6.3 Trigonometric and Polynomial functions

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We can also deal with the case of some mixtures of trigonometric and polynomials

in improper integrals. Expressions such as

Z 1

􀀀1

P(x)

Q(x)

cos ax dx; Z 1

􀀀1

P(x)

Q(x)

sin ax dx

can be integrated.

Since cos x = <(eix); sin x = =(eix), we have the result:

Theorem 6.5 When P and Q are real polynomials with degree of Q at least

two greater than the degree of P, the integral of the function

f(x) =

P(x)

Q(x)

cos(ax)

for a > 0 is given by extending f to the complex plane by putting

f(z) =

P(z) eiaz

Q(z)

whereupon

Z 1

􀀀1

P(x)

Q(x)

cos(ax) dx = 􀀀2_

k

Xj=1

=(Res[f(z); wj])

160 CHAPTER 6. RESIDUES

and

Z 1

��1

P(x)

Q(x)

sin(ax) dx = 2_

k

Xj=1

<(Res[f(z); wj])

where w1; _ _ _; wk are the poles in the top half of the complex plane. 2

Example 6.3.1 Evaluate

Z 1

􀀀1

cos x

x2 + 1

dx

Solution

We have the solution is

􀀀2_ =_Res _ eiz

(z 􀀀 i)(z + i)

; i__

The residue is the coe_cient of 1=(z 􀀀 i) which is, at z = i,

ei(i)

2i

= 􀀀i

e􀀀1

2

So the imaginary part is 􀀀e􀀀1=2 and multiplying by 􀀀2_ gives the _nal value

_

e

A sketch of the graph of this function shows that the result is reasonable.

Exercise 6.3.1 Sketch the graph and estimate the above integral.

We can see immediately from a sketch of the graph that the integral

Z 1

􀀀1

sin x

x2 + 1

dx = 0

From the antisymmetry of the function. This also comes out of the above

example immediately.

The restriction that the degree of P has to be at least two less than the degree

of Q looks sensible for the case of polynomials, but for the case of mixtures

6.4. POLES ON THE REAL AXIS 161

with trigonometric functions, we can do better: the integral Ra

1 1=x, will grow

without bound as a!1, but the integral of (sin x)=x will not, because the

positive and negative bits will partly cancel. A careful argument shows that

we can in practice get away with the degree of P being only at least one less

than the degree of Q. The problems that are associated with this are that

we can run into trouble with the limits. It is essentially the same problem as

that we experience when summing an in_nite series with alternating terms:

grouping the terms di_erently can give you di_erent results. We can therefore

get away with a di_erence of one in the degrees of the polynomials provided

we change the de_nition of the improper integral to impose some symmetry

in the way we take limits, in other words we use the Cauchy version of the

integral, or the Cauchy Principal Value.

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