Problem 11993
(American Mathematical Monthly, Vol.124, August-September 2017) Proposed by C. I. V˘alean (Romania).
Prove that
Z 1
0
ln(1 − x)(ln(1 + x))2
x dx= −π4
240.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
First solution. By letting a = ln(1 − x) and b = ln(1 + x) in the identity 6ab2= (a + b)3+ (a − b)3−2a3, we get
I:=
Z 1
0
ln(1 − x)(ln(1 + x))2
x dx= I1+ I2−2I3
6 where
I1= Z 1
0
(ln(1 − x2))3
x dx=1
2 Z 1
0
ln3(t)
1 − t dt (t = 1 − x2),
I2= Z 1
0
ln
1−x 1+x
3
x dx= 2
Z 1
0
ln3(t)
(1 − t)(1 + t)dt= Z 1
0
ln3(t) 1 − t dt+
Z 1
0
ln3(t)
1 + t dt (t = 1 − x 1 + x), I3=
Z 1 0
(ln(1 − x))3
x dx=
Z 1 0
ln3(t)
1 − t dt (t = 1 − x).
Hence, by the uniform convergence,
I= 1 6
1
2 + 1 − 2
Z 1 0
ln3(t) 1 − t dt+
Z 1
0
ln3(t) 1 + t dt
= 1 6 −1
2
∞
X
n=0
Z 1
0
tnln3(t) dt +
∞
X
n=0
(−1)n Z 1
0
tnln3(t) dt
!
= 1 6
6 2
∞
X
n=0
1
(n + 1)4 −6
∞
X
n=0
(−1)n (n + 1)4
!
= ζ(4)
2 −7ζ(4)
8 = −3ζ(4) 8 = −3
8 ·π4
90 = − π4 240 where we used the fact that for any non-negative integers m, n,
Z 1
0
tnlnm(t) dt = 1
n+ 1tn+1lnm(t)1
0− m n+ 1
Z 1
0
tnlnm−1(t) dt = (−1)mm!
(n + 1)m+1.
Second solution. For x ∈ [0, 1],
(ln(1 + x))2= 2
∞
X
n=1
Hn(−x)n+1 n+ 1 where Hn =Pn
k=1 1
k. Hence, by the uniform convergence, I: =
Z 1
0
ln(1 − x)(ln(1 + x))2
x dx= 2
∞
X
n=1
(−1)n+1Hn
n+ 1 Z 1
0
xnln(1 − x) dx
= 2
∞
X
n=1
(−1)nHnHn+1
(n + 1)2 = 2
∞
X
n=1
(−1)n−1Hn2
n2 −
∞
X
n=1
(−1)n−1Hn
n3
! ,
where we used the fact that Z 1
0
xnln(1 − x) dx = − 1 n+ 1
Z 1 0
ln(1 − x) d(1 − xn+1)
= − 1
n+ 1ln(1 − x)(1 − xn+1)1
0− 1 n+ 1
Z 1 0
1 − xn+1 1 − x dx
= − 1 n+ 1
Z 1
0 n
X
k=0
xkdx= −Hn+1
n+ 1. Since the values of the two following Euler sums are known,
∞
X
n=1
(−1)n−1Hn
n3 =41
16ζ(4) − 2Li4
1 2
−7
4log(2)ζ(3) +1
2log2(2)ζ(2) − 1
12log4(2),
∞
X
n=1
(−1)n−1Hn2
n2 =11
4 ζ(4) − 2Li4
1 2
−7
4log(2)ζ(3) +1
2log2(2)ζ(2) − 1
12log4(2), we may conclude that
I= 2 41
16ζ(4) −11 4 ζ(4)
= −3 8 ·π4
90 = −π4 240.
