Problem 12184
(American Mathematical Monthly, Vol.127, May 2020) Proposed by P. Perfetti (Italy).
Prove
Z ∞
1
ln(x4− 2x2+ 2) x√
x2− 1 dx = π ln(2 +√ 2).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. After letting t =√
x2− 1, we have that Z ∞
1
ln(x4− 2x2+ 2) x√
x2− 1 dx = Z ∞
0
ln(t4+ 1)
t2+ 1 dt = I(π/4) + I(−π/4) where
I(α) :=
Z ∞
0
ln(t2+ 2 sin(α)t + 1) t2+ 1 dt.
Now, for α ∈ (0, π/2), I′(α) =
Z ∞
0
2 cos(α)t
(t2+ 1)(t2+ 2 sin(α)t + 1)dt
= 1
tan(α)
Z ∞
0
dt t2+ 1−
Z ∞
0
dt t2+ 2 sin(α)t + 1
= π
2 tan(α)− 1 sin(α)
π 2 − α
= −π tan(α/2)
2 + α
sin(α). Since
I(0) = Z ∞
0
ln(t2+ 1) t2+ 1 dt =
Z π/2 0
ln(tan2(s) + 1)
tan2(s) + 1 d(tan(s)) = −2 Z π/2
0
ln(cos(s)) ds = π ln(2) it follows
I(α) = I(0) + Z α
0
I′(α) dα = π ln (2 cos(α/2)) + Z α
0
a sin(a)da.
Hence
I(α) + I(−α) = π ln 4 cos2(α/2) = π ln(2(1 + cos(α)).
Finally, for α = π/4 we find
I(π/4) + I(−π/4) = π ln(2 + 2 cos(π/4)) = π ln(2 +√ 2)
and we are done.
Remark. Let J =Rπ/2
0 ln(cos(s)) ds =Rπ/2
0 ln(sin(s)) ds. Then 2J =
Z π/2 0
ln(sin(s) cos(s)) ds = Z π/2
0
ln(sin(2s)) ds − Z π/2
0
ln(2) ds
= 1 2
Z π 0
ln(sin(s)) ds −π ln(2)
2 = J − π ln(2) 2 Hence 2J = −π ln(2).
