# A reason for $64\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\pi^4$ ...

Question: How to show the relation $$J:=\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\frac 1{64}\pi^4$$ (using a "minimal industry" of relations, possibly remaining inside the real analysis)?

So i have found a solution to the problem, it is part of my solution for math.stackexchange.com - questions - 3854736, but not a satisfactory solution. "There should be more", explaining why there is a "clean result" for the integral.

Here, i am not strictly interested in a computational approach. I just want to share this with the community in these days of isolation. Any idea to attack this, or a related integral involving "three log factors" is welcome. (Well, the $$\arctan$$ is a sort of $$\log$$ in a sense that i don't want to define closer, see below.) Computations may be safely done "modulo integrals involving two or one log factor". But an illuminating, short way to show the above formula for $$J$$ would be wonderful.

Motivation: The above relation appeared as i tried to solve the integral posted at the above link:

Calculate $$\displaystyle\int_0^{2\pi} x^2\; \cos x \cdot\operatorname{Li}_2(\cos x)\; dx$$ .

After several simplifications and substitutions, it turns out that the above integral is related to integrals of the shape

• $$\int_0^1\log t\; R(t)\; dt$$ , and
• $$\int_0^1\arctan t\cdot \log t\; R(t)\; dt$$ , and
• $$\int_0^1\arctan^2 t\cdot \log t\; R(t)\; dt$$ ,
• and "similar" expressions.

Here $$R$$ is in each case a (rather simple) rational function. (The more log and/or arctangent factors, the higher the computational complexity.)

I could compute more or less algorithmically most of the the needed integrals to solve the linked problem, all of them but the integral $$K=\int_0^1\arctan^2 t\cdot\log t\cdot\frac2{1-t^2}\; dt\ ,$$ which turned out to be very hard to attack with the methods of real analysis. Computing this integral is more or less equivalent to computing $$J$$, and the question wants $$J$$ instead, since we have a "clean formula", so that some speculation about a "clever substitution" may be accepted.

My solution (for $$K$$) works in complex analysis, the first step is to write $$\int_0^1 =\int_0^i+\int_i^1\ ,$$ then parametrize the first integral using a linear path, the second one using a path on the unit circle.

Some comments: I will say some more words, because the situation is rich in coincidences. Since a numerical evidence is the simplest and shortes way to present (instead of showing how to show), i will use this method to at least list the coincidences. Many equalities below are "equivalent" (modulo computation of integrals of lower complexity) to the formula for $$J$$.

• First of all, a numerical experiment using pari/gp delivers some connection between $$K$$ and a "cousin" of $$J$$:

  ? 2 * intnum( t=0, 1, atan(t)^2 * log(t) / (1-t^2) )
%88 = -0.357038604620289042902893412499686912781214141574556097366337
? real(intnum( t=0, I, (pi/4 - atan(t))^2 * log(t) / (1-t^2) ))
%89 = -0.357038604620289042902893412499686912781214141574556097366337
? intnum( t=0, 1, (pi/4 - atan(t))^2 * log(t) / (1-t^2) )
%90 = -0.357038604620289042902893412499686912781214141574556097366337


In words: \begin{aligned} K &= \int_0^1\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt \\ &= \Re \int_0^i\left(\frac \pi 4-\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt \\ &= \int_0^1\left(\frac \pi 4-\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt \ . \end{aligned} Note the integration margins. What happens if we take the integral on $$[0,i]$$ instead of $$[0,1]$$ in the $$K$$-integral? Numerically:

    ? 2 * real(intnum( t=0, i, atan(t)^2 * log(t) / (1-t^2) ))
%98 = 1.52201704740628808181938019826101736327699352613570971392919
? pi^4/64
%99 = 1.52201704740628808181938019826101736327699352613570971392919


In words: \begin{aligned} K^* &:= \Re\int_0^i\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt \\ &=\frac 1{64}\pi^4 \\ &= -\int_0^1\left(\frac \pi 4+\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt \\ &=-J\ . \end{aligned}

(These observations were leading to the formula for $$K$$ in loc. cit. .)

• One idea is to use partial integration in $$J$$ or $$K$$. Well, we have for $$K$$: \begin{aligned} K &= \int_0^1\arctan^2 t\cdot\log t\;\left(-\log\frac {1-t}{1+t}\right)'\; dt \\ &= \underbrace{\int_0^1\arctan^2 t\cdot\frac 1t\cdot \log\frac {1-t}{1+t}\; dt}_{=2K\text{ (why?)}} \\ &\qquad\qquad+ \underbrace{ \int_0^1 2\arctan t\cdot\frac 1{1+t^2}\cdot \log t\cdot \log\frac {1-t}{1+t}\; dt }_{=-K\text{ (why?)}} \ . \end{aligned}

• Note that $$\arctan$$ is related to the logarithm (over $$\Bbb C$$), we have the relation (around $$0$$) $$\arctan t=\frac 1{2i}\log\frac {1+it}{1-it}\ .$$ The substitution $$t=\frac{1-u}{1+u}$$ and the formula for $$\tan(\arctan 1-\arctan u)$$ are giving: \begin{aligned} K &= \int_0^1\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt \\ &=\int_0^1 \left(\frac\pi2-\arctan u\right)^2\cdot\log\frac {1-u}{1+u}\cdot \frac {du}u\ . \\ &=\int_1^\infty \left(\frac\pi2-\arctan u\right)^2\cdot\log\frac {u-1}{1+u}\cdot \frac {du}u\ . \end{aligned} (Write $$\log t=\frac 12\log t^2$$ to have the same expression under the integral on $$(0,1)$$ and on $$(1,\infty)$$.)

• Note the fact that the factor $$\frac 2{1-t^2}$$ is not "random". It is the right one to make things feasible. It is the derivative of $$\displaystyle -\log\frac{1-t}{1+t}$$, and plugging in $$t=iu$$ into $$\displaystyle \log\frac{1-t}{1+t}$$ leads to an expression related to $$\arctan u$$. And conversely, $$\arctan(iu)$$ is related to such a logarithmic expression in $$u$$.

• When certain products of $\arctan, \log$ yield an integral with simple answer, there need not be an immediate explanation for its simplicity. For example, easier, harder, harder. Nevertheless, there is a systematic and algorithmic way to establish such integral, look at here for details. Nov 14, 2020 at 9:37
• $$\underbrace{\int_0^12\arctan^2 t\cdot\frac 1{t+t^2}\cdot \log t\cdot \log\frac {1-t}{1+t}\; dt}_{=-K\text{ (why?)}}$$ is not correct $$\underbrace{\int_0^12\arctan t\cdot\frac 1{1+t^2}\cdot \log t\cdot \log\frac {1-t}{1+t}\; dt}_{=-K\text{ (why?)}}$$ Nov 25, 2020 at 11:02
• @user178256 Yes, thanks for pointing to the error! Nov 26, 2020 at 12:51

A Mind-blowing Solution by Cornel Ioan Valean

It is this way! Again, it is one of those integrals where it is incredibly difficult to imagine that a simple solution is possible. And yet, it is possible such a simple solution!

First, let the variable change $$\displaystyle t\mapsto \frac{1-t}{1+t}$$ in the main integral that becomes $$\int_0^1 \left(\frac{\pi}{4}+\arctan(t)\right)^2\frac{\log(t)}{1-t^2}\textrm{d}t=\underbrace{-\int_0^1 \frac{\displaystyle\left(\pi/2-\arctan(t)\right)^2 \operatorname{arctanh}(t)}{t}\textrm{d}t}_I}.$$

Now, the magical step is to consider the following result $$\displaystyle\frac{(\pi/2-\arctan(t))^2}{t}=2\int_0^1 \frac{x\operatorname{arctanh}(x)}{t(t^2+x^2)}\textrm{d}x$$ exploited in Cornel's answer here, and then we have that $$\small I=-2\int_0^1 \left(\int_0^1\frac{x\operatorname{arctanh}(x)\operatorname{arctanh}(t)}{t(t^2+x^2)}\textrm{d}x\right)\textrm{d}t=-2\int_0^1 \left(\int_0^1\frac{x\operatorname{arctanh}(x)\operatorname{arctanh}(t)}{t(t^2+x^2)}\textrm{d}t\right)\textrm{d}x$$ $$=-2\left(\underbrace{\int_0^1\frac{\operatorname{arctanh}(t)}{t}\textrm{d}t}_\pi^2/8}\right)^2+\underbrace{2\int_0^1 \left(\int_0^1\frac{t\operatorname{arctanh}(t)\operatorname{arctanh}(x)}{x(x^2+t^2)}\textrm{d}t\right)\textrm{d}x}_-I},$$

whence, by symmetry reasons, we obtain the desired result $$\color{blue}{I=\int_0^1 \left(\frac{\pi}{4}+\arctan (t)\right)^2\frac{\log(t)}{1-t^2}\textrm{d}t=-\frac{\pi^4}{64}.}$$

End of story

Many such powerful auxiliary tools will be presented in the sequel of (Almost) Impossible Integrals, Sums, and Series.

A simple generalization of the main double integral

Based on the symmetry ideas above, there is room for generalizations of the type $$\color{red}{\int_a^b \left(\int_a^b\frac{x f(x)f(y)}{y(y^2+x^2)}\textrm{d}x\right)\textrm{d}y=\frac{1}{2}\left(\int_a^b\frac{f(x)}{x}\textrm{d}x\right)^2},$$ which may be a very strong result for deriving more difficult integrals.

A STUNNING GENERALIZATION (with the $$n$$th power of the inverse hyperbolic tangent) $$\color{purple}{\int_0^1 \frac{\operatorname{arctanh}^n(x)}{x}\Re\biggr\{\operatorname{Li}_{n+1}\left(\frac{1-x^2}{1+x^2}+i\frac{2 x}{1+x^2}\right)\biggr \}\textrm{d}x}$$ $$\color{purple}{=2^{-3 n-1} \left(2^{n+1}-1\right)n!\zeta^2(n+1)},$$ where it is used $$\color{red}{\text{the generalization in red}}$$ and the generalization at the point $$i)$$ in Cornel's post here - more similar results may be obtained with other of those generalizations, and some of them will appear in the sequel of (Almost) Impossible Integrals, Sums, and Series.

• Unique as always (+1) Mar 30, 2022 at 0:26

Considering that $$\big[\tan ^{-1}(t)\big]^2=\sum_{n=1}^\infty (-1)^{n+1}\,a_n\,t^{2n}$$ where $$a_n=\frac 1n\sum_{k=1}^n\frac 1 {2k-1}=\frac{H_{n-\frac{1}{2}}+2 \log (2)}{2 n}$$and using the fact that $$\int_0^1\frac {t^{2n}}{1-t^2}\log(t)=-\frac{1}{4} \zeta \left(2,\frac{2n+1}{2}\right)$$ $$K=2\int_0^1\big[\tan ^{-1}(t)\big]^2\,\frac{\log (t)}{1-t^2}\,dt$$ $$K=\frac 14 \sum_{n=1}^\infty (-1)^n \frac{\zeta \left(2,n+\frac{1}{2}\right) \left(H_{n-\frac{1}{2}}+2 \log (2)\right)}{n}$$ which converges extremely slowly.