Problem 11848
(American Mathematical Monthly, Vol.122, June-July 2015) Proposed by I. Mez˝o (China).
Prove that
1
2πLi2(e−2π) = ln(2π) − 1 −5π 12 −
∞
X
k=1
(−1)kζ(2k) k(2k + 1) .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
By the definition of Bernoulli numbers, we have that
F (x) := 2
∞
X
k=1
(−1)k−1ζ(2k)x2k=
n
X
k=1
B2k
(2πx)2k
(2k)! = 2πx
e2πx− 1 − 1 + πx.
Hence
∞
X
k=1
(−1)k−1ζ(2k) k(2k + 1) = 2
∞
X
k=1
(−1)k−1ζ(2k)
2k − 2
∞
X
k=1
(−1)k−1ζ(2k) 2k + 1 =
Z 1 0
F (x) x dx −
Z 1 0
F (x)dx.
As regards the first integral, we get Z 1
0
F (x)
x dx = π + Z 2π
0
1 et− 1 −1
t
dt = π +ln(1 − e−t) − ln(t)2π
0+= π + ln(1 − e−2π) − ln(2π).
The second one goes as follows Z 1
0
F (x)dx = −1 + π 2 + 1
2π Z 2π
0
t
et− 1dt = −1 + π 2 − 1
2πLi2(e−2π) + ln(1 − e−2π) + π 12. Infact
Z 2π 0
t et− 1dt =
Z 2π 0
te−t 1 − e−tdt =
Z 2π 0
t
∞
X
k=1
e−ktdt =
∞
X
k=1
Z 2π 0
te−ktdt
= −
∞
X
k=1
e−kt
k2 +te−kt k
2π
0
= −Li2(e−2π) + 2π ln(1 − e−2π) +π2 6 . Finally
∞
X
k=1
(−1)k−1ζ(2k)
k(2k + 1) = π + ln(1 − e−2π) − ln(2π) −
−1 +π 2 − 1
2πLi2(e−2π) + ln(1 − e−2π) + π 12
= 1
2πLi2(e−2π) − ln(2π) + 1 + 5π 12.
