Problem 11953
(American Mathematical Monthly, Vol.124, January 2017) Proposed by C. I. V˘alean (Romania).
Calculate
Z ∞
0
Z ∞
0
sin(x) sin(y) sin(x + y) xy(x + y) dx dy.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We have that I :=
Z ∞
0
Z ∞
0
sin(x) sin(y) sin(x + y)
xy(x + y) dx dy = lim
R→+∞
Z R x=0
Z R y=0
sin(x) sin(y) sin(x + y) xy(x + y) dx dy
= lim
R→+∞2 Z R
x=0
Z x y=0
sin(x) sin(y) sin(x + y) xy(x + y) dy
dx
= lim
R→+∞
1 2
Z R x=0
Z x y=0
sin(2x) + sin(2y) − sin(2x + 2y)
xy(x + y) dy
dx.
Then, after letting u = 2x, v = 2y = tu,
I = lim
R→+∞
Z R u=0
Z u v=0
sin(u) + sin(v) − sin(u + v)
uv(u + v) dv
du
= lim
R→+∞
Z R u=0
Z 1 t=0
sin(u) + sin(tu) − sin(u + tu) t(1 + t)u2 dt
du = lim
R→+∞
Z 1
t=0
fR(t) t(1 + t)dt, where
fR(t) :=
Z R u=0
sin(u) + sin(tu) − sin(u + tu)
u2 du =
−sin(u) + sin(tu) − sin(u + tu) u
R u=0
+ Z R
u=0
cos(u) − cos((1 + t)u)
u du + t
Z R u=0
cos(tu) − cos((1 + t)u)
u du.
By using the Frullani’s integral, Z +∞
0
cos(au) − cos(bu)
u dt = ln(b/a) for a, b > 0, it follows that
R→+∞lim fR(t) = f (t) := ln(1 + t) + t ln((1 + t)/t) = (1 + t) ln(1 + t) − t ln(t).
Note that for any t ∈ [0, 1],
|fR(t) − f (t)| ≤ Z ∞
R
3
u2du = 3 R. Hence, by the Dominated Convergence Theorem, we finally obtain
I = Z 1
0
f (t) t(1 + t)dt =
Z 1
0
ln(1 + t)
t dt − [ln(t) ln(1 + t)]10+ Z 1
0
ln(1 + t) t dt
= 2 Z 1
0
ln(1 + t) t dt = 2
∞
X
n=1
(−1)n+1 n
Z 1
0
tn−1dt = 2
∞
X
n=1
(−1)n+1
n2 = ζ(2) = π2 6 .
