Problem 11629
(American Mathematical Monthly, Vol.119, March 2012) Proposed by Olivier Oloa (France).
Let
f (σ) = Z 1
0
xσ
1
log(x) + 1 1 − x
2 dx..
(a) Show that f (0) = log(2π) − 3/2.
(b) Find a closed form expression for f (σ) for σ > 0.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
For n ≥ m and σ > 0 let
Fn,m(σ) = Z 1
0
xσlog(x)n (1 − x)m dx.
Then it is known that F0,0(σ) = 1
σ + 1 , F1,1(σ) = d dσ
Z 1 0
1 − xσ
1 − x dx = (Ψ(σ + 1) + γ)0= −Ψ0(σ + 1) = −Ψ0(σ) + 1 σ2 where Ψ(σ) = Γ0(σ)/Γ(σ). For n > 1, by integration by parts
Fn,n(σ) = 1 n − 1
xσlog(x)n (1 − x)n−1
1
0
− 1
n − 1 Z 1
0
σxσ−1log(x)n+ nxσ−1log(x)n−1
(1 − x)n−1 dx
= 0 − σFn,n−1(σ − 1) + nFn−1,n−1(σ − 1)
n − 1 = −σFn−1,n−10 (σ − 1) + nFn−1,n−1(σ − 1) n − 1
This recurrence relation implies that
F2,2(σ) = σΨ00(σ) + 2Ψ0(σ) , F3,3(σ) = −σ 2
Ψ000(σ) − 3(σ − 1/2)Ψ00(σ) − 3Ψ0(σ).
Hence
f00(σ) = Z 1
0
xσ
1
log(x) + 1 1 − x
2
log(x)2dx
= F0,0(σ) + 2F1,1(σ) + F2,2(σ)
= 1
σ + 1− 2Ψ0(σ) + 2
σ2 + σΨ00(σ) + 2Ψ0(σ).
Therefore,
f0(σ) = Z
f00(σ) dσ = log(σ + 1) − 2Ψ(σ) − 2
σ+ σΨ0(σ) + Ψ(σ) + a and
f (σ) = Z
f0(σ) dσ = (σ + 1) log(σ + 1) − (σ + 1) − 2 log(Γ(σ)) − 2 log(σ) + σΨ(σ) + aσ + b
= (σ + 1) log(σ + 1) − 2 log(Γ(σ + 1)) + σΨ(σ + 1) + (a − 1)σ + b − 2 for some constants a, b.
As σ → +∞ it is easy to verify that f (σ) goes to zero. On the other hand f (σ) = (1 + a)σ − 1/2 − log(2π) + b + O(1/σ)
which implies that a = −1 and b = log(2π) + 1/2. Finally
f (σ) = (σ + 1) log(σ + 1) − 2 log(Γ(σ + 1)) + σΨ(σ + 1) − 2σ + log(2π) − 3/2 and, since f (σ) is continuous, we have that
f (0) = log(2π) − 3/2.
Note that by a similar method, and by using F3,3(σ), it is possible to prove that
Z 1 0
1
log(x) + 1 1 − x
2log(x)
1 − x dx = −5/4 − γ/2 + log(2π)/2.
