I am in the process of proving $$I=\int_0^\infty \frac{\arctan x}{x^4+x^2+1}\mathrm{d}x=\frac{\pi^2}{8\sqrt{3}}-\frac23G+\frac\pi{12}\log(2+\sqrt{3})$$ And I have gotten as far as showing that $$2I=\frac{\pi^2}{4\sqrt{3}}+J$$ Where $$J=\int_0^\infty \log\bigg(\frac{x^2-x+1}{x^2+x+1}\bigg)\frac{\mathrm{d}x}{1+x^2}$$ Then we preform $x=\tan u$ to see that $$J=\int_0^{\pi/2}\log\bigg(\frac{2+\sin2x}{2-\sin2x}\bigg)\mathrm dx$$ Which I have been stuck on for the past while. I tried defining $$k(a)=\int_0^{\pi/2}\log(2+\sin2ax)\mathrm dx$$ Which gives $$J=k(1)-k(-1)$$ Then differentiating under the integral: $$k'(a)=2\int_0^{\pi/2}\frac{x\cos2ax}{2+\sin2ax}\mathrm dx$$ We may integrate by parts with $u=x$ to get a differential equation $$ak'(a)+k(a)=\frac\pi2\log(2+\sin\pi a)$$ With initial condition $$k(0)=\frac\pi2\log2$$ And from here I have no idea what to do.

I also tried tangent half angle substitution, but that just gave me the original expression for $J$.

I'm hoping that there is some really easy method that just never occurred to me... Any tips?


As was pointed out in the comments, I could consider $$P(a)=\frac12\int_0^\pi \log(a+\sin x)\mathrm dx\\\Rightarrow P(0)=-\frac\pi2\log2$$ And $$ \begin{align} Q(a)=&\frac12\int_0^\pi \log(a-\sin x)\mathrm dx\\ =&\frac12\int_0^\pi\log[-(-a+\sin x)]\mathrm dx\\ =&\frac12\int_0^\pi\bigg(\log(-1)+\log(-a+\sin x)\bigg)\mathrm dx\\ =&\frac{i\pi}2\int_0^\pi\mathrm{d}x+\frac12\int_0^\pi\log(-a+\sin x)\mathrm dx\\ =&\frac{i\pi^2}2+P(-a) \end{align} $$ Hence $$J=P(2)-Q(2)=P(2)-P(-2)-\frac{i\pi^2}2$$ So now we care about $P(a)$. Differentiating under the integral, we have $$P'(a)=\frac12\int_0^\pi \frac{\mathrm{d}x}{a+\sin x}$$ With a healthy dose of Tangent half angle substitution, $$P'(a)=\int_0^\infty \frac{\mathrm{d}x}{ax^2+2x+a}$$ completing the square, we have $$P'(a)=\int_0^\infty \frac{\mathrm{d}x}{a(x+\frac1a)^2+g}$$ Where $g=a-\frac1a$. With the right trigonometric substitution, $$P'(a)=\frac1{\sqrt{a^2+1}}\int_{x_1}^{\pi/2}\mathrm{d}x$$ Where $x_1=\arctan\frac1{\sqrt{a^2+1}}$. Then using $$\arctan\frac1x=\frac\pi2-\arctan x$$ We have that $$P'(a)=\frac1{\sqrt{a^2+1}}\arctan\sqrt{a^2+1}$$ So we end up with something I don't know how to to deal with (what a surprise) $$P(a)=\int\arctan\sqrt{a^2+1}\frac{\mathrm{d}a}{\sqrt{a^2+1}}$$ Could you help me out with this last one? Thanks.


$$J=\int_0^{\pi/2}\ln\left(\frac{2+\sin2x}{2-\sin2x}\right)\mathrm dx\overset{2x=t}=\frac12 \int_0^\pi \ln\left(\frac{1+\frac12\sin t}{1-\frac12\sin t}\right)\mathrm dt=\int_0^\frac{\pi}{2}\ln\left(\frac{1+\frac12\sin x}{1-\frac12\sin x }\right)\mathrm dx$$ Now let's consider the following integral: $$I(a)=\int_0^\frac{\pi}{2}\ln\left(\frac{1+\sin a\sin x}{1-\sin a\sin x}\right)dx\Rightarrow I'(a)=2\int_0^\frac{\pi}{2} \frac{\sin a\sin x}{1-\sin^2a\sin^2 x}dx$$ $$=\frac{2}{\sin a}\int_0^\frac{\pi}{2} \frac{\sin x}{\cos^2x +\cot^2 a}dx=\frac{2}{\sin a}\arctan\left(x\tan a\right)\bigg|_0^1=\frac{2a}{\sin a}$$ $$I(0)=0 \Rightarrow J=I\left(\frac{\pi}{6}\right)=2\int_0^\frac{\pi}{6}\frac{x}{\sin x}dx$$ $$=2\int_0^{\frac{\pi}{6}} x \left(\ln\left(\tan \frac{x}{2}\right)\right)'dx=2x \ln\left(\tan \frac{x}{2}\right)\bigg|_0^{\frac{\pi}{6}} -2{\int_0^{\frac{\pi}{6}} \ln\left(\tan \frac{x}{2}\right)dx}=$$ $$\overset{\frac{x}{2}=t}=\frac{\pi}{3}\ln(2-\sqrt 3) -4\int_0^\frac{\pi}{12}\ln (\tan t)dt=\frac{\pi}{3}\ln(2-\sqrt 3) +\frac{8}{3}G$$ $G$ is the Catalan's constant and for the last integral see here.

Also note that there's a small mistake. After integrating by parts you should have: $$2I=\frac{\pi^2}{4\sqrt 3}- \int_0^\infty\frac{(x^2-1)\arctan x}{x^4+x^2+1}dx=\frac{\pi^2}{4\sqrt 3}-\frac12\underbrace{\int_0^\infty \ln\bigg(\frac{x^2-x+1}{x^2+x+1}\bigg)\frac{dx}{1+x^2}}_{=J}$$

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    $\begingroup$ Shouldn't you give justification for differentiating under the integral? $\endgroup$ – user21820 Dec 18 '18 at 9:23
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    $\begingroup$ That is just Feynman's trick, or differentiating under the integral sign. I don't understand what I should justify at all. See here for example for more: medium.com/dialogue-and-discourse/… and here: brilliant.org/wiki/integration-tricks/… $\endgroup$ – Zacky Dec 18 '18 at 9:37
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    $\begingroup$ It is not always valid, and the fact that you don't understand the precise conditions under which you can use it means that you shouldn't be using it! Both your sources are mathematically deficit for the same reason. $\endgroup$ – user21820 Dec 18 '18 at 9:49
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    $\begingroup$ Sure. Please read this pdf by Keith Conrad for counter-examples and a full mathematical explanation. $\endgroup$ – user21820 Dec 18 '18 at 9:56
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    $\begingroup$ Legit!! What a neat solution! Thanks Zacky! $\endgroup$ – clathratus Dec 18 '18 at 19:04


I find that the integral has a closed form given by

$$i = \int\limits_0^{\pi/2}\log\bigg(\frac{2+\sin2x}{2-\sin2x}\bigg)\mathrm dx = \frac{1}{3} \left(8 C-\pi \log \left(2+\sqrt{3}\right)\right) \simeq 1.06346\tag{1}$$


$$C = \sum _{k=1}^{\infty } \frac{(-1)^{k+1}}{(2 k-1)^2} \simeq 0.915966$$

is Catalan's constant.

Heuristic derivation

Notice trivially that because of the symmtery of the integrand the integral can be written as twice the integral from $0$ to $\frac{\pi}{4}$ which we shall utilize in what follows.

The basic idea is the series expansion

$$\log \left(\frac{1+z}{1-z}\right)=2\tanh ^{-1}(z) = 2 \sum _{k=1}^{\infty } \frac{z^{2 k-1}}{2 k-1},|z|<1 \tag{2}$$

The integral is then to be done over the odd powers of the $\sin$ with the result

$$\int_0^{\frac{\pi }{4}} \sin ^{2 k-1}(2 x) \, dx = \frac{\sqrt{\pi } \Gamma (k)}{4 \Gamma \left(k+\frac{1}{2}\right)}\tag{3}$$

Assembling the pieces the sum to be taken to represent $i$ becomes

$$i_s = \sum _{k=1}^{\infty } \frac{\sqrt{\pi } \Gamma (k)}{(2 k-1) 2^{2 k-1} \Gamma \left(k+\frac{1}{2}\right)}\tag{4}$$

and this sum is immediately computed by Mathematica to give the compact result $(1)$.

Let us make the sum more transparent using the chain

$$\frac{\sqrt{\pi } \Gamma (k)}{\Gamma \left(k+\frac{1}{2}\right)}=B\left(\frac{1}{2},k\right)=\int_0^1 \frac{t^{k-1}}{\sqrt{1-t}} \, dt\tag{5}$$

and doing the sum under the integral

$$\sum _{k=1}^{\infty } \frac{t^{k-1}}{(2 k-1) 2^{2 k-1}}=\frac{\tanh ^{-1}\left(\frac{\sqrt{t}}{2}\right)}{\sqrt{t}}\tag{6}$$

leads finally to the integral

$$\int_0^1 \frac{\tanh ^{-1}\left(\frac{\sqrt{t}}{2}\right)}{\sqrt{t} \sqrt{1-t}} \, dt\tag{7}$$

for which Mathematica again quickly gives (1).

But there must be a shorter way ... yes, it is, substituting $\sin (2 x)=\sqrt{t}$ in the original integral gives (7) directly.

  • $\begingroup$ I do truly work better with series. Thank you. $\endgroup$ – clathratus Dec 18 '18 at 18:22

A Possible way: Consider $$I(a)=\int_{0}^{+\infty}\frac{\arctan(ax)}{1+x^2+x^4} dx$$ and $$I'(a)=\int_{0}^{+\infty}\frac{x}{(1+x^2+x^4)(1+x^2a^2)}dx=\int_{0}^{+\infty}\frac{1}{(1+y+y^2)(1+a^2y)}dx$$ and

$$\frac{1}{(1+y+y^2)(1+a^2y)}= \frac{-a^2y-a^2+1}{(a^4-a^2+1)(1+y+y^2)}+\frac{a^4}{(a^4-a^2+1)(ay^2+1)}$$

we can also calculate $I'(a)$ by complex integration (if you've learned that).

Thanks to Dylan for his advice.

  • $\begingroup$ Instead of doing partial fractions wrt $x$, you might consider substituting $u=x^2$ first $\endgroup$ – Dylan Dec 18 '18 at 3:07
  • $\begingroup$ And it is still not so easy. I think there may be a more clever way to solve it. $\endgroup$ – Liu Dec 18 '18 at 3:15
  • $\begingroup$ I meant the first integration. $$\frac12 \int_0^\infty \frac{1}{(1+u+u^2)(1+a^2u)} du$$ and then do partial fractions? $\endgroup$ – Dylan Dec 18 '18 at 3:31
  • $\begingroup$ @Dylan You're right. $\endgroup$ – Liu Dec 18 '18 at 3:34
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    $\begingroup$ If you have to know complex analysis in order to do complex integration, then I can't do it. If complex integration is just normal integration but with a few $i$'s here and there, then I can do that. $\endgroup$ – clathratus Dec 18 '18 at 5:03

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