Problem 12031
(American Mathematical Monthly, Vol.125, March 2018) Proposed by O. Furdui (Romania).
(a) Prove
Z 1 0
Z 1 0
x
1 − xy
dx dy = 1 − γ, where {a} denotes the fractional part of a, and γ is Euler’s constant.
(b) Let k be a nonnegative integer. Prove Z 1
0
Z 1 0
x
1 − xy
k
dx dy = Z 1
0
1 x
k
dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. (a) follows from (b). For k = 1, by letting t = 1/x we have, Z 1
0
Z 1 0
x
1 − xy
dx dy = Z 1
0
1 x
dx =
Z ∞
1
{t}
t2 dt =
∞
X
n=1
Z n+1 n
t − n t2 dt
= lim
N →∞
N −1
X
n=1
ln(n + 1) − ln(n) − 1 n + 1
= lim
N →∞(ln(N ) − HN + 1) = 1 − γ where HN :=PN
n=11/n is the N -th harmonic number.
As regards (b), let t = 1x− y then x = t+y1 ,
∂(x,y)
∂(t,y)
=
1 (t+y)2, Z 1
0
Z 1 0
x
1 − xy
k
dx dy = Z 1
t=0
Z 1 y=1−t
1 t
k
dt dy (y + t)2 +
Z ∞
t=1
Z 1 y=0
1 t
k
dt dy (y + t)2
= Z 1
0
1 t
k 1 − 1
t + 1
dt +
Z ∞
1
1 tk
1 t − 1
t + 1
dt
= Z 1
0
1 t
k
dt − Z 1
0
1 t
k
dt t + 1+
Z ∞
1
dt tk+1(t + 1)
= Z 1
0
1 t
k
dt where the last equality holds because by letting s = 1/t,
Z 1 0
1 t
k
dt t + 1 =
Z ∞
1
{s}k
(1/s + 1)s2ds =
∞
X
n=1
Z n+1 n
(s − n)k s(s + 1)ds
= lim
N →∞
N −1
X
n=1
Z 1 0
sk
1
s + n− 1 s + 1 + n
ds
= lim
N →∞
Z 1 0
sk
N −1
X
n=1
1
s + n− 1 s + 1 + n
ds
= lim
N →∞
Z 1 0
sk
1
s + 1− 1 s + N
= Z 1
0
sk s + 1ds
= Z ∞
1
(1/t)k (1/t + 1)t2dt =
Z ∞
1
dt tk+1(t + 1).