Problem 12228
(American Mathematical Monthly, Vol.128, January 2021) Proposed by H. Grandmontagne (France).
Prove
Z 1 0
(ln(x))2ln 2√
x/(x2+ 1)
x2− 1 dx = 2G2, whereG is Catalan’s constantP∞
n=0(−1)n/(2n + 1)2.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution.
We consider, for x > 0, the digamma function ψ(x) and its derivatives of order m ≥ 1:
ψ(x) = Γ′(x)
Γ(x) = −γ +
∞
X
n=0
1
n + 1− 1 n + x
, ψ(m)(x) = (−1)m+1m!
∞
X
n=0
1 (n + x)m+1. We recall some known values of ψ(x) and ψ(m)(x) up to order 3 at x = 1/2, 1/4, 3/4:
ψ 1 2
= −γ − 2 ln(2), ψ′ 1 2
= π2
2 , ψ′′ 1 2
= −14ζ(3), ψ′′′ 1 2
= π4, ψ 1
4
= −γ − 3 ln(2) −π
2, ψ′ 1 4
= 8G + π2, ψ′′ 1 4
= −56ζ(3) − 2π3, ψ 3
4
= −γ − 3 ln(2) +π
2, ψ′ 3 4
= −8G + π2, ψ′′ 3 4
= −56ζ(3) + 2π3. Moreover
ψ′′′ 1 4
+ ψ′′′ 1 4
= 16π4. We need also the following two identities. Let Hk=Pk
n=1 1
n then, for x > 0, S2(x) :=
∞
X
k=1
Hk
(k + x)2 = (γ + ψ(x)) ψ′(x) −ψ′′(x)
2 , (1)
and
S3(x) :=
∞
X
k=1
Hk
(k + x)3 = −(ψ′(x))2
2 −(γ + ψ(x)) ψ′′(x)
2 +ψ′′′(x)
4 . (2)
Indeed, we have that S2(x) =
∞
X
k=1
Hk
(k + x)2 =
∞
X
k=1
1 (k + x)2
k
X
n=1
1 n =
∞
X
n=1
1 n
∞
X
k=n
1 (k + x)2 =
∞
X
n=1
ψ′(x + n) n
= −
∞
X
n=1
1 n
Z 1 0
tx+n−1ln(t) 1 − t dt = −
Z 1 0
tx−1ln(t) 1 − t
∞
X
n=1
tn n dt =
Z 1 0
tx−1ln(t) ln(1 − t) 1 − t dt
= lim
y→0+
∂2
∂x∂y Z 1
0
tx−1(1 − t)y−1dt = lim
y→0+
∂2
∂x∂y
Γ(x)Γ(y) Γ(x + y)
= lim
y→0+
∂
∂y
Γ(x)Γ(y)
Γ(x + y)(ψ(x) − ψ(x + y))
= lim
y→0+
Γ(x)Γ(y)
Γ(x + y)((ψ(y) − ψ(x + y))(ψ(x) − ψ(x + y)) − ψ′(x + y))
= lim
y→0+
1
y − γ + O(y)
−1
y − γ + O(y) − ψ(x)
−ψ′(x)y −ψ′′(x)
2 y2+ O(y3)
− (ψ′(x) + ψ′′(x)y + O(y2))
= (γ + ψ(x)) ψ′(x) −ψ′′(x) 2 ,
and the proof of (1) is complete. We obtain (2) by differentiating (1) with respect to x.
The given integral can be written as Z 1
0
(ln(x))2ln 2√
x/(x2+ 1)
x2− 1 dx = ln(2)I1+I2
2 − I3= 2G2 where at the last step we applied
I1:=
Z 1 0
ln2(x)
x2− 1dx = −
∞
X
k=0
Z 1 0
ln2(x)x2kdx = −2
∞
X
k=0
1
(2k + 1)3 = −7ζ(3) 4 , I2:=
Z 1 0
ln3(x) x2− 1dx = 6
∞
X
k=0
1
(2k + 1)4 = π4 16,
and, by using (2) with the given evaluations listed at the beginning,
I3:=
Z 1 0
ln2(x) ln(x2+ 1) x2− 1 dx = 2
∞
X
k=0
Z 1 0
ln2(x)x2k
k
X
n=1
(−1)n n dx = 2
∞
X
k=0
1 (2k + 1)3
k
X
n=1
(−1)n n
= −2
∞
X
k=1
Hk
(2k + 1)3+ 2
∞
X
k=1
Hk
(4k + 1)3+ 2
∞
X
k=1
Hk
(4k + 3)3
= −1 4S3
1 2
+ 1
32
S3
1 4
+ S3
3 4
= −1 4
−π4
8 − 14 ln(2)ζ(3) +π4 4
+ 1 32
−(8G + π2)2
2 +(3 ln(2) + π2)(−56ζ(3) − 2π3)
2 −(−8G + π2)2
2 +(3 ln(2) −π2)(−56ζ(3) + 2π3)
2 + 4π4
= −2G2+π4
32−7 ln(2)ζ(3)
4 .