$$I=\int_0^1\frac{\arctan^2x}{1+x}\left(\frac{\ln x}{1-x}+\ln(1+x)\right)dx=-\frac{\pi^4}{512}+\frac{3\pi^2}{128}\ln^22+\frac{\pi}{8}G\ln2-\frac{21}{64}\zeta(3)\ln2$$ This integral was proposed to me by a friend, but without the solution. I tried the integration by parts, but too complex. I tried to find a closed form but without result. $$I=\int_0^1\frac{\arctan^2x}{1±x}\ln{x}dx$$
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Martin Sleziak
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user178256
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Your first integral for $I$ is $−0.103$ but your second is $−0.0606$ or $−0.296$. Are you saying that the integrals are the same because they don't seem to be? – Jam Nov 18 '19 at 20:16
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1(+1) for the question. The integral of the type $\displaystyle \int_0^1\frac{\arctan^2x}{1+x}\ln{x}\textrm{d}x$ can be reduced to simpler integrals to calculate by employing a very similar strategy to the one presented in section $3.24$, the generalization part from pages $142$-$145$, from the book (Almost) Impossible Integrals, Sums, and Series. The integral version with minus sign in denominator can be calculated in a similar style by cleverly employing the PVC idea. – user97357329 Nov 18 '19 at 20:40
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In short,with the strategy I suggested you want to reach the value of $$\int_0^{\infty} \frac{\arctan^2(x)\log(x)}{x(1+x)}\textrm{d}x,$$ and then, by symmetry, to extract the value of the desired integral. Grouping strategically the resulting integrals makes everything easier to do, and from what I saw one doesn't need to calculate immediately every integral that arises in the way. That's all. – user97357329 Nov 18 '19 at 21:15
2 Answers
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Partial solution:
$$\mathcal{I}=\int_0^1\frac{\arctan^2(x)\ln(x)}{1-x^2}dx+\int_0^1\frac{\arctan^2(x)\ln(1+x)}{1+x}dx$$
$$\mathcal{I}=\mathcal{I}_1+\mathcal{I}_2$$
Evaluation of $\mathcal{I}_2$:
Let $x\mapsto\frac{1-x}{1+x}$
$$\mathcal{I}_2=\int_0^1\arctan^2(x)\frac{\ln(1+x)}{1+x}dx=\int_0^1\left(\frac{\pi}{4}-\arctan(x)\right)^2\ \frac{\ln(2)-\ln(1+x)}{1+x}dx$$
$$=\ln(2)\underbrace{\int_0^1\frac{\left(\frac{\pi}{4}-\arctan(x)\right)^2}{1+x}dx}_{x\mapsto (1-x)/(1+x)}-\int_0^1\frac{\left(\frac{\pi}{4}-\arctan(x)\right)^2\ln(1+x)}{1+x}dx$$
if we expand the second integral we get a symmetry
$$2\mathcal{I}_2=\ln(2)\underbrace{\int_0^1\frac{\arctan^2(x)}{1+x}dx}_{IBP}-\frac{\pi^2}{16}\int_0^1\frac{\ln(1+x)}{1+x}dx+\frac{\pi}{2}\underbrace{\int_0^1\frac{\arctan(x)\ln(1+x)}{1+x}dx}_{IBP}$$
$$\small{2\mathcal{I}_2=\frac{\pi^2}{16}\ln^2(2)-2\ln(2)\int_0^1\frac{\arctan(x)\ln(1+x)}{1+x^2}dx-\frac{\pi^2}{32}\ln^2(2)+\frac{\pi^2}{16}\ln^2(2)-\frac{\pi}{4}\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx}$$
$$\mathcal{I}_2=\frac{3\pi^2}{64}\ln^2(2)-\ln(2)\int_0^1\frac{\arctan(x)\ln(1+x)}{1+x^2}dx-\frac{\pi}{8}\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx\tag1$$
For the first integral, we use the equality
$$\int_0^1\frac{\arctan(x)}{1+x^2}\ln\left(\frac{1+x^2}{1+x}\right)dx=\frac{3\pi^2}{64}\ln(2)-\frac{\pi}{8}G$$
which can be proved easily through $x\mapsto\frac{1-x}{1+x}.$
allowing us to write
$$\int_0^1\frac{\arctan(x)\ln(1+x)}{1+x^2}dx=\int_0^1\frac{\arctan(x)\ln(1+x^2)}{1+x^2}dx-\frac{3\pi^2}{64}\ln(2)+\frac{\pi}{8}G$$ where
$$\int_0^1\frac{\arctan(x)\ln(1+x^2)}{1+x^2}dx\overset{x=\tan(\theta)}{=}-2\int_0^{\pi/4} \theta\ln(\cos\theta)d\theta=\frac{21}{64}\zeta(3)+\frac{\pi^2}{16}\ln(2)-\frac{\pi}{4}G$$
where the last result follows from using the Fourier series of $$\ln(\cos x)=-\ln(2)-\sum_{n=1}^\infty \frac{(-1)^n\cos(2nx)}{n}$$
then
$$\int_0^1\frac{\arctan(x)\ln(1+x)}{1+x^2}dx=\frac{21}{64}\zeta(3)+\frac{\pi^2}{64}\ln(2)-\frac{\pi}{8}G\tag2$$
As for the second integral, it is already calculated here where some results of harmonic series were used:
$$\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx=4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2G\ln2$$
but I don't think its a good idea to use this closed form as it has an imaginary term, so we better leave the integral as is hoping it cancels out with one of the sub-integrals of $$\mathcal{I}_1=\int_0^1\frac{\arctan^2(x)\ln(x)}{1-x^2}dx.$$
so by plugging $(2)$ in $(1)$ we get
$$\mathcal{I}_2=\frac{\pi^2}{32}\ln^2(2)-\frac{21}{64}\ln(2)\zeta(3)+\frac{\pi}{8}\ln(2)G-\frac{\pi}{8}\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx$$
Martin Sleziak
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Ali Olaikhan
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@clathratus thank you :) I like these integrals but my favorite ones are those that's related to harmonic series. Any idea how to approach $\mathcal{I}_1$? It seems tough. – Ali Olaikhan Nov 18 '19 at 23:26
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Perhaps using $$\arctan(x)^2=\sum_{n\ge0}(-1)^n x^{2n+2}\sum_{k=0}^{n}\frac{1}{(2k+1)(2n-2k+1)}?$$ Other than that I would have no Idea... – clathratus Nov 18 '19 at 23:31
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Good idea .. by the way the Taylor series of $\arctan^2(x)$ has s neat form. – Ali Olaikhan Nov 18 '19 at 23:45
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1We know that $$\text{arctanh}^2(x)=-\frac12\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}x^{2n}$$
replace $x$ by $ix$ and use the fact that $\text{arctanh}(ix)=i\arctan(x)$ we get $$\arctan^2(x)=\frac12\sum_{n=1}^\infty(-1)^n\frac{H_n-2H_{2n}}{n}x^{2n}$$
– Ali Olaikhan Nov 19 '19 at 01:09 -
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@user178256 thank you. Very nice question by the way. I'm still working on a different approach. – Ali Olaikhan Nov 19 '19 at 13:02
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Continuing on @AliShather's argument, let's focus on the integral \begin{align*} I_1&=\int_0^1\frac{\arctan^2 x\ln x}{1-x^2}dx\\ &\overset{\rm IBP}=-\int_0^1\frac{\arctan^2 x\mathop{\rm artanh} x}{x}+2\underbrace{\frac{\arctan x\mathop{\rm artanh} x \ln x}{1+x^2}}_{x\mapsto \frac{1-x}{1+x}}dx\\ &=-\int_0^1\frac{\arctan^2 x\mathop{\rm artanh} x}{x}-2{\frac{\left(\arctan x-\frac{\pi}{4}\right)\mathop{\rm artanh} x \ln x}{1+x^2}}dx\\ &=-\int_0^1\frac{\arctan^2 x\mathop{\rm artanh} x}{x}dx-\frac{\pi}{4}\int_0^1\frac{\mathop{\rm artanh} x\ln x}{1+x^2}dx\\ &=-A-\frac{\pi}{4}B \end{align*} where $\mathop{\rm artanh}x=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$.
By converting into multiple integrals then integrating back, \begin{align*} A&=\int_{[0,1]^4}\frac{x^2}{\left(1-s^2x^2\right)\left(1+t^2x^2\right)\left(1+u^2x^2\right)}dV\\[3px] &=\mbox{*Partial fractions*}\\[3px] &=-\int_0^1\frac{\mathop{\rm artanh}x}{x}\left(2\arctan x-\arctan\frac{1}{x}\right)\arctan\frac{1}{x}\,dx\\ &=-\frac{\pi^4}{64}+\pi\underbrace{\int_0^1\frac{\arctan x \mathop{\rm artanh}x}{x}dx}_{J} \end{align*} where we used the identity $\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}$.
By $x\mapsto\frac{1-x}{1+x}$, $$J=\int_0^1\frac{\left(\arctan x-\frac\pi 4\right)\ln x}{1-x^2}dx=\frac{\pi^3}{32}+\int_0^1\frac{\arctan x\ln x}{1-x^2}dx$$ Then by using IBP directly to J, \begin{align*} J&=-\int_0^1\frac{\arctan x\ln x}{1-x^2}+\frac{\mathop{\rm artanh}x \ln x}{1+x^2}dx\\&=\frac{\pi^3}{64}-\frac{1}{2}\underbrace{\int_0^1\frac{\mathop{\rm artanh}x \ln x}{1+x^2}dx}_B \end{align*}
By here, $$I_1=\frac\pi 4 B=\frac{3\pi^4}{256}-\frac{G\pi\ln 2}{4}-\frac \pi 2\mathfrak{I}\mathop{\rm Li_3}\frac{1+i}{2}+\frac{\pi^2\ln^2 2}{64}$$
Finally using the closed form of $I_2$ by @AliShather and $I=I_1+I_2$ we should get the desired identity, where of course the closed form of B was not necessary and can be replaced by the partial steps used in the evaluation of B.
Martin Sleziak
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