Problem 11277
(American Mathematical Monthly, Vol.114, March 2007) Proposed by Prithwijit De (Ireland).
Find
Z π/2 φ=0
Z π/2 θ=0
log(2 − sin θ cos φ) sin θ
2 − 2 sin θ cos φ + sin2θ cos2φdθ dφ.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We will show that I =
Z π/2 φ=0
Z π/2 θ=0
log(2 − sin θ cos φ)
1 + (1 − sin θ cos φ)2 · sin θ dθ dφ = π2log(2) 16 .
This is a surface integral over the first octant of the unit sphere with respect to the spherical coordinates. The same integral with respect to the cartesian coordinates is
I = Z 1
x=0
Z √1−x2 y=0
log(2 − x)
1 + (1 − x)2· 1 p1 − x2− y2
dy dx.
After performing the improper integral with respect to y we obtain Z 1
x=0
log(2 − x) 1 + (1 − x)2
"
arctan y
p1 − x2− y2
!#√1−x2 y=0
dx = π 2
Z 1
0
log(2 − x)
1 + (1 − x)2 dx = π 2
Z 1
0
log(1 + x) 1 + x2 dx The Catalan’s constant K ≈ 0.9159655942 has several integral representation. Some of them are the following ones:
K = − Z 1
0
log(x)
1 + x2dx = − Z 1
0
log((1 − x2)/2)
1 + x2 dx = − Z 1
0
log((1 − x)/√ 2) 1 + x2 dx.
Therefore
K = π log(2)
4 −
Z 1
0
log(1 − x) 1 + x2 dx −
Z 1
0
log(1 + x)
1 + x2 dx = π log(2)
8 −
Z 1
0
log(1 − x) 1 + x2 dx and we find that
Z 1
0
log(1 + x)
1 + x2 dx = π log(2)
4 −π log(2)
8 =π log(2)
8 .
Finally
I = π 2
Z 1
0
log(1 + x)
1 + x2 dx = π
2 · π log(2)
8 = π2log(2) 16 .