Problem 11423
(American Mathematical Monthly, Vol.116, March 2009) Proposed by G. Minton (USA).
Show that if n and m are positive integers with n ≥ m and n − m even, then Z +∞
0
(sin x)n xm dx is a rational multiple of π.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
If m > 2 then by integrating by parts Z +∞
0
(sin x)n
xm dx = − 1 m −1
Z +∞
0
(sin x)nd
1 xm−1
= − 1
m −1
(sin x)n xm−1
+∞
0
+ n
m −1 Z +∞
0
(sin x)n−1cos x xm−1 dx
= − n
(m − 1)(m − 2) Z +∞
0
(sin x)n−1cos x d
1 xm−2
= − n
(m − 1)(m − 2)
(sin x)n−1cos x xm−2
+∞
0
+ n
(m − 1)(m − 2) Z +∞
0
(n − 1)(sin x)n−2(cos x)2−(sin x)n−1
xm−2 dx
= n(n − 1) (m − 1)(m − 2)
Z +∞
0
(sin x)n−2
xm−2 dx − n2 (m − 1)(m − 2)
Z +∞
0
(sin x)n xm−2 dx.
Hence it suffices to consider the cases when m = 1, n = 2a − 1, and m = 2, n = 2a for some positive integer a.
Z +∞
0
(sin x)2a−1
x dx = 1
22a−2
a−1
X
k=0
(−1)a+k−12a − 1 k
Z +∞
0
sin((2a − 2k − 1)x)
x dx
= 1
22a−2
a−1
X
k=0
(−1)a+k−12a − 1 k
Z +∞
0
sin x x dx
= π
22a−1
a−1
X
k=0
(−1)a+k−12a − 1 k
=(2a − 3)!!
(2a − 2)!!·π 2. In a similar way it can be seen that
Z +∞
0
(sin x)2a
x2 dx= n Z +∞
0
(sin x)2a−1cos x
x dx=(2a − 3)!!
(2a − 2)!! ·π 2.
