A Challenging Double Integral

Question:

Try to solve the following integral:

\[\int_0^1 \int_0^1 \frac{y^2}{(x^2y^2+1+2y^2)(y^2+1)}\, \mathrm{d}x\mathrm{d}y.\]

I have already solved this integral on Zhihu, and you can see my solution here.

Solution:

First integrate with respect to $x$, and the original integral can be simplified to

\[\begin{aligned} I & =\int_0^1 \frac{y^2\,\mathrm{d}y}{(1+y^2)(1+2y^2)} \int_0^1 \frac{1}{x^2 \frac{y^2}{1+2y^2}+1}\, \mathrm{d}x\\[.3cm] & =\int_0^1 \frac{y^2}{(1+y^2)(1+2y^2)} \cdot \frac{\sqrt{1+2y^2}}{y} \arctan \frac{y}{\sqrt{1+2y^2}}\, \mathrm{d}y\\[.3cm] &= {\color{RoyalBlue}\int_0^1 \frac{y}{(1+y^2)\sqrt{1+2y^2}} \arctan \frac{y}{\sqrt{1+2y^2}}\, \mathrm{d}y }, \quad y\mapsto 1/x \qquad (\star)\\[.3cm] &=\int_1^{\infty} \frac{1}{(1+x^2)\sqrt{2+x^2}} \arctan \frac{1}{\sqrt{2+x^2}}\, \mathrm{d}x\\[.5cm] &=\frac{\pi}{2} \int_1^{\infty} \frac{1}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x - \int_1^{\infty}\frac{\arctan\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}\,\mathrm{d}x\\[.3cm] &=\frac{\pi}{2}\arctan \frac{x}{\sqrt{2+x^2}}\Bigg|_{x=1}^{\infty} - \int_1^{\infty}\frac{\arctan\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}\,\mathrm{d}x \\[.3cm] &=\frac{\pi^2}{24} - \underbrace{\int_1^{\infty} \frac{\arctan \sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}\,\mathrm{d}x}_{J}. \end{aligned}\]

So the question now boils down to the calculation of $J$, which is essentially related to Ahmed’s integral. And

\[J = \int_1^{\infty} \frac{\arctan\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}\,\mathrm{d}x,\]

which can be solved by using Feynman’s trick. Let

\[J(u) = \int_1^{\infty} \frac{\arctan (u\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}},\]

so

\[\begin{aligned} J'(u) &= \int_1^{\infty} \frac{\mathrm{d}x}{(x^2+1)(u^2(x^2+2)+1)}\\[.3cm] &=\frac{1}{1+u^2} \int_1^{\infty}\left(\frac{1}{1+x^2}-\frac{1}{\frac{1+2u^2}{u^2}+x^2}\right)\,\mathrm{d}x \\[.3cm] &=\frac{1}{1+u^2}\left[\arctan x - \frac{u}{\sqrt{1+2u^2}} \arctan \frac{ux}{\sqrt{1+2u^2}}\right] \Bigg|_{x=1}^{x=\infty}\\[.3cm] &=\frac{\pi}{4} \frac{1}{1+u^2} - \frac{\pi}{2} \frac{u}{(1+u^2)\sqrt{1+2u^2}} + \frac{u}{(1+u^2)\sqrt{1+2u^2}}\arctan \frac{u}{\sqrt{1+2u^2}}. \end{aligned}\]

Note that $J(0) = 0$, by integrating with respect to $u$, we can obtain $J$:

\[\begin{aligned} &J(1)-J(0)\\[.3cm] ={}&\int_0^{1}J'(u)\,\mathrm{d}u\\[.3cm] ={}&\frac{\pi}{4}\int_0^{1}\frac{\mathrm{d}u}{1+u^2} - \frac{\pi}{2}\int_0^1 \frac{u\,\mathrm{d}u}{(1+u^2)\sqrt{1+2u^2}} + \int_0^{1}\frac{u}{(1+u^2)\sqrt{1+2u^2}}\arctan \frac{u}{\sqrt{1+2u^2}}\,\mathrm{d}u\\[.3cm] ={}&\frac{\pi}{4}\cdot\frac{\pi}{4} - \frac{\pi}{2}\cdot \frac{\pi}{12} + \int_0^{1}\frac{u}{(1+u^2)\sqrt{1+2u^2}}\arctan \frac{u}{\sqrt{1+2u^2}}\,\mathrm{d}u\\[.3cm] ={}&\frac{\pi^2}{48}+ \underbrace{ {\color{Royalblue}{\int_0^{1}\frac{u}{(1+u^2)\sqrt{1+2u^2}}\arctan \frac{u}{\sqrt{1+2u^2}}\,\mathrm{d}u}} }_{I}. \end{aligned}\]

The result of the blue part in the above equation is exactly the $I$ that we initially calculated. Please refer to equation $(\star)$. Notice that

\[I = \frac{\pi^2}{24} - J,\]

so we can obtain

\[J = \frac{\pi^2}{48} + \frac{\pi^2}{24} - J \quad \Rightarrow \quad J= \frac{\pi^2}{32}.\]

Integrating all the previous results, we can obtain the following result:

\[\color{red}{ \begin{aligned} &\int_0^{1}\frac{x}{(1+x^2)\sqrt{1+2x^2}} \arctan \frac{x}{\sqrt{1+2x^2}}\,\mathrm{d}x = \frac{\pi^2}{96}\\[.3cm] &\int_1^{\infty}\frac{\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x = \frac{\pi^2}{32}\\[.3cm] &\int_0^1 \frac{\arctan \sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x = \frac{5\pi^2}{96}, \quad \color{black}{\text{Ahmed's Integral}} \end{aligned} }\]

And the original integral is

\[\int_0^1 \int_0^1 \frac{y^2}{(x^2y^2+1+2y^2)(y^2+1)}\, \mathrm{d}x\mathrm{d}y = \frac{\pi^2}{96}.\]