# A double integral for $\frac{\pi}{2} \ln 2$.

Show that $$\begin{eqnarray*} I=\int_0^1 \int_0^1 \frac{dx \,dy}{\sqrt{1-x^2y^2}} = \frac{\pi}{2} \ln 2. \end{eqnarray*}$$

My try ... from this question here we have $$\begin{eqnarray*} \int_0^1 \frac{ \sin^{-1}(x)}{x} \,dx = \frac{\pi}{2} \ln 2 . \end{eqnarray*}$$ And from this question here we have $$\begin{eqnarray*} \int_0^1 \ln \left( \frac{1+ax}{1-ax} \right) \frac{dx}{x\sqrt{1-x^2}}=\pi\sin^{-1} a,\qquad |a|\leq 1. \end{eqnarray*}$$ It is easy to show $$\begin{eqnarray*} \int_0^1 \frac{dy}{1+yz}=\frac{1}{z} \ln(1+z). \end{eqnarray*}$$ So we have (with a little tad of algebra) $$\begin{eqnarray*} \frac{\pi}{2} \ln 2 &=& \int_0^1 \frac{ \sin^{-1}(x)}{x}\, dx \\ &=& \frac{1}{\pi} \int_0^1 \int_0^1 \ln \left( \frac{1+xt}{1-xt} \right) \frac{dt}{xt\sqrt{1-t^2}} x\, dx \\ &=& \frac{2}{\pi} \int_0^1 \int_0^1 \int_0^1 \frac{dx \,dy\, dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\ \end{eqnarray*}$$ This suggests we should consider the integral (sub $$t=\sin(\theta)$$) $$\begin{eqnarray*} \frac{2}{\pi} \int_0^1 \frac{ dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\ = \int_0^{\pi/2} \frac{d \theta }{1- x^2 y^2 \sin^2(\theta)}. \end{eqnarray*}$$ Now it is well known (Geometrically expand, integrate term by term & sum the familiar plum) that $$\begin{eqnarray*} \frac{2}{\pi} \int_0^{\pi/2} \frac{d \theta }{1- \alpha \sin^2(\theta)}=\frac{2}{\pi} \frac{1}{\sqrt{1-\alpha}} \end{eqnarray*}$$ and using this we have $$\begin{eqnarray*} \frac{\pi}{2} \ln 2 =\int_0^1 \int_0^1 \frac{dx\, dy}{\sqrt{1-x^2y^2}} . \end{eqnarray*}$$

The above double integral reminds of $$\begin{eqnarray*} \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\int_0^1 \int_0^1 \frac{dx\, dy}{1-x^2y^2} = \frac{\pi^2}{8} \end{eqnarray*}$$ which can be evaluated using the substitution $$x= \frac{\sin u}{\cos v}$$, $$y= \frac{\sin v}{\cos u}$$.

My solution above used some pretty heavy machinery to establish the result. So my question is: is there an easier method ?

$$\int_0^1 \int_0^1 \frac{dxdy}{\sqrt{1-x^2y^2}}\overset{xy=t}=\int_0^1 \frac{1}{y}\int_0^y \frac{1}{\sqrt{1-t^2}}dtdy=\int_0^1 \frac{\arcsin y}{y}dy$$ $$\overset{IBP}=-\int_0^1 \frac{ \ln y}{\sqrt{1-y^2}}dy\overset{y=\sin x}=-\int_0^\frac{\pi}{2}\ln( \sin x)dx=\frac{\pi}{2}\ln 2$$ See here for the above integral.
Through series expansions: $$\iint_{(0,1)^2}\frac{dx\,dy}{\sqrt{1-x^2 y^2}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\iint_{(0,1)^2}x^{2n}y^{2n}\,dx\,dy=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(2n+1)^2}=\int_{0}^{1}\frac{\arcsin(x)}{x}\,dx$$ and this is $$\int_{0}^{\pi/2}x\cot(x)\,dx\stackrel{\text{IBP}}{=}\int_{0}^{\pi/2}\log(\sin x)\,dx$$ which is well known to be $$\frac{\pi}{2}\log(2)$$. It is possible to exploit symmetry, derivatives of the Beta function, Fourier (or Fourier-Legendre) series and probably much more. For instance, the identity $$\prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\frac{2n}{2^n}$$ and just Riemann sums.
Why not go head-on? It seems to work here. That is, write the integral as $$\int_0^1\int_0^1\frac {\mathrm d x \mathrm d y}{\sqrt{1-x^2y^2}}=\int_0^1\int_0^1\frac {\mathrm d x \mathrm d y}{y \sqrt{\frac{1}{y^2}-x^2}}.$$ Setting $$x=\frac1y\sin\phi$$ as usual helps us evaluate the first integral, or you can simply note that it has the form of an $$\arcsin.$$ If you make this substitution, the first integral becomes $$\int_0^{\arcsin y}\frac1y\mathrm d \phi=\frac{\arcsin y}{y}.$$ Thus, the integral reduces to $$\int_0^1\frac{\arcsin y}{y}\mathrm d y,$$ which may be done by parts. Forgetting limits for now, this begins as $$\arcsin y\log y-\int\frac{\log y}{\sqrt{1-y^2}}\mathrm dy,$$ etc.