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How do I prove the equality of the two integrals: $$\int_{0}^{1}{\dfrac{x \arctan(x)}{1+x^4}} \ dx=\int_{0}^{1}{\dfrac{x^3 \text{arctanh}(x)}{1+x^4}} \ dx.$$

This question is based on observation, both the integrals equate to almost $0.211\cdots$

Also the first integral has been evaluated to be $$\dfrac{\pi^2}{32}-\dfrac{\ln^2{\sqrt{2}+1}}{8}$$From the question: Find the value of the integral: $\int_{0}^{1}\dfrac{x}{1+x^4}\arctan(x)\,\mathrm{d}x$

Some of the methods outlined in the answers to the question are too advanced and out of my reach for now

user
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Eater of soap The VIIth
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  • Are you specifically trying to convert from one integral to the other? Or do you just want to show that the integral of $(x\arctan x-x^3\operatorname{artanh}x)/\cdots$ is zero? – user170231 Nov 28 '23 at 14:38
  • @user170231 that was a pretty great idea, as wolfram confirms here – Ninad Munshi Nov 28 '23 at 14:43
  • Very nice, but why is this so? Can someone prove it? I have a proof but it relies heavily on the Basel problem... – Eater of soap The VIIth Nov 28 '23 at 15:00
  • Despite me answering the question you linked, I don't know how to prove the two integrals equal each other off the top of my head. :P You could type out your proof involving the Basel problem so that other people might get some ideas. – Accelerator Nov 28 '23 at 16:18
  • It gets very convoluted and I don't have the skill to write it down but essentially squaring the Dirichlet L-function $\dfrac{\pi}{2\sqrt{2}}$ results in the summation of the squares of the odd numbers and also a residue. Since $\pi^2/8=\pi^2/8$, The residue must be $0$. The residue is also the subtraction of the two integrals that I mentioned here. – Eater of soap The VIIth Nov 28 '23 at 16:45
  • Hello @Accelerator , what I am not getting is how we'll prove that $x\arctan{x}$=$x^3\text{arcth}{x}$? Isn't it like proving $\int_0^{\pi/2} \cos{x} , dx = \int_0^1 dx = 1$ – Lucky Chouhan Nov 28 '23 at 16:57
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    I think you missed something. In fact$$\int_{0}^{i}{\dfrac{x^3 \arctan(x)}{1+x^4}} \ dx\overset{x\to xi}=\int_{0}^{1}{\dfrac{i^3x^3 \arctan(xi)}{1+(xi)^4}} \ idx=i\int_{0}^{1}{\dfrac{x^3 \arctanh(x)}{1+x^4}} \ dx.$$ – xpaul Nov 28 '23 at 17:00
  • @LuckyChouhan That's not what one would have to show. $\int_0^1(f(x)-g(x)),dx=0$ does not imply $f(x)-g(x)=0$. – user170231 Nov 28 '23 at 17:19
  • Oh yeah @user170231 so sorry! So if $\int (f(x)-g(x))dx = 0$ then it means that only area of $f(x)$ and $g(x)$ is equal. – Lucky Chouhan Nov 28 '23 at 18:42

2 Answers2

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If you do not bother evaluating the second integral directly, we could still have elementary approach, let $x\mapsto\frac{1-x}{1+x}$ $$ \begin{aligned} \int_{0}^{1} \frac{x^3}{1+x^4}\operatorname{artanh}x\,\mathrm{d}x &=-\frac1{2} \int_{0}^{1} \frac{x^3}{1+x^4}\ln\left(\frac{1-x}{1+x}\right) \mathrm{d}x\\ &=-\frac1{2} \int_{0}^{1} \frac{(1-x)^3}{(1+6x^2+x^4)(1+x)}\ln x\,\mathrm{d}x\\ &=-\frac1{2} \left(\int_{0}^{1} \frac{\ln x}{1+x}\,\mathrm{d}x - \int_{0}^{1} \frac{3x+x^3}{1+6x^2+x^4}\ln x\,\mathrm{d}x\right) \end{aligned} $$ where integrating by parts $$ \begin{aligned} \int_{0}^{1} \frac{3x+x^3}{1+6x^2+x^4}\ln x\,\mathrm{d}x &=-\frac1{4}\int_{0}^{1} \frac{\ln(1+6x^2+x^4)}{x}\,\mathrm{d}x\\ &=-\frac1{8}\int_{0}^{1} \frac{\ln(1+6x+x^2)}{x}\,\mathrm{d}x \end{aligned} $$ recalling for $a>0$ that $$ \int_{0}^{1} \frac{\ln((x+a)(x+\frac1{a}))}{x}\,\mathrm{d}x = -\left(\operatorname{Li}_2(-a)+\operatorname{Li}_2\left(-\frac1{a}\right)\right) = \frac{\pi^2}{6}+\frac{\ln^2a}{2} $$ plug-in $a=3-2\sqrt2=(\sqrt2-1)^2$ or $a=3+2\sqrt2=(\sqrt2+1)^2$ we have $$ \int_{0}^{1} \frac{x^3}{1+x^4}\operatorname{artanh}x\,\mathrm{d}x = \frac{\pi^2}{32}-\frac{\ln^2(\sqrt2+1)}{8} $$ where the primary part of our integral, namely $$ \int_{0}^{1} \frac{\ln(1+6x+x^2)}{x}\,\mathrm{d}x $$ happen to coincide with the equivalent form of some typical integrals $$ \int_{0}^{1} \frac{x}{1+x^4}\arctan x\,\mathrm{d}x = \frac1{4} \int_{0}^{\pi/2}\arctan(\sin x)\,\mathrm{d}x = \frac1{4} \left(\frac{\pi^2}{8}-\frac{\ln^2(\sqrt2+1)}{2} \right) $$ which you may have more insight by yourself.

Nanayajitzuki
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\begin{align}J&=\int_0^1 \frac{x\arctan x}{1+x^4}dx\\ &=\int_0^1 \frac{x}{1+x^4}\int_0^1 \left(\frac{x}{1+t^2x^2}dt\right)dx\\ &=2\left(\int_0^1 \frac{x^2}{1+x^4}dx\right)\left(\int_0^1 \frac{1}{1+t^4}dt\right)-\int_0^1\frac{t}{1+t^4}\int_0^1 \left(\frac{t}{1+t^2x^2}dx\right)dt\\ &=2\left(\underbrace{\int_0^1 \frac{x^2}{1+x^4}dx}_{=B}\right)\left(\underbrace{\int_0^1 \frac{1}{1+t^4}dt}_{=A}\right)-J\\ J&=\boxed{AB}\\ K&=\int_0^1 \frac{x^3\text{arctanh }x}{1+x^4}dx\\ &=\int_0^1 \frac{x^3}{1+x^4}\left(\int_0^1 \frac{x}{1-t^2x^2}dt\right)dx\\ &=-A^2-B^2+\int_0^1 \frac{1}{t(1+t^4)}\left(\int_0^1 \frac{t}{1-t^2x^2}dx\right)dt\\ &=-A^2-B^2+\underbrace{\int_0^1 \frac{\text{arctanh }t}{t}dt}_{u=\frac{1-t}{1+t}}-K\\ K&=-\frac{A^2}{2}-\frac{B^2}{2}-\frac{1}{2}\underbrace{\int_0^1 \frac{\ln u}{1-u^2}du}_{=-\frac{\pi^2}{8}}\\ &=\boxed{-\frac{A^2}{2}-\frac{B^2}{2}+\frac{\pi^2}{16}} \end{align} On the other hand, \begin{align}2J-2K&=2AB+A^2+B^2-\frac{\pi^2}{8}\\ &=(A+B)^2-\frac{\pi^2}{8}\\ &=\left(\underbrace{\int_0^1 \frac{1+x^2}{1+x^4}dx}_{=Z}\right)^2-\frac{\pi^2}{8}\\ Z&=\int_0^1 \frac{1+\frac{1}{x^2}}{\left(\frac{1}{x}-x\right)^2+2}dx\\ &\overset{u=\frac{1}{x}-x}=\int_0^\infty \frac{1}{2+u^2}du\\ &=\frac{\pi }{2\sqrt{2}}\\ \end{align} Therefore, \begin{align}Z^2&=\frac{\pi^2}{8}\\ 2J-2K&=0\\ &\boxed{J=K} \end{align}

FDP
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