application of sine integral at infinity∗
pahio†
2013-03-22 2:30:35
For finding the value of the improper integral
Z ∞
sin ax
dx := f (a)
(a > 0)
2)
x(1+x
0
(1)
we first use the partial fraction representation
1
1
x
= −
.
x(1+x2 )
x 1+x2
Thus we may write
Z
f (a) =
0
∞
sin ax
dx −
x
∞
Z
0
x sin ax
dx.
1 + x2
π
But by the entry sine integral at infinity, the first integral equals . When we
2
check
Z ∞
Z ∞
cos ax
x sin ax
f 0 (a) =
dx, f 00 (a) = −
dx,
2
1+x
1+x2
0
0
we see that there is the linear differential equation
f (a) =
π
+ f 00 (a)
2
(2)
i.e.
π
f 00 − f = − ,
2
satisfied by the sought function a 7→ f (a). We have the initial conditions
Z
f (0) =
∞
0 dx = 0,
f 0 (0) =
Z
0
0
∞
dx
=
1+x2
.∞
0
arctan x =
π
.
2
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1
Therefore the general solution
f (a) = C1 ea + C2 e−a +
π
2
of (2) requires that C1 = 0, C2 = π2 , and consequently the sought integral f (a)
has the value
Z ∞
sin ax
π
dx = (1 − e−a )
(3)
2)
x(1+x
2
0
2