§ 3. Some Applications of Cauchy’s Method
69
A quicker second method consists in integrating over the same closed
contours as in the absolute convergence case. Start with the relation f (z) =
c/z + O(1/z
2 ) where c = − Res(f, ∞) – careful with the sign ! – and observe
that the integral of O(1/z
2 ) over the half-circle approaches 0. Therefore, the
integral of f along the latter approaches the same limit as that of c/z; it is
calculated by setting z = Re
it , and so dz/z = idt, and as integration is over
(0, π), the result is equal to πic = −πi Res(f, ∞). Taking into account the
poles in the interior of the integration contour,
R
f (x)dx − πi Res(f, ∞) = 2πi
Im(a) > 0
. Res(f, a)
This again leads to (8).
(iii) Absolutely convergent Fourier transforms. For t real, let us now consider the Fourier integral
ˆ
f (t) =
f (x)e(−tx)dx =
f (x) exp(−2πitx)dx ,
(8.9)
where f = p/q is again a rational fraction without any real roots; here too
the integral is absolutely convergent if n = d
◦ (q)− d
◦ (p) ≥ 2.
Suppose first that t > 0. The function
g(z) = f (z)e(−tz)
is holomorphic on C deprived of the roots of q; f (z) ∼ c/z
n for large |z|,
though |e(−tz)| = exp(2πty) is ≤ 1 on the half-plane Im(z) ≤ 0. Hence, if g
is integrated over the contour formed by the interval [−R, R] followed by the
lower half-circle of radius R, its contribution is O(1/R
n−1 ) for large R, and
so approaches 0. As the integration contour is followed clockwise, the index
of a point in its interior is −1 and the residue theorem shows that
ˆ
f (t) = −2πi
Im(a)<0
Res
z=a
[f (z)e(−tz)] (t ≥ 0) ,
(8.10’)
the notation for the residue being self-explanatory. For t ≤ 0, we use the
upper half-circle on which e(−tz) is bounded, and so
ˆ
f (t) = +2πi
Im(a)>0
Res
z=a
[f (z)e(−tz)] (t ≤ 0) .
(8.10”)
For t = 0, we recover the results of section (i).
Suppose for example that
f (x) =
x
2 + w
2
−1 =
1
2iw
1
x − iw
−
1
x + iw
(8.11)
69
A quicker second method consists in integrating over the same closed
contours as in the absolute convergence case. Start with the relation f (z) =
c/z + O(1/z
2 ) where c = − Res(f, ∞) – careful with the sign ! – and observe
that the integral of O(1/z
2 ) over the half-circle approaches 0. Therefore, the
integral of f along the latter approaches the same limit as that of c/z; it is
calculated by setting z = Re
it , and so dz/z = idt, and as integration is over
(0, π), the result is equal to πic = −πi Res(f, ∞). Taking into account the
poles in the interior of the integration contour,
R
f (x)dx − πi Res(f, ∞) = 2πi
Im(a) > 0
. Res(f, a)
This again leads to (8).
(iii) Absolutely convergent Fourier transforms. For t real, let us now consider the Fourier integral
ˆ
f (t) =
f (x)e(−tx)dx =
f (x) exp(−2πitx)dx ,
(8.9)
where f = p/q is again a rational fraction without any real roots; here too
the integral is absolutely convergent if n = d
◦ (q)− d
◦ (p) ≥ 2.
Suppose first that t > 0. The function
g(z) = f (z)e(−tz)
is holomorphic on C deprived of the roots of q; f (z) ∼ c/z
n for large |z|,
though |e(−tz)| = exp(2πty) is ≤ 1 on the half-plane Im(z) ≤ 0. Hence, if g
is integrated over the contour formed by the interval [−R, R] followed by the
lower half-circle of radius R, its contribution is O(1/R
n−1 ) for large R, and
so approaches 0. As the integration contour is followed clockwise, the index
of a point in its interior is −1 and the residue theorem shows that
ˆ
f (t) = −2πi
Im(a)<0
Res
z=a
[f (z)e(−tz)] (t ≥ 0) ,
(8.10’)
the notation for the residue being self-explanatory. For t ≤ 0, we use the
upper half-circle on which e(−tz) is bounded, and so
ˆ
f (t) = +2πi
Im(a)>0
Res
z=a
[f (z)e(−tz)] (t ≤ 0) .
(8.10”)
For t = 0, we recover the results of section (i).
Suppose for example that
f (x) =
x
2 + w
2
−1 =
1
2iw
1
x − iw
−
1
x + iw
(8.11)