Problem 11400
(American Mathematical Monthly, Vol.115, December 2008) Proposed by P. Bracken (USA).
Letζ be the Riemann zeta function. Evaluate
∞
X
n=1
ζ(2n) n(n + 1) in closed form.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
For 0 ≤ x < 1
∞
X
n=1
ζ(2n) n xn=
∞
X
n=1
1 n
∞
X
k=1
√ x k
2n
=
∞
X
k=1
∞
X
n=1
1 n
√ x k
2n
= − log
∞
Y
k=1
1 −
√ x k
2!
= − log sin(π√ x) π√
x
.
Hence, because of uniform convergence of the series, we get the desired closed form
∞
X
n=1
ζ(2n) n(n + 1) =
Z 1
0
∞
X
n=1
ζ(2n) n xn = −
Z 1
0
log sin(π√ x) π√
x
dx
= −
Z 1
0
log sin(π√
x) dx + Z 1
0
log (π) dx +1 2
Z 1
0
log (x) dx
= − 2 π2
Z π 0
log (sin(t)) t dt + log(π) −1 2
= log(2) + log(π) −1
2 = log(2π) − 1
2 ≈ 1.337877.
Note that the nontrivial integral can be computed by exploiting symmetry:
I = Z π
0
log (sin(t)) t dt = Z π
0
log (sin(π − t)) (π − t) dt = π Z π
0
log (sin(t)) dt − I and
J =
Z π 0
log (sin(t)) dt = Z π
0
log (2 sin(t/2) cos(t/2)) dt
= π log(2) + 2 Z π/2
0
log sin(t) dt + 2 Z π/2
0
log cos(t) dt
= π log(2) + 2 Z π/2
0
log sin(t) dt + 2 Z π/2
0
log sin(π/2 − t) dt = π log(2) + 2J
therefore J = −π log(2) and I = πJ/2 = −π2log(2)/2.
