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

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.

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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.

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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 $u$.

Answer 1

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 $$\int_0^1 \left(\frac{\pi}{4}+\arctan(t)\right)^2\frac{\log(t)}{1-t^2}\textrm{d}t=\underbrace{-\int_0^1 \frac{\displaystyle\left(\pi/2-\arctan(t)\right)^2 \operatorname{arctanh}(t)}{t}\textrm{d}t}_{\displaystyle I}.$$

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 $$\small I=-2\int_0^1 \left(\int_0^1\frac{x\operatorname{arctanh}(x)\operatorname{arctanh}(t)}{t(t^2+x^2)}\textrm{d}x\right)\textrm{d}t=-2\int_0^1 \left(\int_0^1\frac{x\operatorname{arctanh}(x)\operatorname{arctanh}(t)}{t(t^2+x^2)}\textrm{d}t\right)\textrm{d}x$$ $$=-2\left(\underbrace{\int_0^1\frac{\operatorname{arctanh}(t)}{t}\textrm{d}t}_{\displaystyle \pi^2/8}\right)^2+\underbrace{2\int_0^1 \left(\int_0^1\frac{t\operatorname{arctanh}(t)\operatorname{arctanh}(x)}{x(x^2+t^2)}\textrm{d}t\right)\textrm{d}x}_{\displaystyle -I},$$

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.

Answer 2

This not an answer

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.