Problem 12145
(American Mathematical Monthly, Vol.126, November 2019) Proposed by T. Amdeberhan and V. Moll (USA).
Prove
Z ∞
0
cos(t) sin(√ 1 + t2)
√1 + t2 dt= π 4.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. By letting t = sinh(x), by using the Werner’s formula 2 cos(α) sin(β) = sin(α + β) − sin(α − β), and finally by letting s = e±x, we have
Z ∞
0
cos(t) sin(√ 1 + t2)
√1 + t2 dt= Z ∞
0
cos(sinh(x)) sin(cosh(x)) dx
= 1 2
Z ∞
0
(sin(sinh(x) + cosh(x)) − sin(sinh(x) − cosh(x))) dt
= 1 2
Z ∞
0
sin(ex) + sin(e−x)) dx
= 1 2
Z ∞
1
sin(s) s ds+1
2 Z 1
0
sin(s) s ds
= 1 2
Z ∞
0
sin(s) s ds=1
2 · π 2 = π
4.
