Problem 11697
(American Mathematical Monthly, Vol.120, February 2013)
Proposed by Moubinool Omarjee (France).
Let n and q be integers, with 2n > q ≥ 1. Let
f (t) = Z
Rq
e−t(x2n1 +···+x2nq )
1 + x2n1 + · · · + x2nq dx1· · · dxq. Prove that limt→+∞tq/2nf (t) = n−q(Γ(1/2n))q.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
First we remind the definition of the Gamma function for x > 0, Γ(x) = R∞
0 tx−1e−tdt. For i = 1, . . . , q, let x2ni = ui, then dxi= 2n1 (ui)2n1−1dui and
f (t) = 2q Z
[0,+∞)q
e−t(x2n1 +···+x2nq )
1 + x2n1 + · · · + x2nq dx1· · · dxq
= n−q Z
[0,+∞)q
i=1ui2n1−1e−tui 1 + u1+ · · · + uq
du1· · · duq
= n−q Z
[0,+∞)q q
Y
i=1
ui2n1−1e−tui
Z ∞
0
e−s(1+u1+···+uq)ds
du1· · · duq
= n−q Z ∞
0
e−s Z
[0,+∞)q q
Y
i=1
u
1 2n−1
i e−(s+t)uidu1· · · duq
! ds
= n−q Z ∞
0
e−s
Z +∞
0
u2n1−1e−(s+t)udu
q ds
= n−q Z ∞
0
e−s
1
(s + t)1/2n Z +∞
0
r2n1−1e−rdr
q ds
= n−q Z ∞
0
e−s
Γ(1/2n) (s + t)1/2n
q
ds = n−q(Γ(1/2n))q Z ∞
0
e−s (s + t)q/2nds.
Hence
t→+∞lim tq/2nf (t) = n−q(Γ(1/2n))q lim
t→+∞
Z ∞
0
t s + t
q/2n
e−sds = n−q(Γ(1/2n))q
because
0 ≤
t s + t
q/2n
≤ 1, lim
t→+∞
t s + t
q/2n
= 1 and Z ∞
0
e−sds = Γ(1) = 1.
It seems that the condition 2n > q is not necessary.
