Problem 12199
(American Mathematical Monthly, Vol.127, August-September 2020) Proposed by S. Sharma (India).
Prove
Z ∞
0
x sinh(x)
3 + 4 sinh2(x)dx = π2 24.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. By letting e−x= t, we have that Z ∞
0
x sinh(x)
3 + 4 sinh2(x)dx = −1 2
Z 1 0
log(t)(1/t − t) 3 + (1/t2+ t2− 2)
dt t = −1
2 Z 1
0
log(t)(1 − t2) t4+ t2+ 1 dt
= −1 2
Z 1 0
log(t)(1 − t2)2
1 − t6 dt = −1 2
Z 1 0
log(t)(1 − 2t2+ t4)
∞
X
k=0
t6kdt
=1
2 −
Z 1 0
log(t)
∞
X
k=0
t6kdt + 2 Z 1
0
log(t)
∞
X
k=0
t6k+2dt − Z 1
0
log(t)
∞
X
k=0
t6k+4dt
!
=1
2 −
∞
X
k=0
Z 1 0
log(t)t6kdt + 2
∞
X
k=0
Z 1 0
log(t)t6k+2dt −
∞
X
k=0
Z 1 0
log(t)t6k+4dt
!
=1 2
∞
X
k=0
1
(6k + 1)2 − 2
∞
X
k=0
1 (6k + 3)2 +
∞
X
k=0
1 (6k + 5)2
!
=1 2
∞
X
k=0
1
(2k + 1)2 − 3
∞
X
k=0
1 (6k + 3)2
!
=1 2
1 − 3
32
∞ X
k=0
1
(2k + 1)2 =1 3
∞
X
k=0
1
(2k + 1)2 =π2 24
where the interchange of summation and integration is justified by Fubini’s theorem and we applied Z 1
0
log(t)tndt = − Z ∞
0
xe−(n+1)xdx = − 1 (n + 1)2
Z ∞
0
se−sds = − 1 (n + 1)2.
