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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

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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 .



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