Problem 11924
(American Mathematical Monthly, Vol.123, August-September 2016) Proposed by C. I. V˘alean (Romania).
Calculate
Z π/2
0
{tan x}
tan x dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let tn= arctan(n), then
I :=
Z π/2
0
{tan x}
tan x dx = Z π/2
0
1 −⌊tan x⌋
tan x
dx = π 2 −
∞
X
n=1
n Z tn+1
tn
dx tan x
= π 2 −
∞
X
n=1
n (ln(sin tn+1) − ln(sin tn)) .
Since sin(arctan t) = √ 1
1+1/t2 when t > 0, it follows that for N ≥ 2,
N −1
X
n=1
n (ln(sin tn+1) − ln(sin tn)) =
N −1
X
n=1
[(n + 1) ln(sin tn+1) − n ln(sin tn)] −
N −1
X
n=1
ln(sin tn+1)
= N ln(sin tN) −
N −1
X
n=0
ln(sin tn+1)
= N ln 1
p1 + 1/N2
!
− ln
N
Y
n=1
1 p1 + 1/n2
! . Hence, by taking the limit as N → +∞, we obtain
I = π 2 −1
2ln
∞
Y
n=1
1 + 1
n2
!
= π 2 −1
2ln sin(πi) πi
= π 2 −1
2ln sinh(π) π
=1 2ln
2π 1 − e−2π
, where we used the sine product formula
sin(πz) = πz
∞
Y
n=1
1 − z2
n2
.
