Problem 11275
(American Mathematical Monthly, Vol.114, February 2007) Proposed by M. Becker (USA).
Find
Z ∞
y=0
Z ∞
x=y
(x − y)2log((x + y)/(x − y)) xy sinh(x + y) dx dy.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We change the variables: letting u = x + y and v = (x − y)/(x + y), the integral becomes
I = Z 1
v=0
Z ∞
u=0
−(vu)2log(v) (u2(1 − v2)/4) sinh(u)
∂(x, y)
∂(u, v)
du dv =
Z 1
v=0
−2v2log(v) 1 − v2 dv
·
Z ∞ u=0
u sinh(u)du
.
The first integral:
Z 1
v=0
−2v
2log(v) 1 − v2 dv = 2
Z 1
v=0
log(v) dv − 2 Z 1
v=0
log(v)
1 − v2dv = −2 − 2 Z 1
v=0
log(v) 1 − v2dv.
The second integral: let t = e−u then Z ∞
u=0
u
sinh(u)du = −2 Z 0
t=1
log(t) t−1− t ·
− 1 t
dt = −2 Z 1
t=0
log(t) 1 − t2dt.
Finally we find
2 Z 1
0
log(t) 1 − t2dt =
Z 1
0
log(t) 1 − t dt +
Z 1
0
log(t)
1 + t dt = −Dilog(0) + Z 1
0
log(t) d(log(1 + t))
= −Dilog(0) + [log(t) log(1 + t)]10− Z 1
0
log(1 + t)
t dt
= −Dilog(0) + Dilog(2) = −π2 6 −
π2
12 = −π2 4 where
Dilog(x) :=
Z x 1
log(t) 1 − t dt =
Z 1−x 0
log(1/(1 − s))
s ds
and
Dilog(0) :=
Z 1
0
1 s
∞
X
k=1
sk k ds =
∞
X
k=1
1 k2 =π2
6 and Dilog(2) :=
Z −1
0
1 s
∞
X
k=1
sk k ds =
∞
X
k=1
(−1)k
k2 = −π2 12 Hence
I =
−2 + π2
4
· π2
4 = π2(π2− 8) 16 .
