# Closed form of $\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx$

Can a closed form solution for the following integral be found: $$\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx\,?$$

I have tried all the standard tricks such as integration by parts, various substitutions, and parametric differentiation (Feynman's trick), but all to no avail.

An attempt is letting $$f(t):=\int_0^\infty\,\arctan^2\left(\frac{2tx}{1+x^2}\right)\,\text{d}x\,.$$ Therefore, $$f'(t)=\int_0^\infty\,\frac{8x^2(x^2+1)}{\big(x^4+2(2t^2+1)x^2+1\big)^2}\,\left(1+x^2-4tx\arctan\left(\frac{2tx}{1+x^2}\right)^{\vphantom{a^2}}\right)\,\text{d}x\,.$$ This doesn't seem to go anywhere. Help!

• I would suggest looking into using residues to solve integrals of this sort. Perhaps it is not applicable in your case (it's been a year since I've used them), but it is worthwhile to know of them if you often perform such integrations. Jun 4 '20 at 4:57

$$I=\int_0^\infty \arctan^2 \left (\frac{2x}{x^2 + 1} \right ) dx\overset{IBP}=4\int_0^\infty \frac{x(x^2-1)\arctan\left(\frac{2x}{x^2+1}\right)}{x^4+6x^2+1}dx$$ We have that: $$4\int\frac{x(x^2-1)}{x^4+6x^2+1}dx=(\sqrt 2 +1)\ln(x^2+(\sqrt 2+1)^2)-(\sqrt 2-1)\ln(x^2+(\sqrt 2-1)^2)$$ $$\frac{d}{dx}\arctan\left(\frac{2x}{x^2+1}\right)=\frac12\left(\frac{\sqrt 2+1}{x^2+(\sqrt 2+1)^2}-\frac{\sqrt 2-1}{x^2+(\sqrt 2-1)^2}\right)$$ Thus integrating by parts again and simplifying we obtain: $$I=\int_0^\infty \frac{(\sqrt 2+1)^2 \ln(x^2+(\sqrt 2+1)^2)}{x^2+(\sqrt 2+1)^2}dx+\int_0^\infty \frac{(\sqrt 2-1)^2 \ln(x^2+(\sqrt 2-1)^2)}{x^2+(\sqrt 2-1)^2}dx$$ $$-\int_0^\infty \frac{\ln(x^2+(\sqrt 2-1)^2)}{x^2+(\sqrt 2+1)^2}dx-\int_0^\infty \frac{\ln(x^2+(\sqrt 2+1)^2)}{x^2+(\sqrt 2-1)^2)}dx$$ From here we have the following result: $$\int_0^\infty \frac{\ln(x^2+a^2)}{x^2+b^2}dx=\frac{\pi}{b}\ln(a+b), \ a,b>0$$ So using this result and with some algebra everything simplifies to: $$\boxed{\int_0^\infty \arctan^2 \left (\frac{2x}{x^2 + 1} \right ) dx=2\pi \ln(1+\sqrt 2)-\sqrt 2\pi \ln 2}$$

• Nice solution! (+1) Using the same technique, we also get $$\int_{0}^{\infty}\arctan^2\left(\frac{2rt}{1+t^2}\right)\,\mathrm{d}t=2\pi\left(r\log\left(\sqrt{r^2+1}+r\right)-\sqrt{r^2+1}\log\sqrt{r^2+1}\right).$$ Jun 4 '20 at 10:38

Here is a solution based on Fubini's theorem.

According to an addition formula $$\begin{equation*} \arctan\left(\dfrac{2x}{1+x^2}\right) = \arctan((\sqrt{2}+1)x)-\arctan((\sqrt{2}-1)x) . \end{equation*}$$ Furthermore $$\begin{equation*} \arctan x=\mathrm{sign}(x)\dfrac{\pi}{2}-\arctan\dfrac{1}{x} . \end{equation*}$$ Consequently $$\begin{equation*} \arctan\left(\dfrac{2x}{1+x^2}\right) = \arctan\dfrac{\sqrt{2}+1}{x}-\arctan\dfrac{\sqrt{2}-1}{x}=\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{x}{x^2+s^2}\, ds . \end{equation*}$$ Via Fubini's theorem we get $$\begin{gather*} \int_{0}^{\infty}\arctan^2\left(\dfrac{2x}{1+x^2}\right)\, dx = \int_{0}^{\infty}\left(\arctan\dfrac{\sqrt{2}+1}{x}-\arctan\dfrac{\sqrt{2}-1}{x}\right)^2\, dx=\\[2ex] \int_{0}^{\infty}\left(\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{x}{x^2+s^2}\, ds\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{x}{x^2+t^2}\, dt\right)\, dx=\\[2ex] \int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\int_{0}^{\infty}\dfrac{x^2}{(x^2+s^2)(x^2+t^2)}\, dx\right)\, ds\right)\, dt=\\[2ex] \dfrac{\pi}{2}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{1}{s+t}\, ds\right)\, dt=\\[2ex] \dfrac{\pi}{2}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\ln(t+\sqrt{2}+1)-\ln(t+\sqrt{2}-1)\right)\, dt=\\[2ex] 2\pi\ln(\sqrt{2}+1)-\sqrt{2}\pi\ln 2. \end{gather*}$$

Remark. Since $$\begin{equation*} \arctan\left(\dfrac{2x\sinh\alpha}{1+x^2}\right)=\arctan\left(\dfrac{e^{\alpha}}{x}\right)-\arctan\left(\dfrac{e^{-\alpha}}{x}\right) = \int_{e^{-\alpha}}^{e^{\alpha}}\dfrac{x}{x^2+s^2}\, ds \end{equation*}$$ the $$@$$Sangchul Lee's generalization can be proved in the same way.

Here is another solution with a generalization:

Let $$r=\sinh\alpha$$ and $$s=\sinh\beta$$. Then \begin{aligned} &\int_{0}^{\infty} \arctan\left(\frac{2rx}{1+x^2}\right)\arctan\left(\frac{2sx}{1+x^2}\right) \, \mathrm{d}x\\ &= \pi \left( \alpha \sinh\beta+\beta\sinh\alpha+(\cosh\alpha+\cosh\beta)\log\left(\frac{e^{\alpha}+e^{\beta}}{1+e^{\alpha+\beta}}\right) \right) \end{aligned} \tag{*}

Proof. Let $$J = J(\alpha,\beta)$$ denote the right-hand side of $$\text{(*)}$$. Then

$$J(0, \beta) = 0, \qquad J_{\alpha}(\alpha, 0) = 0, \qquad J_{\alpha\beta} = \pi \left( \frac{1+\cosh\alpha\cosh\beta}{\cosh\alpha + \cosh\beta} \right).$$

Now let $$I = I(\alpha, \beta)$$ denote the left-hand side of $$\text{(*)}$$. Then by the substitution $$x=\tan(\theta/2)$$, we get

$$I = \frac{1}{2}\int_{0}^{\pi} \frac{\arctan(\sinh\alpha\sin\theta)\arctan(\sinh\beta\sin\theta)}{1+\cos\theta} \, \mathrm{d}\theta.$$

From this, we easily check that $$I$$ also satisfies

$$I(0, \beta) = 0, \qquad I_{\alpha}(\alpha, 0) = 0.$$

Moreover,

\begin{align*} \require{cancel} I_{\alpha\beta} &= \frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta(1-\cos\theta)}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\theta \\ &= \frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\theta \\ &\quad - \cancelto{0}{\frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\sin\theta}\\ &= \frac{1}{2}\int_{0}^{\infty} \frac{\cosh\alpha\cosh\beta (1 + t^2)}{(t^2 + \cosh^2\alpha)(t^2 + \cosh^2\beta)} \, \mathrm{d}t \tag{t=\cot\theta} \end{align*}

It is not hard to check that the last integral is equal to $$J_{\alpha\beta}$$. Therefore we get $$I = J$$.