24.13 DEFINITE INTEGRALS USING CONTOUR INTEGRATION
The proof of the lemma is straightforward once it has been observed that, for0 ≤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∫π0e−mRsinθdθ=2MR∫π/ 20e−mRsinθdθ.Thus, using (24.71),
IΓ≤ 2 MR∫π/ 20e−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−adx, 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−adx−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−adx=−πsinma, as required,P
∫∞
−∞sinmx
x−adx=πcosma, as a bonus.24.13.3 Integrals of multivalued functionsWe 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