6.5 More Complicated Functions

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We can do something for functions which contain square roots and the like.

It should, after all, be possible to use the same ideas for:

Z 1

0

px

x2 + 1

dx

The problem we face immediately is that the square root function is well

de_ned on the associated Riemann surface, and we have to worry about the

problem of the so called 'multi-valued functions'.

6.5. MORE COMPLICATED FUNCTIONS 165

-R R

Figure 6.3: A contour for zq, 0 < q < 1.

We may do this in a variety of ways, but it is convenient to look at the

function zq for 0 < q < 1, and to observe that if we de_ne this for r > 0 and

for 0 < _ < 2_, then

zq = eq log(rei_ ) = eq log reqi_ = rq (cos q_ + i sin q_)

It is clear that this is analytic and one-one on the region

r > 0; 0 < _ < 2_

We are, in e_ect, introducing a branch cut along the positive real axis. Put

q = 1=2 for the branch of the square root function given in the _rst paragraph

of section 2.3.1. The square root will pull the plane with the positive real axis

removed back to the top half plane, just as the square took the top half plane

and wrapped it around to the plane with the positive real axis removed.

Suppose we take a contour which avoids the branch cut as shown in _gure 6.3

but which encloses all the poles in the positive half plane of some rational

function (ratio of polynomials)

P(z)

Q(z)

I shall divide up the total contour c into four parts:

166 CHAPTER 6. RESIDUES

1. OC, the outer (almost) circle, going from Rei_ for _ some small positive

real number, to Rei(2_􀀀_).

2. IC, the inner semi-circle which has some small radius, r, and which is

centred on the origin.

3. T, the top line segment which is just above the positive real axis and

4. B, the line segment just below the real axis.

Now we consider the function

f(z) = zqP(z)

Q(z)

It is clear that if there are k poles w1; _ _ _; wk in the entire plane, none of

which are on the positive real axis, then for some value of R and r,

Ic

f(z) dz = 2_i

k

Xj=1

Res[f;wj]

We can approximate the left hand side by

IRS1

f(z) dz + Z R

0

xq P(x)

Q(x)

dx 􀀀 Z R

0

xq eqi2_ P(x)

Q(x)

dx + ZIC

f(z) dz

Now as the semi-circle SC gets smaller, the radius goes down, and the value

of zq also goes down; provided that P(z)

Q(z) does not have a pole of order greater

than one, i.e. provided Q(z) has no zero of order greater than one at the

origin, then the reducing size of the circle ensures that the last term goes to

zero.

Similarly, if the degree of Q is at least two more than the degree of P, the

integral over OC will also go to zero.

This gives us the result:

Z 1

0

xq P(x)

Q(x)

=

2_i

1 􀀀 eqi2_

k

Xj=1

Res[f;wj]

This is the idea of the proof of:

6.5. MORE COMPLICATED FUNCTIONS 167

Theorem 6.8 For any polynomials P(x) and Q(x) with the degree of Q at

least two more than the degree of P, and for any real q : 0 < q < 1, then

provided Q has a zero of at most one at the origin and no zeros on the positive

reals, and if the zeros of Q, i.e. the poles of P=Q are w1; _ _ _; wk, we have

that

Z 1

0

xq P(x)

Q(x)

=

1

1 􀀀 eqi2_

k

Xj=1

Res[f;wj]

2

It is easy enough to use this result:

Example 6.5.1 Evaluate

Z 1

0

px

1 + x2 dx

Solution

There are two poles of the complex function

pz

1 + z2

one at i and one at 􀀀i. The residues are, respectively

pi

2i

;

p

􀀀i

􀀀2i

which sum to

1

2p2

[(1 􀀀 i) 􀀀 (􀀀1 􀀀 i)] = 􀀀i

p2

We take care to choose the square roots for the given branch cut. I have

chosen to take the square roots in the top half of the plane in both cases.

We multiply by 2_i and divide by 1 􀀀 ei_ = 2 to get

_

p2

Exercise 6.5.1 Solve the above problem by putting x = y2 and using the

earlier method. Con_rm that you get the same answer.

168 CHAPTER 6. RESIDUES

The same ideas can be used to handle other integrals. Some of the regions to

be integrated over and the functions used are far from obvious, and a great

deal of ingenuity and experience is generally required to tackle new cases.

I am reluctant to show you some of the special tricks which work (and which

you wouldn't have thought of in a million years 4 ) because your only possible

response is to ask if you should learn it for an exam. And knowing special

tricks isn't much use in general. Nor do you have the time to spend on

acquiring the general expertise. So if you should ever be told that some

particularly foul integral can be evaluated by contour integration, you can

demand to be told the function and the contour, and you can check it for

yourself, but it is unlikely that you will hit upon some of the known special

results in the time you have available. So I shall stop here, but warn you

that there are many developments which I am leaving out.

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