Problem 12221
(American Mathematical Monthly, Vol.127, December 2020) Proposed by N. Batir (Turkey).
Prove
Z 1 0
log(x6+ 1)
x2+ 1 dx = π log(6) 2 − 3G, whereG is the Catalan’s constantP∞
k=0(−1)k/(2k + 1)2.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. By letting x = 1/t, I :=
Z 1
0
log(x6+ 1) x2+ 1 dx =
Z +∞
1
log(1/t6+ 1) t2+ 1 dt =
Z +∞
1
log(t6+ 1) t2+ 1 dt − 6
Z +∞
1
log(t) t2+ 1dt.
We have Z ∞
1
log(t)
t2+ 1dt = − Z 1
0
log(x)
1 + x2dx = − Z 1
0 log(x) d(arctan(x)) = [− log(x) arctan(x)]10+ Z 1
0
arctan(x)
x dx
= 0 + Z 1
0
∞
X
k=0
(−1)kx2k 2k + 1 dx =
∞
X
k=0
(−1)k (2k + 1)2 = G.
Hence, since x6+ 1 = (x2+ 1)(x2+√
3x + 1)(x2−√
3x + 1), we find I = 1
2 Z 1
0
log(x6+ 1) x2+ 1 dx + 1
2 Z +∞
1
log(t6+ 1) t2+ 1 dt − 3
Z +∞
1
log(t) t2+ 1dt = 1
2 Z +∞
0
log(x6+ 1)
x2+ 1 dx − 3G
=J(0) + J(√
3) + J(−√ 3)
2 − 3G,
where, for |a| < 2, J(a) =
Z +∞
0
log(x2+ ax + 1) x2+ 1 dx =
Z π/2 0
log
1
cos2(t)+ a tan(t)
dt
= Z π/2
0
log 1 +a
2sin(2t) dt − 2
Z π/2 0
log(cos(t)) dt
= Z π/2
0
log 1 +a
2sin(t)
dt + π log(2).
In the last step we appliedRπ/2
0 log(cos(t)) dt = −π2log(2) which follows from Z π/2
0
log(cos(t)) dt = Z π/2
0
log(sin(t)) dt =1 2
Z π/2 0
log(sin(2t)/2) dt = 1 2
Z π/2 0
log(sin(t)) dt −π 4log(2).
By using the Wallis’ formulaRπ2
0 sin2n(x) dx =π241n 2nn, J(a) + J(−a) =
Z π/2 0
log
1 − a2
4 sin2(t)
dt + 2π log(2)
= −
∞
X
n=1
(a2/4)n n
Z π/2 0
sin2n(t) dt + 2π log(2)
= −π 2
∞
X
n=1
a2n 16nn
2n n
+ 2π log(2) = π log 1 +p1 − a2/4 2
!
+ 2π log(2).
Finally,
I = J(0) + J(√
3) + J(−√ 3)
2 − 3G = π log(2) + π log 34 + 2π log(2)
2 − 3G = π log(6)
2 − 3G.
