Problem 12127
(American Mathematical Monthly, Vol.126, August-September 2019) Proposed by O. Furdui and A. Sintamarian (Romania).
Calculate
Z 1 0
Li2(1) − Li2(x) 1 − x
2
dx.
whereLi2(z) =P∞
k=1zk/k2 is the dilogarithm function.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Since Li2(1) − Li2(x) = Li2(1 − x) + ln(x) ln(1 − x), the given integral I can be written as I =
Z 1 0
Li2(1 − x) + ln(x) ln(1 − x) 1 − x
2
dx = Z 1
0
Li2(x) + ln(x) ln(1 − x) x
2
dx
= Z 1
0
Li22(x) x2 dx + 2
Z 1 0
Li2(x) ln(x) ln(1 − x)
x2 dx +
Z 1 0
ln2(x) ln2(1 − x)
x2 dx.
The first integral:
Z 1 0
Li22(x) x2 dx =
∞
X
n=1
1 n2
Z 1 0
xn−2Li2(x) dx =
∞
X
n=1
1 n2
∞
X
n=1
1 k2
Z 1 0
xn+k−2dx
=
∞
X
n=1
∞
X
k=1
1 n2k2(n + k − 1)
= ζ(3) + ζ(2)
∞
X
n=2
1 n2(n − 1) −
∞
X
n=2
Hn−1
n2(n − 1)2. The second integral:
Z 1 0
Li2(x) ln(x) ln(1 − x)
x2 dx =
∞
X
n=1
1 n2
Z 1 0
xn−2ln(x) ln(1 − x) dx = −
∞
X
n=1
1 n2
∞
X
k=1
1 k
Z 1 0
xn+k−2ln(x)) dx
=
∞
X
n=1
∞
X
k=1
1 n2k(n + k − 1)2
= ζ(3) − ζ(2)
∞
X
n=2
1 n2(n − 1) +
∞
X
n=2
Hn−(2)1
n2(n − 1)+
∞
X
n=2
Hn−1
n2(n − 1)2. The third integral:
Z 1 0
ln2(x) ln2(1 − x) x2 dx = 2
∞
X
n=1
Hn
n + 1 Z 1
0
xn−1ln2(x) dx = 4
∞
X
n=1
Hn
(n + 1)n3. Hence, by partial fraction decomposition,
I = 3ζ(3) − ζ(2)
∞
X
n=2
1
n2(n − 1)+ 2
∞
X
n=2
Hn−(2)1
n2(n − 1)+
∞
X
n=2
Hn−1
n2(n − 1)2 + 4
∞
X
n=1
Hn (n + 1)n3
= 3ζ(3) − ζ(2)(2 − ζ(2)) + (2ζ(3) − 3ζ(4)/2) + (−2ζ(2) + 3ζ(3)) + (4ζ(2) − 8ζ(3) + 5ζ(4))
= ζ(2)2+7ζ(4) 2 = π4
15
where we used the known multiple zeta values
∞
X
n=2
Hn−1
n2 = ζ(2, 1) = ζ(3),
∞
X
n=2
Hn−1
n3 = ζ(3, 1) = ζ(4) 4 ,
∞
X
n=2
Hn−(2)1
n2 = ζ(2, 2) = 3ζ(4) 4 .
