Problem 11564
(American Mathematical Monthly, Vol.118, April 2011) Proposed by Albert Stadler (Switzerland).
Prove that
Z ∞ 0
e−x(1 − e−6x)
x(1 + e−2x+ e−4x+ e−6x+ e−8x)dx = log 3 +√ 5 2
! .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We will prove a more general result: if a, b, c > 0 and d = a + b + 2c then Z ∞
0
(1 − e−ax)(1 − e−bx)e−cx
x(1 − e−dx) dx = log sin(π(a + c)/d) sin(πc/d)
. By letting a = 2, b = 6, and c = 1, then the integral is equal to
log sin(3π/10) sin(π/10)
= log 3 +√ 5 2
! .
By expanding the integrand, we obtain Z ∞
0
(1 − e−ax)(1 − e−bx)e−cx x(1 − e−dx) dx =
Z ∞ 0
X
n≥1
(−1)n((a + b)n− an− bn)xn−1 n!
X
k≥0
e−(kd+c)xdx
=X
n≥1
(−1)n((a + b)n− an− bn) n!
X
k≥0
Z ∞ 0
xn−1e−(kd+c)xdx
=X
k≥0
X
n≥1
(−1)n((a + b)n− an− bn) n(kd + c)n
=X
k≥0
log
1 + a kd + c
+ log
1 + b kd + c
− log
1 + a + b kd + c
=X
k≥0
log (kd + a + c)(kd + b + c) (kd + c)(kd + a + b + c)
= logY
k≥0
(kd + a + c)(kd + b + c) (kd + c)(kd + a + b + c)
= log
Γ(c/d)Γ((a + b + c)/d) Γ((a + c)/d)Γ((b + c)/d)
where in the last step we used the Weirstarss product zY
k≥1
1 + z k
e−kz = 1 Γ(z)eγz. By the reflection property, if z is not an integer then
Γ(z)Γ(1 − z) = π sin(πz). Hence, since d = a + b + 2c, it follows that
Γ(c/d)Γ((a + b + c)/d)
Γ((a + c)/d)Γ((b + c)/d) = sin(π(a + c)/d) sin(πc/d) .
