a solution with only the use of integrals would be nice. $\endgroup$
    – FDP
    Commented Dec 30, 2019 at 23:26
  • $\begingroup$ Yes @FDP I'm thinking about a different way because I dont like using the generation function . $\endgroup$ Commented Dec 31, 2019 at 2:24

2 Answers 2


Different approach

From here we have

$$\frac{\arctan x}{1+x^2}=\frac12\sum_{n=1}^\infty(-1)^n\left(H_n-2H_{2n}\right)x^{2n-1}$$

multiply both sides by $\ln^2x$ then integrate from $x=0$ to $x=1$ to get

$$\int_0^1\frac{\arctan x\ln^2x}{1+x^2}dx=\frac12\sum_{n=1}^\infty(-1)^n(H_n-2H_{2n})\int_0^1x^{2n-1}\ln^2x\ dx$$ $$=\sum_{n=1}^\infty(-1)^n\frac{H_n-2H_{2n}}{(2n)^3}=\frac18\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}-2\sum_{n=1}^\infty(-1)^n\frac{H_{2n}}{(2n)^3}$$


where $\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}$$=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$

and $\sum_{n=1}^\infty(i)^n\frac{H_n}{n^3}$ can be evaluated using the generating function

\begin{align} \sum_{n=1}^\infty\frac{H_n}{n^3}y^n&=\operatorname{Li}_4\left(\frac{y}{y-1}\right)-\frac12\operatorname{Li}_2^2\left(\frac{y}{y-1}\right)+2\operatorname{Li}_4(y)-\operatorname{Li}_4(1-y)-\ln(1-y)\operatorname{Li}_3(y)\\ &\quad +\frac12\ln^2(1-y)\operatorname{Li}_2(y)+\frac12\operatorname{Li}_2^2(y)+\frac16\ln^4(1-y)-\frac16\ln y\ln^3(1-y)\\ &\quad+\frac12\zeta(2)\ln^2(1-y)+\zeta(3)\ln(1-y)+\zeta(4) \end{align}

now set $y=i$ and consider the real part.


A slightly different way through the harmonic sum forest, but which quickly leads to @Ali Shather harmonic sum result, is to integrate by parts first. Doing so, we have \begin{align} I &= \int_0^1 \frac{1}{1 + x^2} \cdot \log^2 x \arctan x \, dx\\ &= -\int_0^1 \frac{\log^2 x \arctan x}{1 + x^2} \, dx - 2 \int_0^1 \frac{\log x \arctan^2 x}{x}, \end{align} or $$I = -\int_0^1 \frac{\log x \arctan^2 x}{x} \, dx.$$ Next, making use of the Cauchy product for $\arctan^2 x$, namely, $$\arctan^2 x = \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n} \left (H_{2n} - \frac{1}{2} H_n \right ) x^{2n},$$ leads to \begin{align} I &= \sum_{n = 1}^\infty \frac{(-1)^n}{n} \left (H_{2n} - \frac{1}{2} H_n \right ) \int_0^1 x^{2n - 1} \log x \, dx\\ &= \frac{1}{8} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^3} - 2 \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{(2n)^3}, \end{align}
the same point Ali Shather arrives at in his earlier solution to the problem.

Granted, a solution that only makes use of integrals would be much nicer.

  • $\begingroup$ nice omegadot. by the way, have you tried evaluating $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^3}$ without using the tedious calculations of taking the real part of the polylogarithms? $\endgroup$ Commented Dec 31, 2019 at 9:03
  • $\begingroup$ @Ali Shather No, I haven't found a way to evaluate that particular Euler sum that does not involve finding the real part of a bunch of polylogarithms. Having an alternative method at hand that is simpler and more direct would be worth knowing though. $\endgroup$
    – omegadot
    Commented Jan 1, 2020 at 6:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .