$\int_{-\infty}^{\infty}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}}dx=\zeta(2)$ Without Feynman Integration

Question

How do I find $$\int_{-\infty}^{\infty}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}}dx=\zeta(2)$$ without Feynman integration? I saw this video, which gives $$\int_{0}^{1}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}}dx=\frac{5\pi^2}{96}$$ Via Feynman integration, but I would like to know another method.

Answer 1

Notice that $\frac{\arctan x}{x} = \int_{0}^{1} \frac{dy}{1+x^2y^2}$. So

\begin{align*} \int_{-\infty}^{\infty} \frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}} \, dx &= \int_{-\infty}^{\infty} \int_{0}^{1} \frac{1}{(x^2+1)(1+(x^2+2)y^2)} \, dydx. \end{align*}

Notice that

$$ \frac{1}{(x^2+1)(1+(x^2+2)y^2)} = \frac{1}{(x^2+1)(y^2+1)} - \frac{y^2}{(y^2+1)(1+(x^2+2)y^2)}. $$

So interchanging the order of integration,

\begin{align*} \int_{-\infty}^{\infty} \frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}} \, dx &= \int_{0}^{1} \left( \frac{\pi}{y^2+1} - \frac{\pi y}{(y^2+1)\sqrt{2y^2+1}} \right) \, dy \\ &= \pi \left[ \arctan(y) - \arctan\sqrt{2y^2+1} \right]_{0}^{1} \\ &= \frac{\pi^2}{6}. \end{align*}

Answer 2

This integral can also be evaluated via complex residues.

First, we rewrite

$$\begin{align*} \mathcal I &= \int_{-\infty}^\infty \frac{\arctan \sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}} \, dx \\ &= 2 \int_0^\infty \frac{\arctan \sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}} \, dx \\ &= 2 \underbrace{\int_{\sqrt2}^\infty \frac{\arctan x}{\left(x^2-1\right) \sqrt{x^2-2}} \, dx}_{=:I} & \left[x\mapsto\sqrt{x^2-2}\right] \end{align*}$$

Next, introduce the complex function

$$\begin{align*} f(z) &= \frac{\arctan z}{\left(z^2-1\right) \sqrt{z^2-2}} \\ &= -\frac i2 \frac{\log\left\lvert\frac{i-z}{i+z}\right\rvert + i \arg \left(\frac{i-z}{i+z}\right)}{\left(z^2-1\right) \sqrt{\left\lvert z^2-2\right\rvert} \cdot e^{\tfrac i2 \left[\arg\left(z-\sqrt2\right)+ \arg\left(z+\sqrt2\right)\right]}} \end{align*}$$

where we use the principal branches for $\arctan z$ and $\sqrt{z+\sqrt2}$ such that $\arg\left(\dfrac{i-z}{i+z}\right)\in(-\pi,\pi)$ and $\arg\left(z+\sqrt2\right)\in(-\pi,\pi)$; for $\sqrt{z-\sqrt2}$ we take a branch cut along the positive real axis on $\left[\sqrt2,\infty\right)$ so that $\arg\left(z-\sqrt2\right)\in(0,2\pi)$. We integrate $f(z)$ along an indented circular contour in the counterclockwise direction:

The integrals along the circular portions will all vanish in their respective limits (proof omitted). The remaining contributions are outlined below.

• Around the cut $\left[\sqrt2,\infty\right)$:

$$\begin{align*} \int_A f(z) \, dz &= \int_{\sqrt2+\varepsilon}^R f(x+i\varepsilon) \, dx \\ &\to \int_{\sqrt2}^\infty \frac{\arctan x}{\left(x^2-1\right) \sqrt{x^2-2} \cdot e^{\tfrac i2\left[0+0\right]}} \, dx \\ \int_{A'} f(z) \, dz &= \int_R^{\sqrt2+\varepsilon} f(x-i\varepsilon) \, dx \\ &\to - \int_{\sqrt2}^\infty \frac{\arctan x}{\left(x^2-1\right) \sqrt{x^2-2} \cdot e^{\tfrac i2 \left[2\pi+0\right]}} \, dx \\[2ex] \implies \int_{A\cup A'} f(z) \, dz &\to 2I \end{align*}$$

• Around the cut $i[1,\infty)$:

$$\begin{align*} \int_B f(z) \, dz &= i \int_{1+\varepsilon}^R f(-\varepsilon+ix) \, dx \\ &\to -\frac 12 \int_1^\infty \frac{\log\left\lvert\frac{1-x}{1+x}\right\rvert - i \pi}{\left(x^2+1\right) \sqrt{x^2+2} \cdot e^{\tfrac i2\left[\tfrac\pi2+\tfrac\pi2\right]}} \, dx \\ \int_{B'} f(z) \, dz &= i \int_R^{1+\varepsilon} f(\varepsilon+ix) \, dx \\ &\to \frac 12 \int_1^\infty \frac{\log\left\lvert\frac{1-x}{1+x}\right\rvert + i \pi}{\left(x^2+1\right) \sqrt{x^2+2} \cdot e^{\tfrac i2\left[\tfrac\pi2+\tfrac\pi2\right]}} \, dx \\[2ex] \implies \int_{B\cup B'} f(z) \, dz &\to \pi \int_1^\infty \frac{dx}{\left(x^2+1\right) \sqrt{x^2+2}} \, dx = \frac{\pi^2}{12} \end{align*}$$

• Around the cut $\left(-\infty,-\sqrt2\right]$, a similar analysis to the $A/A'$ pair indicates that $$\int_{C\cup C'} f(z) \, dz \to 2I;$$ the only difference here is that $\arg\left(z-\sqrt2\right)+\arg\left(z+\sqrt2\right)\to-\pi+\pi$ along $C$, and $\to\pi+\pi$ along $C'$.

• Around the cut $i[-1,-\infty)$, in the same vein as $B/B'$, $$\int_{D\cup D'} f(z) \, dz \to \frac{\pi^2}{12},$$ with $\arg\left(z-\sqrt2\right)+\arg\left(z+\sqrt2\right)\to\dfrac{3\pi}2-\dfrac\pi2$ along both paths.

Now, $f(z)$ has two simple poles at $z=\pm1$, with residues

$$\underset{z=\pm1}{\operatorname{Res}} f(z) = -\frac{i\pi}8$$

so by the residue theorem,

$$i2\pi \left(-\frac{i\pi}8 - \frac{i\pi}8\right) = 2I + \frac{\pi^2}{12} + 2I + \frac{\pi^2}{12} \\ \implies I = \frac{\pi^2}{12} \implies \boxed{\mathcal I = \frac{\pi^2}6}$$

Answer 3

\begin{align*} & 2\int_0^\infty \frac{\arctan \sqrt{x^2 + 2}}{(x^2+1)\sqrt{x^2 + 2} } \ dx\\ =& \ 2\int_0^\infty \int_0^x \frac{x}{(x^2+1)[x^2+(x^2+2)y^2]}dy\ dx\\ =&\ 2\int_0^\infty \int_y^\infty \frac{x}{(x^2+1) [x^2+(x^2+2)y^2]}dx\ dy\\ =&\int_0^\infty \frac1{y^2-1}\ln\frac{y^2(3+y^2)}{(1+y^2)^2} \overset{y=\frac{1-t}{1+t}}{dy}\\ =&\ \frac12\int^1_{-1}\frac1{t}\bigg( \ln\frac {1+t^2}{(1-t)^2} + \ln\frac {1+t^2}{1+t+t^2}\bigg) dt\\ =&\ \frac12 \int^1_{-1}\bigg( \int_{\pi/2}^\pi +\int_{\pi/2}^{\pi/3} \bigg)\frac{2\sin\theta}{1+2t\cos\theta+t^2}d\theta \ dt\\ =&\ \frac\pi2\bigg( \int_{\pi/2}^\pi +\int_{\pi/2}^{\pi/3} \bigg)d\theta =\frac{\pi^2}{6} \end{align*}