Problem 11805
(American Mathematical Monthly, Vol.121, December 2014) Proposed by G. Glebov and S. Fraser (Canada).
(a) Show that
∞
X
k=0
(−1)k (3k + 1)3 +
∞
X
k=0
(−1)k
(3k + 2)3 = 5π3√ 3 243 ,
∞
X
k=0
(−1)k (3k + 1)3 −
∞
X
k=0
(−1)k
(3k + 2)3 = 13 18ζ(3).
(b) Prove that
ζ(3) = 9 13
Z 1
0
(ln x)2
x3+ 1dx −18 13
∞
X
k=0
(−1)k (3k + 2)3.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
(a) We note that
A :=
∞
X
k=0
(−1)k (3k + 1)3 =
∞
X
k=0
1 (6k + 1)3 −
∞
X
k=0
1 (6k + 4)3, B :=
∞
X
k=0
(−1)k (3k + 2)3 =
∞
X
k=0
1 (6k + 2)3 −
∞
X
k=0
1 (6k + 5)3. Hence
A − B = −Li3(−1) +
∞
X
k=1
1 6k3−
∞
X
k=0
1
(6k + 3)3 = −26
27Li3(−1) =13 18ζ(3).
Moreover, by letting ω = eiπ/3 we have
Li3(ω) =
∞
X
k=1
ωk k3 =
∞
X
k=1
1 6k3 −
∞
X
k=0
1
(6k + 3)3+ ω
∞
X
k=0
1
(6k + 1)3 − ω
∞
X
k=0
1 (6k + 4)3
− ω
∞
X
k=0
1
(6k + 2)3 + ω
∞
X
k=0
1 (6k + 5)3
= −ζ(3)
36 + ωA − ωB = −ζ(3)
36 +A − B 2 +i√
3(A + B)
2 .
It is known that polylogarithms satisfy the following identity Lin(e2πix) + (−1)nLin(e−2πix) = −(2πi)n
n! Bn(x)
where Bn is the Bernoulli polynomial of degree n ≥ 1. For n = 3 and x = 1/6, it becomes 2iIm (Li3(ω)) = Li3(ω) − Li3(ω) = 4π3i
3 B3(1/6) =5π3i 81 . Hence
A + B = 2√ 3
3 Im(Li3(ω)) = 5π3√ 3 243 .
(b) If |z| = 1 and z 6= 1 then Z 1
0
(ln x)2
x − z dx = −1 z
Z 1 0
(ln x)2
1 − (x/z)dx = −1 z
Z 1 0
(ln x)2
∞
X
k=0
xk
zk dx = −
∞
X
k=0
1 zk+1
Z 1 0
(ln x)2xkdx
= −
∞
X
k=0
1 zk+1
xk+1 (ln x)2
k + 1 − 2 ln x
(k + 1)2 + 2 (k + 1)3
1
0
= − 2
∞
X
k=0
1
zk+1(k + 1)3 = −2Li3(z).
Hence
Z 1 0
(ln x)2 x3+ 1dx = 1
3 Z 1
0
(ln x)2
x + 1 dx − ω 3
Z 1 0
(ln x)2
x − ω dx −ω 3
Z 1 0
(ln x)2 x − ω dx
=2
3(−Li3(−1) + ωLi3(ω) + ωLi3(ω))
=ζ(3) 2 +2
3Re (Li3(ω)) +2√ 3
3 Im (Li3(ω))
=ζ(3) 2 +2
3
−ζ(3)
36 +A − B 2
+ (A + B)
=13ζ(3) 18 +5√
3π3 243
which is equivalent to the required equality.
