Problem 12107
(American Mathematical Monthly, Vol.126, April 2019) Proposed by C. I. V˘alean (Romania).
Prove
Z 1 0
Z 1 0
1
√1 + x2p1 + y2(1 − x2y2)dx dy = G whereG is the Catalan’s constantP∞
n=1(−1)n−1/(2n − 1)2.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We have that Z 1
0
Z 1 0
dx dy
√1 + x2p1 + y2(1 − x2y2) = Z π/4
0
Z π/4 0
cos(u) cos(v) cos2(u) − sin2(v)dv
!
du (x = tan(u), y = tan(v))
=1 2
Z π/4 0
log cos(u) + sin(v) cos(u) − sin(v)
π/4 v=0
du
=1 2
Z π/4 0
log cos(u) +√12 cos(u) −√12
! du = 1
4 Z π/4
−π/4
log cos(u) +√12 cos(u) −√12
! du
=1 4
Z π/2 0
log cos(s) + sin(s) + 1 cos(s) + sin(s) − 1
ds
u = s −π 4
=1 2
Z 1 0
log
1+t t(1−t)
1 + t2 dt (t = tan(s/2))
= −1 2
Z 1 0
log(t) 1 + t2dt −1
2 Z 1
0
log
1−t1+t
1 + t2 dt
= −1 2
Z 1 0
log(t) 1 + t2dt −1
2 Z 1
0
log(r) 1 + r2 dr
r = 1 − t 1 + t
= − Z 1
0
log(t)
1 + t2dt = [− log(t) arctan(t)]10+ Z 1
0
arctan(t) t dt
= Z 1
0
∞
X
n=1
(−1)n−1t2n−2 2n − 1 =
∞
X
n=1
(−1)n−1 2n − 1
Z 1 0
t2n−2dt
= X∞ n=1
(−1)n−1 (2n − 1)2 = G
and we are done.
