Problem 11709
(American Mathematical Monthly, Vol.120, May 2013) Proposed by Moubinool Omarjee (France).
Find
I = Z +∞
x=0
1 x
Z x y=0
cos(x − y) − cos(x)
y dy dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We have that I =
Z +∞
x=0
cos(x) x
Z x y=0
cos(y) − 1 y
dy dx + Z +∞
x=0
sin(x) x
Z x y=0
sin(y) y dy
dx
= Z +∞
y=0
1 − cos(y)
y Ci(y) dy + Z +∞
x=0
sin(x)
x Si(x) dx where
Ci(y) = − Z +∞
x=y
cos(x)
x dx and Si(x) = Z x
y=0
sin(y) y dy.
Now
Z +∞
x=0
sin(x)
x Si(x) dx = 1
2Si(x)2+∞
x=0=1 2
Z +∞
x=0
sin(x) x dx
2
= π2 8 . Moreover, for t > 0, by letting
F (t) = Z +∞
y=0
1 − cos(y)
y Ci(ty) dy, we obtain
Z +∞
y=0
1 − cos(y)
y Ci(ty) dy = F (1) = Z 1
+∞
F0(t) dt = Z 1
+∞
ln
1 − 1
t2
dt 2t = 1
4 Z 1
0
ln(1 − s)
s ds =π2 24 because
F0(t) = Z +∞
y=0
1 − cos(y)
y ·cos(ty) t dy
= 1 2t
Z +∞
y=0
2 cos(ty) − cos((t + 1)y) − cos((t − 1)y)
y dy
= 1 2t
Z +∞
y=0
cos(ty) − cos((t + 1)y)
y dy + 1
2t Z +∞
y=0
cos(ty) − cos((t − 1)y)
y dy
= 1
2tln t + 1 t
+ 1
2tln t − 1 t
= 1 2tln
1 − 1
t2
. Hence
I =π2 24+π2
8 =π2 6 .