Problem 11937
(American Mathematical Monthly, Vol.123, November 2016) Proposed by J. C. Sampedro (Spain).
Lets be a complex number such that Re(s) > 0. Prove Z 1
0
Z 1
0
(xy)s−1−y
(1 − xy) log(xy)dxdy = Γ′(s) Γ(s).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We show a more general result:
Let s be a complex number such that Re(s) > 0 and let n be a positive integer. Then Z 1
0
Z 1
0
(xy)s−1−yn
(1 − xy) log(xy)dxdy = Γ′(s)
Γ(s) −log(n!) n .
Let u = xy, then x = u/y, dx = du/y and Z 1
0
Z 1
0
(xy)s−1−yn
(1 − xy) log(xy)dxdy = Z 1
u=0
Z 1
y=u
us−1−yn (1 − u) log(u)
dudy y
= Z 1
u=0
1 (1 − u) log(u)
Z 1
y=u
us−1
y −yn−1
dy
du
= Z 1
u=0
1 (1 − u) log(u)
us−1log(y) −yn n
1
y=u
du
= − Z 1
0
us−1
(1 − u)+ 1 log(u)
du +
Z 1
0
1 − un n(1 − u)−1
du log(u)
= − Z 1
0
us−1
(1 − u)+ 1 log(u)
du − 1
n
n−1
X
k=1
Z 1
0
uk−1 log(u) du
=Γ′(s)
Γ(s) −log(n!) n ,
where in the last step we used a known integral representation of the digamma function (see 4.281.4 in Table of integrals series and products by Gradshteyn and Ryzhik),
Z 1
0
us−1
(1 − u)+ 1 log(u)
du = −Γ′(s)
Γ(s) for Re(s) > 0 and the fact that for t ≥ 0,
f (t) :=
Z 1
0
ut−1
log(u)du = log(t + 1) which can be easily obtained by noting that f (0) = 0 and
f′(t) = Z 1
0
utdu = 1 t + 1.
