proofs

44 Three times π 2 /6
v
1
1
2
y
v
1
u
x
This evaluation also shows that the double integral (over a positive function
with a pole at x = y = 1) is finite. Note that the computation is also easy
and straightforward if we read it backwards — thus the evaluation of ζ(2)
leads one to the double integral I.
The second way to evaluate I comes from a change of coordinates: in the
new coordinates given by u := y+x
2
and v := y−x
2
the domain of integration
is a square of side length 2√ 1 2, which we get from the old domain by
first rotating it by 45 ◦ and then shrinking it by a factor of √ 2. Substitution
of x = u − v and y = u + v yields
1
1 − xy = 1
1 − u 2 + v 2 .
To transform the integral, we have to replace dxdy by 2 du dv, to compensate
for the fact that our coordinate transformation reduces areas by a
constant factor of 2 (which is the Jacobi determinant of the transformation;
see the box on the next page). The new domain of integration, and the
function to be integrated, are symmetric with respect to the u-axis, so we
just need to compute two times (another factor of 2 arises here!) the integral
over the upper half domain, which we split into two parts in the most
natural way:
I 1
1
2
I 2
1
u
Using
∫
I = 4
1/2( u
0
∫
0
) ∫
dv
1
1 − u 2 + v 2 du + 4
1/2
( 1−u ∫
0
)
dv
1 − u 2 + v 2 du.
∫
dx
a 2 + x 2 = 1 a arctan x + C , this becomes
a
∫1/2
( )
1
I = 4 √ arctan u
√ du
1 − u
2 1 − u
2
+ 4
0
∫ 1
( )
1 1 − u
√ arctan √ du.
1 − u
2 1 − u
2
1/2
These integrals can be simplified and finally evaluated by substituting u =
sin θ resp. u = cosθ. But we proceed more directly, by computing that the
derivative of g(u) := arctan ( )
√ u
1−u 2 is g ′ (u) = 1 , while the deriva-
√
1−u 2
tive of h(u) := arctan ( (
√1−u
1−u 2) √ = arctan
1−u
1+u)
is h ′ (u) = − 1 √
2
. 1−u 2
So we may use ∫ b
a f ′ (x)f(x)dx = [ 1
2 f(x)2] b
a = 1 2 f(b)2 − 1 2 f(a)2 and get
I = 4
∫ 1/2
0
= 2
[g(u) 2] 1/2
g ′ (u)g(u)du + 4
0
∫ 1
1/2
− 4
[h(u) 2] 1
1/2
−2h ′ (u)h(u)du
1
= 2g( 1 2 )2 − 2g(0) 2 − 4h(1) 2 + 4h( 1 2 )2
= 2 ( )
π 2 (
6 − 0 − 0 + 4 π
) 2
6
= π2
6 . □