§ 3. Some Applications of Cauchy’s Method
67
If all the roots of q are simple,
p(z)/q(z) = [p(a) + p
(a)(z − a) + . . .] / [q
(a)(z − a) + . . .]
=
p(a)
q (a)(z − a)
[1+?(z − a) + . . .]
in a neighbourhood of such a root a since the quotient of two power series
starting with 1 is a power series of the same type. Hence the formula
Res(p/q, a) = p(a)/q
(a)
that allows us to compute the integral.
Let us for example choose the function f (z) = p(z)/(1 + z
2n ), where p
is a polynomial of degree ≤ 2n − 2; its poles are (at most) the roots of the
equation z
2n = −1 = exp(πi), i.e. the 2n points
ω k = ω
2k+1
where ω = exp(πi/2n) , 0 ≤ k ≤ 2n − 1 .
The roots with positive imaginary parts are obtained for 0 < k < n − 1 and
the residue at ω k is equal to
p(ω k )/2nω
2n−1
k
= −p(ω k )ω k /2n .
So
p(x)
1 + x 2n dx = −
πi
n
n−1
0
ω k p (ω k ) .
For p(x) = x
m−1 , the sum
ω
m
k =
0 ≤ k ≤ n−1
ω
(2k+1)m = ω
m
ω
2mk = ω
m 1 − ω
2mn
1 − ω 2m =
(−1)
m − 1
ω m − ω −m ,
needs to be calculated so that the integral sought equals 0 if m is even (obvious!) and π/n sin(mπ/2n) if m is odd.
(ii) Semi-convergent integrals of rational functions. The previous method
can also be applied if d
◦ (q) = d
◦ (p) + 1. But some precautions are called
for since the extended integral over R is no longer absolutely convergent.
Nonetheless, we can set
R
f (x)dx = lim
+R
−R
f (x)dx = lim
R
0
[f (x) + f (−x)] dx .
(8.5)
This limit exists since
f (z) = c/z + O(1/z
2 ) at infinity