Problem 12207
(American Mathematical Monthly, Vol.127, October 2020) Proposed by O. Furdui and A. Sintamarian (Romania).
Letf : [0, 1] → R be a continuous function satisfyingR1
0 f (x) dx = 1. Evaluate
n→∞lim n ln(n)
Z 1 0
xnf (xn) ln(1 − x) dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let t = xn, then x = t1/n, dx = t1/n−1dt/n and n
ln(n) Z 1
0
xnf (xn) ln(1 − x) dx = − Z 1
0
f (t)un(t) dt where
un(t) = −t1/nln(1 − t1/n) ln(n) . For all t ∈ (0, 1),
n→∞lim un(t) = − lim
n→∞
ln(1 − eln(t)/n)
ln(n) = − lim
n→∞
ln(− ln(t)/n)
ln(n) = − lim
n→∞
ln(− ln(t)) − ln(n)
ln(n) = 1.
Moreover, for n ≥ 3,
0 ≤ un(t) ≤ −ln(1 − t1/n)
ln(n) ≤ −ln((1 − t)/n)
ln(n) ≤ −ln(1 − t) + 1 andR1
0(− ln(1 − t) + 1) dt = 2.
Therefore, by the Dominated Convergence Theorem,
n→∞lim n ln(n)
Z 1 0
xnf (xn) ln(1 − x) dx = − lim
n→∞
Z 1 0
f (t)un(t) dt = − Z 1
0
f (t) dt = −1.
