Problem 11926
(American Mathematical Monthly, Vol.123, August-September 2016) Proposed by O. Furdui (Romania).
Letk be an integer, k ≥ 2. Find
Ik :=
Z +∞
0
ln |1 − x|
x1+1/k dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We have that Z +∞
1
ln |1 − x|
x1+1/k dx = Z +∞
1
ln(x − 1) x1+1/k dx =
Z 0 1
ln(1/t − 1) (1/t)1+1/k d(1/t)
= Z 1
0
ln(1 − t) t1−1/k dt −
Z 1 0
ln t t1−1/k dt.
Now
Z 1 0
ln t
t1−1/k dt =h
−k2t1/k+ kt1/kln ti1
t=0+= −k2. Moreover if α ∈ (0, 2) then
− Z 1
0
ln(1 − t) tα dt =
Z 1 0
P
j≥1 tj
j
tα dt =X
j≥1
1 j
Z 1 0
tj−α dt =X
j≥1
1 j(j + 1 − α),
where we can swap the sum with the integral by monotone convergence theorem: for t ∈ (0, 1) and for any N ≥ 1,
0 ≤
N
X
j=1
tj
j ≤ − ln(1 − t) and t−αP
j≥1 tj
j = −ln(1−t)tα is integrable on (0, 1). Hence
Ik = k2+ Z 1
0
ln(1 − t) t1−1/k dt +
Z 1 0
ln(1 − t) t1+1/k dt
= k2−X
j≥1
1
j(j +1k)−X
j≥1
1 j(j −1k)
= k2+ 2X
j≥1
1
1
k2 − j2 = kπ cot(π/k) where in the last step we used the known identity
π cot(πz) =1
z + 2zX
j≥1
1 z2− j2
which holds for z 6∈ Z.
