24.13 DEFINITE INTEGRALS USING CONTOUR INTEGRATION
The proof of the lemma is straightforward once it has been observed that, for
0 ≤θ≤π/2,
1 ≥
sinθ
θ
≥
2
π
. (24.71)
Then, since on Γ we have|exp(imz)|=|exp(−mRsinθ)|,
IΓ≤
∫
Γ
|eimzf(z)||dz|≤MR
∫π
0
e−mRsinθdθ=2MR
∫π/ 2
0
e−mRsinθdθ.
Thus, using (24.71),
IΓ≤ 2 MR
∫π/ 2
0
e−mR(2θ/π)dθ=
πM
m
(
1 −e−mR
)
<
πM
m
;
hence, asR→∞,IΓtends to zero sinceMtends to zero.
Find the principal value of
∫∞
−∞
cosmx
x−a
dx, forareal,m> 0.
Consider the function (z−a)−^1 exp(imz); although it has no poles in the upper half-plane
it does have a simple pole atz=a, and further|(z−a)−^1 |→0as|z|→∞. We will use a
contour like that shown in figure 24.15 and apply the residue theorem. Symbolically,
∫a−ρ
−R
+
∫
γ
+
∫R
a+ρ
+
∫
Γ
=0. (24.72)
Now asR→∞andρ→0 we have
∫
Γ→0, by Jordan’s lemma, and from (24.68) and
(24.69) we obtain
P
∫∞
−∞
eimx
x−a
dx−iπa− 1 =0, (24.73)
wherea− 1 is the residue of (z−a)−^1 exp(imz)atz=a, which is exp(ima). Then taking the
real and imaginary parts of (24.73) gives
P
∫∞
−∞
cosmx
x−a
dx=−πsinma, as required,
P
∫∞
−∞
sinmx
x−a
dx=πcosma, as a bonus.
24.13.3 Integrals of multivalued functions
We have discussed briefly some of the properties and difficulties associated with
certain multivalued functions such asz^1 /^2 or Lnz. It was mentioned that one
method of managing such functions is by means of a ‘cut plane’. A similar
technique can be used with advantage to evaluate some kinds of infinite integral
involving real functions for which the corresponding complex functions are multi-
valued. A typical contour employed for functions with a single branch point