Use the linear substitution
x = u − v , y = u + v . x=u-v,\qquad y=u+v. x = u − v , y = u + v . The inverse transformation is
u = x + y 2 , v = y − x 2 , u=\frac{x+y}{2},
\qquad
v=\frac{y-x}{2}, u = 2 x + y , v = 2 y − x , and
∣ ∂ ( x , y ) ∂ ( u , v ) ∣ = ∣ det ( 1 − 1 1 1 ) ∣ = 2. \left|\frac{\partial(x,y)}{\partial(u,v)}\right|
=
\left|
\det
\begin{pmatrix}
1&-1\\
1&1
\end{pmatrix}
\right|
=2. ∂ ( u , v ) ∂ ( x , y ) = det ( 1 1 − 1 1 ) = 2. The square 0 ≤ x , y ≤ 1 0\le x,y\le1 0 ≤ x , y ≤ 1 ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 2 , ± 1 2 ) (\tfrac12,\pm\tfrac12) ( 2 1 , ± 2 1 ) ( 1 , 0 ) (1,0) ( 1 , 0 ) u v uv uv
{ 0 ≤ u ≤ 1 2 , − u ≤ v ≤ u , 1 2 ≤ u ≤ 1 , u − 1 ≤ v ≤ 1 − u . \begin{cases}
0\le u\le\frac12,\quad -u\le v\le u,\\[4pt]
\frac12\le u\le1,\quad u-1\le v\le1-u.
\end{cases} { 0 ≤ u ≤ 2 1 , − u ≤ v ≤ u , 2 1 ≤ u ≤ 1 , u − 1 ≤ v ≤ 1 − u . In the new coordinates,
x y = ( u − v ) ( u + v ) = u 2 − v 2 , 1 − x y = 1 − u 2 + v 2 . xy=(u-v)(u+v)=u^2-v^2,
\qquad
1-xy=1-u^2+v^2. x y = ( u − v ) ( u + v ) = u 2 − v 2 , 1 − x y = 1 − u 2 + v 2 . The integrand is even in v v v
S = 4 ∫ 0 1 / 2 ∫ 0 u 1 1 − u 2 + v 2 d v d u + 4 ∫ 1 / 2 1 ∫ 0 1 − u 1 1 − u 2 + v 2 d v d u . \begin{aligned}
S
&=
4\int_0^{1/2}\int_0^u
\frac{1}{1-u^2+v^2}\,dv\,du\\
&\quad+
4\int_{1/2}^{1}\int_0^{1-u}
\frac{1}{1-u^2+v^2}\,dv\,du.
\end{aligned} S = 4 ∫ 0 1/2 ∫ 0 u 1 − u 2 + v 2 1 d v d u + 4 ∫ 1/2 1 ∫ 0 1 − u 1 − u 2 + v 2 1 d v d u . For fixed u u u
∫ 0 a d v 1 − u 2 + v 2 = 1 1 − u 2 arctan ( a 1 − u 2 ) . \int_0^a\frac{dv}{1-u^2+v^2}
=
\frac1{\sqrt{1-u^2}}
\arctan\left(\frac{a}{\sqrt{1-u^2}}\right). ∫ 0 a 1 − u 2 + v 2 d v = 1 − u 2 1 arctan ( 1 − u 2 a ) . Each piece now needs one trigonometric substitution. The first uses
u = sin θ u=\sin\theta u = sin θ u = cos θ u=\cos\theta u = cos θ
For the first piece, let u = sin θ u=\sin\theta u = sin θ 0 ≤ θ ≤ π / 6 0\le\theta\le\pi/6 0 ≤ θ ≤ π /6
u 1 − u 2 = tan θ , \frac{u}{\sqrt{1-u^2}}=\tan\theta, 1 − u 2 u = tan θ , and hence
S 1 = 4 ∫ 0 1 / 2 1 1 − u 2 arctan ( u 1 − u 2 ) d u = 4 ∫ 0 π / 6 θ d θ = π 2 18 . \begin{aligned}
S_1
&=
4\int_0^{1/2}
\frac1{\sqrt{1-u^2}}
\arctan\left(\frac u{\sqrt{1-u^2}}\right)\,du\\
&=4\int_0^{\pi/6}\theta\,d\theta
=\frac{\pi^2}{18}.
\end{aligned} S 1 = 4 ∫ 0 1/2 1 − u 2 1 arctan ( 1 − u 2 u ) d u = 4 ∫ 0 π /6 θ d θ = 18 π 2 . For the second piece, let u = cos θ u=\cos\theta u = cos θ 0 ≤ θ ≤ π / 3 0\le\theta\le\pi/3 0 ≤ θ ≤ π /3 u u u 1 / 2 1/2 1/2 1 1 1 θ \theta θ π / 3 \pi/3 π /3 0 0 0 d u = − sin θ d θ du=-\sin\theta\,d\theta d u = − sin θ d θ
1 − u 1 − u 2 = 1 − cos θ sin θ = tan θ 2 . \frac{1-u}{\sqrt{1-u^2}}
=\frac{1-\cos\theta}{\sin\theta}
=\tan\frac\theta2. 1 − u 2 1 − u = sin θ 1 − cos θ = tan 2 θ . Therefore
S 2 = 4 ∫ 0 π / 3 θ 2 d θ = π 2 9 . \begin{aligned}
S_2
&=4\int_0^{\pi/3}\frac\theta2\,d\theta\\
&=\frac{\pi^2}{9}.
\end{aligned} S 2 = 4 ∫ 0 π /3 2 θ d θ = 9 π 2 . Adding the two pieces,
S = S 1 + S 2 = π 2 18 + π 2 9 = π 2 6 . S=S_1+S_2
=\frac{\pi^2}{18}+\frac{\pi^2}{9}
=\frac{\pi^2}{6}. S = S 1 + S 2 = 18 π 2 + 9 π 2 = 6 π 2 .