Mathematical Methods for Physicists: A concise introduction - Site Map

FUNCTIONS OF A COMPLEX VARIABLE
only (2 i†=5 lies inside C, and residue at this pole is
lim ‰z …2 i†=5Š 2
z!…2i†=5 …1 2i†z 2 ‡ 6iz 1 2i
Then
Z 2
0
ˆ
lim
z!…2i†=5
I
d
3 2cos ‡ sin ˆ
C
2
2…1 2i†z ‡ 6i ˆ 1
2i
by L'Hospital's rule:
2dz
…1 2i†z 2 ˆ 2i…1=2i† ˆ:
‡ 6iz 1 2i
Fourier integrals of the form
Z 1
1
sin mx
f …x† dx
cos mx
If f …x† is a rational function satisfying the assumptions stated in connection with
improper integrals of rational functions, then the above integrals may be evaluated
in a similar way. Here we consider the corresponding integral
I
f …z†e imz dz
C
over the contour C as that in improper integrals of rational functions (Fig. 6.16),
and obtain the formula
Z 1
1
f …x†e imx dx ˆ 2i X Res‰ f …z†e imz Š …m > 0†; …6:45†
where the sum consists of the residues of f …z†e imz at its poles in the upper halfplane.
Equating the real and imaginary parts on each side of Eq. (6.45), we obtain
Z 1
1
Z 1
1
f …x† cos mxdx ˆ2 X Im Res‰ f …z†e imz Š;
f …x† sin mxdx ˆ 2 X Re Res‰ f …z†e imz Š:
…6:46†
…6:47†
To establish Eq. (6.45) we should now prove that the value of the integral over
the semicircle in Fig. 6.16 approaches zero as r !1. This can be done as
follows. Since lies in the upper half-plane y 0andm > 0, it follows that
From this we obtain
je imz jˆje imx j e my j j ˆ e my 1 …y 0; m > 0†:
j f …z†e imz jˆjf …z† j je imz jjf …z† j …y 0; m > 0†;
which reduces our present problem to that of an improper integral of a rational
function of this section, since f …x† is a rational function satisfying the assumptions
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