§ 3. Some Applications of Cauchy’s Method
107
a < R in order to integrate the power series term by term over (0, a); the
contribution from ]0, a] is then equal to
f
(n) (0)a
n+s /n!(s + n), a series
that converges even better than the power series of f at x = a since |s + n|
increases indefinitely. The residue of the function at s = −n can be immediately calculated since a
n+s = 1 at this point; Hence we again get f
(n) (0)/n!.
This result is, as it should be, independent of the chosen point a.
Through a change of of variable x → 1/x, which leaves the measure d
∗ x
invariant, the integral
Γ
+
f (s) =
+∞
1
f (x)x
s d
∗ x
(13.5”)
reduces to the preceding case. However,
+∞
1
x
s d
∗ x = −1/s if Re(s) < 0 .
Hence if there is a bounded expansion
f (x) = b 1 x
−v1 + . . . + b n x
−vn + O
x
−vn+1
(13.6”)
at infinity, with v 1 < . . . < v n < v n+1 , integral (5”), a priori defined and
holomorphic on Re(s) < v 1 , can be analytically extended to the half-plane
Re(s) < Re(v n+1 ), with simple poles at the points v k and residues equal to the
coefficients b k . In particular, if there is an unbounded asymptotic expansion
f (x) ≈
b n x
−un , x −→ +∞ , with lim v n = +∞ ,
(13.8”)
it can be extended to all of C, excepting at some simple poles.
For example, choose the function f (x) = x/(1 + x
2 ) = f (x
−1 );f (x) ∼ x in
the neighbourhood of 0. This gives the convergence condition Re(s) + 1 > 0,
and f (x) ∼ 1/x at infinity, which in turn gives the convergence condition Re(s) − 1 < 0; the Mellin integral, therefore, converges on the strip
| Re(s)| < 1. For x ≤ a < 1, f (x) =
(−1)
n x
2n+1 , which is much better
than an asymptotic expansion. Γ
−
f (s), a priori defined for Re(s) > −1, can
be analytically extended to all of C, with simple poles at the points −2n − 1
and residues equal to (−1)
n . Similarly,
f (x) =
(−1)
n x
−2n−1
at infinity. So Γ
+
f (s), a priori defined for Re(s) < 1, can be extended to C,
with simple poles at the points 2n+1. However, Γ f (s) = Γ
−
f (s)+Γ
+
f (s) on the
strip | Re(s)| < 1 where these three functions are defined and holomorphic.
Since the right hand term is meromorphic on C, it follows that Γ f can be
analytically extended to all of C, its singularities being simple poles at the
points 2n + 1, n ∈ Z, with residues equal to (−1)
n . In fact, we will show in
n
◦ 15 that
107
a < R in order to integrate the power series term by term over (0, a); the
contribution from ]0, a] is then equal to
f
(n) (0)a
n+s /n!(s + n), a series
that converges even better than the power series of f at x = a since |s + n|
increases indefinitely. The residue of the function at s = −n can be immediately calculated since a
n+s = 1 at this point; Hence we again get f
(n) (0)/n!.
This result is, as it should be, independent of the chosen point a.
Through a change of of variable x → 1/x, which leaves the measure d
∗ x
invariant, the integral
Γ
+
f (s) =
+∞
1
f (x)x
s d
∗ x
(13.5”)
reduces to the preceding case. However,
+∞
1
x
s d
∗ x = −1/s if Re(s) < 0 .
Hence if there is a bounded expansion
f (x) = b 1 x
−v1 + . . . + b n x
−vn + O
x
−vn+1
(13.6”)
at infinity, with v 1 < . . . < v n < v n+1 , integral (5”), a priori defined and
holomorphic on Re(s) < v 1 , can be analytically extended to the half-plane
Re(s) < Re(v n+1 ), with simple poles at the points v k and residues equal to the
coefficients b k . In particular, if there is an unbounded asymptotic expansion
f (x) ≈
b n x
−un , x −→ +∞ , with lim v n = +∞ ,
(13.8”)
it can be extended to all of C, excepting at some simple poles.
For example, choose the function f (x) = x/(1 + x
2 ) = f (x
−1 );f (x) ∼ x in
the neighbourhood of 0. This gives the convergence condition Re(s) + 1 > 0,
and f (x) ∼ 1/x at infinity, which in turn gives the convergence condition Re(s) − 1 < 0; the Mellin integral, therefore, converges on the strip
| Re(s)| < 1. For x ≤ a < 1, f (x) =
(−1)
n x
2n+1 , which is much better
than an asymptotic expansion. Γ
−
f (s), a priori defined for Re(s) > −1, can
be analytically extended to all of C, with simple poles at the points −2n − 1
and residues equal to (−1)
n . Similarly,
f (x) =
(−1)
n x
−2n−1
at infinity. So Γ
+
f (s), a priori defined for Re(s) < 1, can be extended to C,
with simple poles at the points 2n+1. However, Γ f (s) = Γ
−
f (s)+Γ
+
f (s) on the
strip | Re(s)| < 1 where these three functions are defined and holomorphic.
Since the right hand term is meromorphic on C, it follows that Γ f can be
analytically extended to all of C, its singularities being simple poles at the
points 2n + 1, n ∈ Z, with residues equal to (−1)
n . In fact, we will show in
n
◦ 15 that