6.1 Trigonometric Integrals

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This is something more than just a trick, because it gives a practical method

for solving some very nasty de_nite integrals of trigonometric functions.

What we do is to transform them into path integrals and use residues to

evaluate them. An example will make this clear:

Example 6.1.1 Evaluate

Z 2_

0

d_

3 cos _ + 5

We are going to transform into an integral around the unit circle S1. In this

case we have

z = ei_; dz = iz d_; 1=z = e􀀀i_

From which we deduce that

cos _ =

z + 1=z

2

Substituting in the given integral we get

􀀀i IS1

dz

z[ 3

2 (z + 1=z) + 5]

This can be rewritten

􀀀i IS1

dz

3

2z2 + 3

2 + 5z

= 􀀀2iIS1

dz

(3z + 1)(z + 3)

= 􀀀

2i

3 IS1

dz

(z + 1=3)(z + 3)

The pole at z = 􀀀3 is outside the unit circle so we evaluate the residue at

􀀀1=3 and we know the coe_cient there is 1=(z + 3) = 3=8. The integral is

therefore

􀀀2_i

2i

3

(3=8) = _=2

154 CHAPTER 6. RESIDUES

The substitution for sin _ = 1

2i (z 􀀀 1=z) is obvious.

It is clear that we can reduce a trigonometric integral from 0 to 2_ to a

rational function in a great many cases, and thus use the residue theory to

get a result. This is (a) cute and (b) useful.

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