Problem 12189
(American Mathematical Monthly, Vol.127, June-July 2020) Proposed by H. Katsuura (USA).
Evaluate
Z 1 0
(k + 1)xk−Pk j=0xjk xk(k+1)−1 dx , wherek is a positive integer.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let 0 < t < 1, then Z t
0
(k + 1)xk−Pk j=0xjk xk(k+1)−1 dx =
Z t 0
1
xk(k+1)−1d(xk+1) − Z t
0
1 xk−1dx
= Z tk+1
0
dx xk−1 −
Z t 0
dx xk−1 =
Z t tk+1
dx 1 − xk
= 1 k
−ln(1 − xk) xk−1
t
tk+1
+k − 1 k
Z t tk+1
−ln(1 − xk) xk dx.
Now, we note that
0 ≤ −ln(1 − xk)
xk ≤ −2kln(1 − x) ∀x ∈ [1/2, 1), and − Z 1
1/2
ln(1 − x) = 1 + ln(2) 2 imply thatR1
1/2
−ln(1−xk)
xk dx is finite and therefore
t→1lim− Z t
tk+1
−ln(1 − xk)
xk dx = 0.
Moreover, by letting t = 1 − s with s → 0+,
−ln(1 − xk) xk−1
t
tk+1
= −ln(1 − (1 − s)k)
(1 − s)k−1 +ln(1 − (1 − s)k(k+1)) (1 − s)k2−1
= −ln(ks + o(s))
1 + o(1) +ln(k(k + 1)s + o(s)) 1 + o(1)
= (− ln(ks) + ln(k(k + 1)s)) (1 + o(1))
= ln(k + 1) + o(1) → ln(k + 1).
Therefore, we may conclude that Z 1
0
(k + 1)xk−Pk j=0xjk
xk(k+1)−1 dx = lim
t→1−
Z t 0
(k + 1)xk−Pk j=0xjk
xk(k+1)−1 dx = ln(k + 1)
k .
