By the log-convexity of the gamma function on the positive real axis, it follows that for t log(Γ(2

Problem 11975

(American Mathematical Monthly, Vol.124, April 2017) Proposed by I. Mez˝o (China).

Letx be a real number in [0, 1). Prove that (1 − γ)^x

1 − x ≤ Z 1

0

Γ^x(t) dt ≤ 1 1 − x whereΓ is the gamma function and γ is the Euler-Mascheroni constant.

Solution proposed by Moubinool Omarjee, Lyc´ee Henri IV, Paris, France, and Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, Italy.

Solution. By the log-convexity of the gamma function on the positive real axis, it follows that for t ∈ [0, 1],

0 = log(Γ(2)) = log(Γ(t(t + 1) + (1 − t)(t + 2)) ≤ t log(Γ(t + 1)) + (1 − t) log(Γ(t + 2)) that is, since Γ(t + 2) = (t + 1)Γ(t + 1),

1 ≤ Γ^t(t + 1)Γ^1−t(t + 2) = (t + 1)^1−tΓ(t + 1) and

m := min

t∈[0,1]

1

(t + 1)^1−t ≤ Γ(t + 1).

On the other hand, log-convexity implies convexity and therefore for t ∈ [0, 1], Γ(t + 1) ≤ max{Γ(0 + 1), Γ(1 + 1)} = 1.

Hence, putting all together we get

m ≤ Γ(t + 1) = tΓ(t) ≤ 1 and, for x ∈ [0, 1), we have

m^x

t^x ≤ Γ^x(t) ≤ 1 t^x. Finally, by integration with respect to t over [0, 1], we obtain

m^x 1 − x = m^x

Z 1

0

1 t^xdt ≤

Z 1

0

Γ^x(t) dt ≤ Z 1

0

1

t^xdt = 1 1 − x

which is a more general result because m ≈ 0.815154 > 1/2 > (1 − γ).