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Problem d: Use equation (13.14) to show that for the pole at z = i the residue is given

by: Res = 1=2i.

13.3. APPLICATION TO SOME INTEGRALS 167

Problem e: Use the residue theorem (13.12) to deduce that

Z1 1

I= dx = : (13.23)

;1 1 + x2

This value is identical to the value obtained at the beginning of this section by using that

the primitive of 1=(1 + x2 ) is equal to arctan x. Note that the analysis is very systematic

and that we did not need to \know" the primitive function.

In the treatment of this problem there is no reason why the contour should be closed

in the upper half plane. The estimate (13.21) hold equally well for a semicircle in the

lower half-plane.

Problem f: Redo the analysis of this section when you close the contour in the lower half

plane. Use that now the pole at z = contributes and take into account that the

;i

sense of integration now is in the clockwise direction rather than the anti-clockwise

direction. Show that this leads to the same result (13.23) that was obtained by

closing the contour in the upper half-plane.

In the evaluation of the integral (13.17) there was a freedom whether to close the

contour in the upper half-plane or in the lower half-plane. This is not always the case. To

see this consider the integral

Z 1 cos (x =4)

J= dx : (13.24)

;

;1 1 + x2

Since exp ix = cos x + i sin x this integral can be written as

Z 1 ei(x; =4) !

J= dx (13.25)

;1 1 + x2

<

where ) again denotes the real part. We want to evaluate this integral by closing this

<(

integration path with a semicircle either in the upper half-plane or in the lower half-plane.

Due to the term exp (ix) in the integral we now have no real choice in this issue. The

decision whether to close the integral in the upper half-plane or in the lower half-plane

is dictated by the requirement that the integral over the semicircle vanishes as R ! 1.

This can only happen when the integral vanishes (faster than 1=R) as R Let z be a

! 1.

point in the complex plane on the semicircle CR that we use for closing the contour. On

the semicircle z can be written as z = R exp i'. In the upper half-plane 0 ' < and

'<2 .

for the lower half-plane

Problem g: Use this representation of z to show that

eiz = e;Rsin' : (13.26)

Problem h: Show that in the limit R this term only goes to zero when z is in the

!1

upper half-plane.

This means that the integral over the semicircle only vanishes when we close the contour

in the upper half-plane. Using steps similar as in (13.21) one can show that the integral

over a semicircle in the upper half-plane vanishes as R ! 1.

CHAPTER 13. COMPLEX INTEGRATION

168

Problem i: Take exactly the same steps as in the derivation of (13.23) and show that

Z 1 cos (x =4)

J= 1 + x2 dx = 2 e : (13.27)

;

;1

p

Problem j: Determine the integral R;1 sin(x; 2=4) dx without doing any additional calcu-

1

1+x

=4).

lations. Hint look carefully at (13.25) and spot sin (x ;

13.4 Response of a particle in syrup

Up to this point, contour integration has been applied to mathematical problems. How-

ever, this technique does have important application in physical problems. In this section

we consider a particle with mass m on which a force f(t) is acting. The particle is sus-

pended in syrup, this damps the velocity of the particle and it is assumed that this damping

force is proportional to the velocity v(t) of the particle. The equation of motion of the

particle is given by

m dv + v = f(t) : (13.28)

dt

where is a parameter that determines the strength of the damping of the motion by the

uid. The question we want to solve is: what is the velocity v(t) for a given force f(t)?

We will solve this problem by using a Fourier transform technique. The Fourier trans-

form of v(t) is denoted by V (!). The velocity in the frequency domain is related to the

velocity in the time domain by the relation:

Z1

V (!)e;i!t d!

v(t) = (13.29)

;1

1 Z 1 v(t)e+i!t dt :

and its inverse

V (!) = 2 (13.30)

;1

You may be used to seeing a di erent sign in the exponents and by seeing the factor 2

in a di erent place, but as long as the exponents in the forward and backward Fourier

transform have opposite sign and the product of the scale factors in the forward and

backward Fourier transform is equal to 1=2 the di erent conventions are equally valid

and lead to the same nal result. This issue is treated in detail in section (11.5). The

force f(t) is Fourier transformed using the same expressions in the frequency domain it

is denoted by F(!).

Problem a: Use the de nitions of the Fourier transform to show that the equation of

motion (13.28) is in the frequency domain given by

(!) + V (!) = F(!) : (13.31)

;i!mV

Comparing this with the original equation (13.28) we can see immediately why the Fourier

transform is so useful. The original expression (13.28) is a di erential equation while

expression (13.31) is an algebraic equation. Since algebraic equations are much easier to

solve than di erential equations we have made considerable progress.

13.4. RESPONSE OF A PARTICLE IN SYRUP 169

Problem b: Solve the algebraic equation (13.31) for V (!) and use the Fourier transform

(13.29) to derive that

i Z 1 F(!)e;i!t d! :

v(t) = m (13.32)

;1 ! + im

Now we have an explicit relation between the velocity v(t) in the time-domain and the

force F(!) in the frequency domain. This is not quite what we want since we want to nd

the relation of the velocity with the force f(t) in the time domain.

Problem c: Use the inverse Fourier transform (13.30) (but for the force) to show that

i Z 1 f(t0 ) Z 1 e;i!(t;t0 ) d! dt0 :

v(t) = 2 m (13.33)

;1 ;1 ! + im

This equation looks messy, but we will simplify it by writing it as

i Z 1 f(t0)I(t t0 ) dt0

v(t) = m (13.34)

;1

;

with Z

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