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I've tried series summation and contour integral, but neither works. Perhaps there's something related to number theory here. The result should be $$\frac{3\pi}{8}\log2-\frac\pi6\log(2+\sqrt{3})$$but I have no idea how to obtain it.

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    Most integrals do not have a nice "closed form". Is there any reason you believe this one should have one? Neither Wolfram Alpha, not Maple is able to produce something nice here, so I would suspect that there is no closed form. – Simon Jan 18 '22 at 15:28
  • Of course a numerical solution (around $0.127...$) can be obtained easily – Simon Jan 18 '22 at 15:29
  • I was informed that the result should be $\frac{3\pi}{8}\log2-\frac\pi6\log(2+\sqrt{3})$, but I wonder how to obtain that. –  Jan 18 '22 at 15:40
  • Interesting, what's the connection with number theory? If you could expand on that in a comment or edit the post it would be nice. – Sarvesh Ravichandran Iyer Jan 18 '22 at 15:41
  • Two observations: Integration by parts helps a bit (you get a rational function times $\log(1+x)$. The other is that $\log(2+\sqrt 3) = \log(2) + \log(\cos(\pi/6 + \sin(\pi/6))$, and $f(x) = \log(\cos(x) + \sin(x))$ differentiates to $\tan(\pi/4 - x)$. – preferred_anon Jan 18 '22 at 16:34
  • I disagree with the latter observation. There may be some computatioin errors. –  Jan 18 '22 at 16:46
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    From where did you get this integral? (as you mentioned that someone informed you about the answer I guess there's a source for this). // Another question is about whether you mean $\arctan (x^3)$ or $\arctan^3 (x)$? (although I tend to believe it's the first one). – Zacky Jan 18 '22 at 17:27
  • A friend of mine asked me. He is a mathematics undergraduate. I don't think it matters much but as you wish. –  Jan 18 '22 at 17:36
  • It very much matters. If the source is not a reliable one, then we should be skeptical of there existing a closed form for the integral. – Angel Jan 18 '22 at 17:43
  • I guess you're right. By the way, I do consider the result is reliable, since there's a given one and it fits well numerically. –  Jan 18 '22 at 17:46
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    Sure, the result is correct and a good start was already given in the comments. You just have to integrate by parts and arrive at: $$I=\frac{\pi}{4}\ln 2 +\int_0^1 \frac{\ln(1+x)}{1+x^2}dx-\int_0^1 \frac{(1+x^2)\ln(1+x)}{1-x^2+x^4}dx$$ The first integral appeared many times here on MSE and it equals to $\frac{\pi}{8}\ln 2$ and the second one can be found here. – Zacky Jan 18 '22 at 17:51
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    Thanks for your guidance! –  Jan 18 '22 at 17:58
  • $$\int_0^1\frac{\arctan(x^3)}{1+x},\mathrm{d}x=\log(2)\arctan(1)-\log(1)\arctan(0)-\int_0^1\frac{3x^2\log(1+x)}{1+x^6},\mathrm{d}x$$ $$=\frac{\pi}4\log(2)-\int_0^1\frac{3x^2\log(1+x)}{1+x^6},\mathrm{d}x$$ is the integration by parts. – Angel Jan 18 '22 at 18:28

1 Answers1

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Firstly, set this integral as $$I=\int_0^1\frac{\arctan\left(x^3\right)}{1+x}\mathrm{~d}x.$$ Then, we have $$ \begin{align} I&=\int_0^1\frac{\arctan\left(x^3\right)}{1+x}\mathrm{~d}x\\ &=\int_0^1\arctan\left(x^3\right)\mathrm{~d}\ln(1+x)\\ &=\frac{\pi\ln2}{4}-3\int_0^1\frac{x^2\ln(1+x)}{1+x^6}\mathrm{~d}x\\ &=\frac{\pi\ln2}{4}+\color{red}{\int_0^1\frac{\ln(1+x)}{1+x^2}\mathrm{~d}x}\\ &\quad-\int_0^1\frac{\left(1+x^2\right)\ln(1+x)}{1-x^2+x^4}\mathrm{~d}x\\ &=\frac{\pi\ln2}{4}+\color{red}{\frac{\pi\ln2}{8}}-\frac{1}{2}\int_0^1\frac{\ln(1+x)}{1-\sqrt{3}x+x^2}\mathrm{~d}x\\ &\quad-\frac{1}{2}\int_0^1\frac{\ln(1+x)}{1+\sqrt{3}x+x^2}\mathrm{~d}x\\ &=\frac{3\pi\ln2}{8}-\frac{1}{2}(J_1+J_2). \end{align} $$ Where the red integral could be calculated as $$ \begin{align} &\quad\underbrace{\int_0^1\frac{\ln(1+x)}{1+x^2}\mathrm{~d}x}_{x\to\frac{1-x}{1+x}}\\ &=\int_0^1\frac{\ln2-\ln(1+x)}{1+x^2}\mathrm{~d}x\\ &=\frac{\ln2}{2}\int_0^1\frac{1}{1+x^2}\mathrm{~d}x\\ &=\frac{\pi\ln2}{8}. \end{align} $$ Next, we might as well let $$ J_1(a)=\int_0^1\frac{\ln(1+ax)}{1-\sqrt{3}x+x^2}\mathrm{~d}x.$$ Where $a\gt-1.$ Then, we get $$ \begin{align} J_1^\prime(a)&=\int_0^1\frac{x}{(1+ax)\left(1-\sqrt{3}x+x^2\right)}\mathrm{~d}x\\ &=\frac{1}{2\left(1+\sqrt{3}a+a^2\right)}\int_0^1\frac{2x-\sqrt{3}}{1-\sqrt{3}x+x^2}\mathrm{~d}x\\ &\quad-\frac{a}{1+\sqrt{3}a+a^2}\int_0^1\frac{1}{1+ax}\mathrm{~d}x\\ &\quad+\frac{2a+\sqrt{3}}{2\left(1+\sqrt{3}a+a^2\right)}\int_0^1\frac{1}{1-\sqrt{3}x+x^2}\mathrm{~d}x\\ &= \left.\frac{1}{2\left(1+\sqrt{3}a+a^2\right)}\ln\left[\frac{1-\sqrt{3}x+x^2}{(1+ax)^2}\right]\right|_0^1\\ &\quad+\frac{2(2a+\sqrt{3}\,)}{1+\sqrt{3}a+a^2}\int_0^1\frac{1}{1+(2x-\sqrt{3}\,)^2}\mathrm{~d}x\\ &=\frac{\ln(2-\sqrt{3}\,)-2\ln(1+a)}{2\left(1+\sqrt{3}a+a^2\right)}\\ &\quad+\frac{2a+\sqrt{3}}{1+\sqrt{3}a+a^2}\left.\arctan(2x-\sqrt{3}\,)\right|_0^1\\ &=\frac{\ln(2-\sqrt{3}\,)}{2\left(1+\sqrt{3}a+a^2\right)}+\frac{5\pi(2a+\sqrt{3}\,)}{12\left(1+\sqrt{3}a+a^2\right)}\\ &\quad-\frac{\ln(1+a)}{1+\sqrt{3}a+a^2}. \end{align} $$ Note that$ J_1(0)=0$, $J_1=J_1(1)$, hence $$ \begin{align} J_1&=J_1(1)=\int_0^1J_1^\prime(a)\mathrm{~d}a+J_1(0)\\ &=\frac{\ln(2-\sqrt{3}\,)}{2}\int_0^1\frac{1}{1+\sqrt{3}a+a^2}\mathrm{~d}a\\ &\quad+\frac{5\pi}{12}\int_0^1\frac{2a+\sqrt{3}}{1+\sqrt{3}a+a^2}\mathrm{~d}a\\ &\quad-\underbrace{\int_0^1\frac{\ln(1+a)}{1+\sqrt{3}a+a^2}\mathrm{~d}a}_{a\to x}\\ &=2\ln(2-\sqrt{3}\,)\int_0^1\frac{1}{1+(2a+\sqrt{3}\,)^2}\mathrm{~d}a\\ &\quad+\frac{5\pi}{12}\ln\left(1+\sqrt{3}a+a^2\right)\left.\right|_0^1\\ &\quad-\int_0^1\frac{\ln(1+x)}{1+\sqrt{3}x+x^2}\mathrm{~d}x\\ &=\frac{\pi\ln(2-\sqrt{3}\,)+5\pi\ln(2+\sqrt{3}\,)}{12}\\ &\quad-\int_0^1\frac{\ln(1+x)}{1+\sqrt{3}x+x^2}\mathrm{~d}x\\ &=\frac{\pi\ln(2+\sqrt{3}\,)}{3}-\int_0^1\frac{\ln(1+x)}{1+\sqrt{3}x+x^2}\mathrm{~d}x. \end{align} $$ After that, note that $$J_1(1)=\int_0^1\frac{\ln(1+x)}{1-\sqrt{3}x+x^2}\mathrm{~d}x$$, $$J_2=\int_0^1\frac{\ln(1+x)}{1+\sqrt{3}x+x^2}\mathrm{~d}x.$$ Thus $$ \begin{align} &\quad\int_0^1\frac{\ln(1+x)}{1-\sqrt{3}x+x^2}\mathrm{~d}x+\int_0^1\frac{\ln(1+x)}{1+\sqrt{3}x+x^2}\mathrm{~d}x\\ &=J_1+J_2=\int_0^1\frac{\left(1+x^2\right)\ln(1+x)}{1-x^2+x^4}\mathrm{~d}x\\ &=\frac{\pi\ln(2+\sqrt{3}\,)}{3}. \end{align} $$ Finally, we obtain $$ \begin{align} I&=\frac{3\pi\ln2}{8}-\frac{1}{2}(J_1+J_2)\\ &=\frac{3\pi\ln2}{8}-\frac{\pi}{6}\ln(2+\sqrt{3}\,). \end{align} $$

Nicholas Jet
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