Find $\int _0^1\frac{12\arctan ^2 x\ln (\frac{(1-x)^2}{1+x^2})-\ln ^3(\frac{(1-x)^2}{1+x^2})}{x}\:dx$

Question

I want to find and prove that: $$\int _0^1\frac{12\arctan ^2\left(x\right)\ln \left(\frac{\left(1-x\right)^2}{1+x^2}\right)-\ln ^3\left(\frac{\left(1-x\right)^2}{1+x^2}\right)}{x}\:dx=\frac{9 \pi ^4}{16}$$ I tried to split the integral: $$=12\int _0^1\frac{\arctan ^2\left(x\right)\ln \left(\frac{\left(1-x\right)^2}{1+x^2}\right)}{x}\:dx+12\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x^2\right)}{x}\:dx$$ $$-6\int _0^1\frac{\ln \left(1-x\right)\ln ^2\left(1+x^2\right)}{x}\:dx+\int _0^1\frac{\ln ^3\left(1+x^2\right)}{x}\:dx-8\int _0^1\frac{\ln ^3\left(1-x\right)}{x}\:dx$$ But the first $3$ integrals are very tough, expanding some terms yield very complicated series. I also attempted to integrate by parts but it was rather messy and I'd prefer not to put it here. I have faith that there must be a trick for finding this one because of its closed forms simplicity.

Note $2$: I found this integral in a certain mathematics group, I've ask for hints but received none.

Answer 1

Integrals of such form (i.e. combination of $\arctan x, \log x, \log(1\pm x), \log(1+x^2$)) have systematic way of calculation. If you randomly write down such an integral, there is likely no slick way.

However, for this particular one, there is something behind. Writing $a=\arctan x, b=\log(\frac{(1-x)^2}{1+x^2})$, I will simultaneously prove OP's question as well as the following generalizations $$\begin{aligned}I_3 &= \color{red}{\int_0^1 \frac{12a^2b-b^3}{x} dx = \frac{9\pi^4}{16}} \\ I_5 &= \int_0^1 \frac{-80 a^4 b+40 a^2 b^3-b^5}{x} dx = \frac{33 \pi ^6}{8} \\ I_7 &= \int_0^1 \frac{448 a^6 b-560 a^4 b^3+84 a^2 b^5-b^7}{x} dx = \frac{2193 \pi ^8}{32} \end{aligned}$$

Here numerator of $I_n$ is real part of $(2ai-b)^n$. Write $$I_n=\int_0^1 \frac{\Re[(2ai-b)^n]}{x}dx = -2^n \int_0^1 \frac{\Re[(\log(1-x)-\log(1+ix))^n]}{x}dx $$ then for $n$ odd, $\frac{I_n}{\pi^{n+1}} \in \mathbb{Q}$. More precisely, if $n=2m+1$,

$$\tag{*}I_n = 2^{2n-2}3\pi^{n+1}(-1)^{m+1} \sum_{r=0}^m \binom{n}{2r+1} \left(\frac{3}{8}\right)^{2r} \frac{1}{n-2r}B_{n-2r}\left(\frac{3}{8}\right)$$

here $B_n(x)$ is Bernoulli polynomial, from which above cited values of $I_n$ follows.

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Proof of $(*)$: Integrate $\frac{(\log(1-z)-\log(1+iz))^n}{z}$ around quarter circle in the first quadrant, the integral from $0$ to $i$ is $$\begin{aligned}2^n \int_0^i \frac{(\log(1-x)-\log(1+ix))^n}{x}dx &= 2^n \int_0^1 \frac{(\log(1-ix)-\log(1-x))^n}{x}dx \\ &= -2^n \int_0^1 \frac{(\log(1-x)-\log(1-ix))^n}{x}dx \end{aligned}$$ so the real part of $\int_0^i$ is $-I_n$. Hence, if $C$ is the quarter circle, $$\begin{aligned} 2I_n &= 2^n\Re \int_C \frac{(\log(1-z)-\log(1+iz))^n}{z}dz \\ &= -2^n\Im \int_0^{\pi/2} (\log(1-e^{ix})-\log(1+ie^{ix}))^n dx \end{aligned}$$ Now $$\log(1-e^{ix})-\log(1+ie^{ix}) = \log\left(\frac{\sin(x/2)}{\sin(\pi/4-x/2)}\right)-\frac{3i\pi}{4}$$ Hence if we write $n=2m+1$, $$I_n = 2^n \sum_{r=0}^m \binom{2m+1}{2r+1} (-1)^r\left(\frac{3\pi}{4}\right)^{2r+1}\int_0^{\pi/4} \log^{2(m-r)}\left(\frac{\sin x}{\sin(\pi/4-x)}\right) dx$$ As a lesser known fact, integrals like those in the summand are always rational multiple of $\pi^{2(m-r)+1}$: $$\int_0^{\pi/4}\log^{2n}\left(\frac{\sin x}{\sin(\pi/4-x)}\right) dx = \pi^{2n+1}\frac{(-1)^{n+1}2^{2n+1}}{2n+1}B_{2n+1}\left(\frac{3}{8}\right)$$ QED.

Answer 2

This is not an answer.

I had a look at $$\int _0^1\frac{\arctan ^2\left(x\right)\log \left(\frac{\left(1-x\right)^2}{1+x^2}\right)}{x}\,dx$$ and used $$\log \left(\frac{\left(1-x\right)^2}{1+x^2}\right)=-4\sum_{n=1}^\infty \frac{ \sin ^2\left(\frac{n\pi }{4}\right)}{n}\,x^n$$ I have not been able to compute $$I_n=\int_0^1 x^{n-1}\,\arctan ^2\left(x\right)\,dx$$ for general $n$ but they look interesting $$\left( \begin{array}{cc} n & I_n \\ 1 & -C+\frac{\pi ^2}{16}+\frac{1}{4} \pi \log (2) \\ 2 & -\frac{\pi }{4}+\frac{\pi ^2}{16}+\frac{\log (2)}{2} \\ 3 & \frac{1}{3}+\frac{C}{3}-\frac{\pi }{6}+\frac{\pi ^2}{48}-\frac{1}{12} \pi \log (2) \\ 4 & \frac{1}{12}+\frac{\pi }{12}-\frac{\log (2)}{3} \\ 5 & -\frac{4}{15}-\frac{C}{5}+\frac{\pi }{10}+\frac{\pi ^2}{80}+\frac{1}{20} \pi \log (2) \\ 6 & -\frac{13}{180}-\frac{13 \pi }{180}+\frac{\pi ^2}{48}+\frac{23 \log (2)}{90} \\ 7 & \frac{73}{315}+\frac{C}{7}-\frac{2 \pi }{21}+\frac{\pi ^2}{112}-\frac{1}{28} \pi \log (2) \\ 8 & \frac{29}{420}+\frac{19 \pi }{420}-\frac{22 \log (2)}{105} \end{array} \right)$$ Hard to find a pattern.