Show that : $\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$ I found it with help of Wolfram alpha .
$$\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$$
This integral admits an antiderivative (I can add it if so) so I can conclude that this result is true .I would like to see a proof with the use of complex integration or a 'real' proof . For that I think we can put $y=\frac{2}{x^2+1}$ and use integration by parts (perhaps?).For the proof with complex integration(if it exists I'm new for it )  can you detail the solution ?
Thanks!
 A: Here's a brute-force method:
$$
\begin{aligned}
 \int{\arctan}\left( \frac{2}{x^2+1} \right) \,dx&=\int{\mathrm{arccot}}\left( \frac{1}{2}+\frac{x^2}{2} \right) \,dx\\
 &=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) +\int{\frac{x^2}{\frac{5}{4}+\frac{x^2}{2}+\frac{x^4}{4}}}\,dx\\
 &=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) -\frac{1}{2}\int{\frac{\sqrt{5}-x^2}{\frac{5}{4}+\frac{x^2}{2}+\frac{x^4}{4}}}\,dx+\frac{1}{2}\int{\frac{\sqrt{5}+x^2}{\frac{5}{4}+\frac{x^2}{2}+\frac{x^4}{4}}}\,dx\\
&=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) \\&+\frac{\displaystyle \int{\frac{\sqrt{2\left( -1+\sqrt{5} \right)}+2x}{-\sqrt{5}-\sqrt{2\left( -1+\sqrt{5} \right)}x-x^2}}\,dx}{\sqrt{2\left( -1+\sqrt{5} \right)}}+\frac{\displaystyle \int{\frac{\sqrt{2\left( -1+\sqrt{5} \right)}-2x}{-\sqrt{5}+\sqrt{2\left( -1+\sqrt{5} \right)}x-x^2}}\,dx}{\sqrt{2\left( -1+\sqrt{5} \right)}}\\&+\int{\frac{1}{\sqrt{5}-\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2}}\,dx+\int{\frac{1}{\sqrt{5}+\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2}}\,dx\\
&=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) \\&+\frac{\log \left( \sqrt{5}-\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2 \right)}{\sqrt{2\left( -1+\sqrt{5} \right)}}-\frac{\log \left( \sqrt{5}+\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2 \right)}{\sqrt{2\left( -1+\sqrt{5} \right)}}\\
&+\sqrt{\frac{2}{\sqrt{5}+1}} \arctan\left(\frac{2 x+\sqrt{2 \left(\sqrt{5}-1\right)}}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right)\\&-\sqrt{\frac{2}{\sqrt{5}+1}} \arctan \left(\frac{\sqrt{2 \left(\sqrt{5}-1\right)}-2 x}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right)
\\&\overset{\text{def}}{=} F(x)
\end{aligned}
$$
One can easily see that
$$
\underset{x\to \infty }{\text{lim}}x \text{ arccot}\left(\frac{x^2}{2}+\frac{1}{2}\right)=0
$$
and
$$
\begin{aligned}
&\lim_{x\to\infty} \frac{\log \left(x^2-\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}\right)}{\sqrt{2 \left(\sqrt{5}-1\right)}}-\frac{\log \left(x^2+\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}\right)}{\sqrt{2 \left(\sqrt{5}-1\right)}} \\&=\lim_{x\to\infty}\frac{\log \left(\frac{x^2-\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}}{x^2+\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}}\right)}{\sqrt{2 \left(\sqrt{5}-1\right)}}\\
&=0
\end{aligned}
$$
Also it is easy to see that
$$
\underset{x\to \infty }{\text{lim}}\arctan\left(\frac{\sqrt{2 \left(\sqrt{5}-1\right)}-2 x}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right)=-\dfrac{\pi}{2}
$$
and
$$
\underset{x\to \infty }{\text{lim}}\arctan\left(\frac{2 x+\sqrt{2 \left(\sqrt{5}-1\right)}}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right) = \dfrac{\pi}2
$$
Thus, combine the above results, from Newton-Leibniz formula, we get
$$
\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx= F(\infty) - F(0)=\sqrt{\frac{2}{\sqrt{5}+1}} \pi = \color{blue}{\pi\sqrt{\phi^{-1}}}
$$
As desired.
A: $$I=\int_{0}^{\infty} \tan^{-1}\frac{2}{x^2+1} dx$$
Let us do it by parts taking 1 as second function:
$$I=x\tan^{-1}\frac{2}{1+x^2}|_{0}^{\infty}+4\int_{0}^{\infty}\frac{x^2}{x^4+2x^2+5}dx$$
$$\implies I=4\int_{0}^{\infty} \frac{dx}{x^2+5/x^2+2}$$ $$=2\int_{0}^{\infty} \left(\frac{(1+\sqrt{5}/x^2)}{(x-\sqrt{5}/x)^2+2+2\sqrt{5}}+\frac{(1-\sqrt{5}/x^2)}{(x+\sqrt{5}/x)^2+2-2\sqrt{5}}\right)dx$$
Let $x-\sqrt{5}/x=u$ and $x+\sqrt{5}/x=v$, then the limits in the second integral being identical it vanishes, we get
$$I=2\int_{-\infty}^{\infty} \frac{du}{u^2+2+2\sqrt{5}}=\frac{2}{\sqrt{2+2\sqrt{5}}} \tan^{-1}\frac{u}{\sqrt{2+2\sqrt{5}}}|_{-\infty}^{\infty}=\pi\sqrt{\frac{2}{1+\sqrt{5}}}=\pi \phi^{-1/2}.$$
A: Partial answer: When $y > 0$, we have that $\arctan (1/y)= \frac{\pi}{2} - \arctan y$, so the integral becomes
$$
\int_0^{+\infty} \frac{\pi}{2} - \arctan\left(\frac{1+x^2}{2}\right) dx
$$
Using integration by parts, we get to
$$
\int_{-\infty}^{+\infty} \dfrac{2 x^2}{4+(x^2+1)^2} dx
$$
This last integral can be computed by elementary means or complex integration.
A: To get things started, the integral definition of the arctangent, switching the order of integration, and a change of variables gives us
$$\begin{align}
\int_0^\infty\arctan\left(2\over x^2+1\right)\,dx
&=\int_0^\infty\int_0^{2/(x^2+1)}{1\over1+y^2}\,dy\,dx\\
&=\int_0^2\int_0^\sqrt{(2/y)-1}{1\over1+y^2}dx\,dy\\
&=\int_0^2{\sqrt{(2/y)-1}\over1+y^2}\,dy\\
&=\int_\infty^0{\sqrt{z^2}\over1+\left(2\over1+z^2 \right)^2}\cdot{-4z\,dz\over(1+z^2)^2}\\
&=4\int_0^\infty{z^2\over(1+z^2)^2+4}\,dz\\
&=2\int_{-\infty}^\infty{z^2\over z^4+2z^2+5}\,dz
\end{align}$$
The integral is now ripe for residues. The singularities are at $z=\pm\sqrt{-1\pm\sqrt{1-5}}=\pm\sqrt{-1\pm2i}$. Some straightforward algebra shows that the ones in the upper half plane are at
$$z=\pm{1\over\sqrt\phi}+i\sqrt\phi$$
Calling these $c_+$ and $c_-$ and noting that $c_\pm^2+1=\pm2i$, the residue theorem tells us
$$\begin{align}
2\int_{-\infty}^\infty{z^2\over z^4+2z^2+5}\,dz
&=4\pi i\left({c_+^2\over4c_+^3+4c_+}+{c_-^2\over4c_-^3+4c_-} \right)\\
&=4\pi i\left( {c_+\over8i}+{c_-\over-8i}\right)\\
&={\pi\over2}(c_+-c_-)\\
&={\pi\over\sqrt\phi}
\end{align}$$
A: $$I=\int_{0}^{\infty} \tan^{-1}\frac{2}{x^2+1} dx$$
Let us do it by parts taking 1 as second function:
$$I=x\tan^{-1}\frac{2}{1+x^2}|_{0}^{\infty}+4\int_{0}^{\infty}\frac{x^2}{x^4+2x^2+5}dx$$
$$\implies I=2\int_{-\infty}^{\infty} \frac{dx}{x^2+5/x^2+2}=2\int_{-\infty}^{\infty} \frac{dx}{(x-\sqrt{5}/x)^2+2+2\sqrt{5}}=2\int_{-\infty}^{\infty} \frac{dx}{x^2+2+2\sqrt{5}}$$ $$=\frac{2}{\sqrt{2+2\sqrt{5}}} \tan^{-1}\frac{x}{\sqrt{2+2\sqrt{5}}}|_{-\infty}^{\infty}=\pi\sqrt{\frac{2}{1+\sqrt{5}}}=\pi \phi^{-1/2}.$$
See the very interesting Glasser's property of integrals:
A simple proof for Glasser: $\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx, a>0$
