Problem 11902
(American Mathematical Monthly, Vol.123, April 2016) Proposed by C. I. Vˇalean (Romania).
Prove that
Z 1
0
Z 1
0
Z 1
0
x y
ny z
o nz x
o2
dx dy dz
= 1 − ζ(2) 2 −ζ(3)
2 +7ζ(6)
48 +ζ(2)ζ(3)
18 +ζ(3)2
18 +ζ(3)ζ(4) 12 .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
By symmetry, we have that Z 1
0
Z 1
0
Z 1
0
x y
ny z
o nz x
o2
dx dy dz = 3I1+ 3I2
where I1=
Z Z Z
0≤x≤y≤z≤1
x y y z
nz x
o2
dx dy dz and I2= Z Z Z
0≤x≤z≤y≤1
x y
ny z
o nz x
o2
dx dy dz.
Now, for m ≥ 4, let
F (m) = Z +∞
1
{u}2 um du =
∞
X
k=1
Z k+1 k
(u − k)2 um du =
∞
X
k=1
Z k+1 k
1
um−2 − 2k um−1 + k2
um
du
= 1
(m − 3)− 2 (m − 2)
∞
X
k=1
k
km−2 − k (k + 1)m−2
+ 1
(m − 1)
∞
X
k=1
k2
km−1− k2 (k + 1)m−1
= 1
(m − 3)−2ζ(m − 2)
(m − 2) +2ζ(m − 2) − ζ(m − 1)
(m − 1) = 1
(m − 3)− 2ζ(m − 2)
(m − 1)(m − 2)−ζ(m − 1) (m − 1). Hence
I1= Z 1
z=0
Z z y=0
Z y x=0
x z
nz x
o2
dx dy dz = Z 1
z=0
Z z x=0
x z
nz x
o2Z z y=x
dy
dx dz
= Z 1
z=0
Z z x=0
x z
nz x
o2
(z − x)dx dz u=z/x= Z 1
z=0
Z +∞
u=1
{u}2 u2
z − z u
z u2du dz
= Z +∞
u=1
{u}2 u4
1 − 1
u
du
Z 1
z=0
z2dz = F (4) − F (5)
3 .
Moreover,
I2= Z 1
y=0
Z y z=0
Z z x=0
x y
ny z
o nz x
o2
dx dz dy = Z 1
z=0
Z 1
y=z
Z z x=0
x y
ny z
o nz x
o2
dx dy dz
u=z/x,v=y/z
=
Z 1 z=0
Z 1/z v=1
Z +∞
u=1
{u} {v}
uv
2
· z2
u2du dv dz = F (4) Z 1
z=0
Z 1/z v=1
{v}2z2 v2 dv dz
= F (4) Z +∞
v=1
{v}2 v2
Z 1/v z=0
z2dz
!
dv = F (4) · F (5)
3 .
Finally, since ζ(2) = π2/6, ζ(4) = π4/90, and ζ(6) = π6/945 = 4ζ(2)ζ(4)/7, we have that 3I1+ 3I2= F (4) − F (5) + F (4) · F (5)
=
1 − ζ(2) 3 −ζ(3)
3
− 1 2−ζ(3)
6 −ζ(4) 4
+
1 −ζ(2) 3 −ζ(3)
3
· 1 2−ζ(3)
6 −ζ(4) 4
= 1 − ζ(2) 2 −ζ(3)
2 +7ζ(6)
48 +ζ(2)ζ(3)
18 +ζ(3)2
18 +ζ(3)ζ(4) 12 .
