lightning overview of contour integration€¦ · lightning overview of contour integration...

Lightning Overview of ContourIntegration

Wednesday, 23 October 2013

The prettiest math I know, and extremely useful.

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1. Overview

This handout summarizes key results in the theory of functions of a complexvariable, leading to the inverse Laplace transform. I suspect that this materialwas covered in depth in Math 115. I make no pretense at rigor, but I’ll do my bestto ensure that I don’t say anything false. The main points are:

1. The Cauchy-Riemann conditions, which must be satisfied by a differen-tiable function f (z) of a complex variable z.

2.∮

f (z)d z = 0 if f (z) is continuous and differentiable everywhere insidethe closed path.

3. If f (z) blows up inside a closed contour in the complex plane, then thevalue of

∮f (z)d z is 2πi times the sum of the residues of all the poles inside

the contour.4. The Fourier representation of the delta function is

∫ ∞−∞ e i kx dk = 2πδ(x).

5. The inverse Laplace transform equation.

2. Cauchy-Riemann

For the derivative of a function of a complex variable to exist, the value of thederivative must be independent of the manner in which it is taken. If z = x+i y isa complex variable, which we may wish to represent on the x y plane, and f (z) =u(x, y)+ i v(x, y), then the value of d f

d z should be the same at a point z whetherwe approach the point along a line parallel to the x axis or a line parallel to the yaxis (or any other direction, for that matter). Therefore,

d f

d z= ∂u

∂x+ i

∂v

∂y(along x)

d f

d z= ∂u

∂(i y)+ i

∂v

∂(i y)= ∂v

∂y− i

∂u

∂y(along y)

Comparing real and imaginary parts, we obtain the Cauchy-Riemann equations: For a function f (z) of the complex vari-able z = x + i y having real part u(x, y)and imaginary part v(x, y), the Cauchy-Riemann equations are necessary andsufficient for the existence of the deriva-tive f ′(z).

For a function f (z) of the complex vari-able z = x + i y having real part u(x, y)and imaginary part v(x, y), the Cauchy-Riemann equations are necessary andsufficient for the existence of the deriva-tive f ′(z).

Physics 111 1 of 8 Peter N. Saeta

4. RESIDUES

∂u

∂x= ∂v

∂yand

∂v

∂x=−∂u

∂y(1)

3. Integrals around closed paths

Stokes’s theorem holds that∮

F ·d l =Ò(∇×F) ·dA, indicating that the integral of

the curl of a vector function over an enclosed area is equal to the line integral ofthe function over the boundary to the area. The right-hand rule is used to orientthe contour with respect to the outward normal to the area.

The z component of the curl is (∇× F)z = ∂Fy

∂x − ∂Fx∂y . By Stokes’s theorem, the

integral of this quantity over a closed area in the x y plane must be equal to theline integral around the area,Ï (

∂Fy

∂x− ∂Fx

∂y

)d x d y =

∮(Fx d x +Fy d y)

Now consider the integral of a function of a complex variable around a closedcontour on the complex plane,∮

f (z)d z =∮

(u + i v)(d x + i d y) =∮

(u d x − v d y)+ i∮

(u d y + v d x)

We can now use Stokes’s theorem to show that each of the integrals on the rightvanishes. For the first integral, let Fx = u and Fy =−v . Then∮

(u d x − v d y) =Ï (

−∂v

∂x− ∂u

∂y

)d x d y

but the integrand vanishes by the Cauchy-Riemann equation. Similarly for thesecond integral, let Fx = v and Fy = u. Then∮

(v d x +u d y) =Ï (

∂u

∂x− ∂v

∂y

)d x d y = 0

Therefore, the integral of f (z) around a closed contour vanishes, provided that You may thus blithely distort the con-tour of integration on the complexplane without changing the value of theintegral, provided you don’t cross a pole.

You may thus blithely distort the con-tour of integration on the complexplane without changing the value of theintegral, provided you don’t cross a pole.

f (z) is everywhere differentiable inside the contour.

4. Residues

Suppose that f (z) blows up at a point z0 inside a closed contour. Although I willnot justify this, suppose further that in the neighborhood of z0 the function maybe expanded in a Laurent series, which is a generalization of Taylor’s series:

f (z) =∞∑

n=−∞an(z − z0)n


5. FOURIER REPRESENTATION OF δ(x)

Unlike Taylor’s series, the Laurent series includes both negative and positivepowers. Since the integral around any closed contour that doesn’t contain asingularity vanishes, we are free to deform the contour to include a detour thatcomes within ε of z0, then goes around a circle of radius ε centered on z0 andthen heads right back out from where it came, as illustrated in Fig. 1. The contri-butions along the path in and then out precisely cancel, and we are left with thecontribution around the small circle.

x

є

y

Figure 1: Deforming a contour toremove a pole from the interior ofa region. Only the a−1ζ

−1 term inthe Laurent expansion of f (z) con-tributes to the integral around thecircle of infinitesimal radius ε; thecoefficient a−1 is called the residueof the pole.

Let ζ = z − z0 = εe iθ, where ε is a constant.Then∮

f (z)d z =∮ ( ∞∑

n=−∞anζ

n)

dζ=∞∑

n=−∞

∮anζ

n dζ

where I have blithely interchanged the orderof integration and summation. See the mathdepartment for the details. We now must eval-

uate∮ζn dζ, but unless n =−1,

∫ζn dζ= ζn+1

n+1 .This is a single-valued function of ζ, whichmeans that when we evaluate at ζ = εe iθ forθ = 2π and θ = 0, we get the same value.Hence, the integral vanishes.

When n = −1 we have to proceed more care-fully:∮

dζ

ζ= ln(εe iθ)

∣∣∣2π

0= (lnε+ iθ)|2π0 = 2πi (2)

So, the only term in the infinite series thatcontributes is the one with n =−1, and it con-tributes 2πi times the coefficient a−1. This coefficient is called the residue of thepole. Thus, we have the residue theorem:∮

f (z)d z = 2πi∑

residues inside the contour (3)

where the sum of residues means the sum of the residues of all poles lying withinthe contour. If the contour runs through a pole, then we get half the residue, as The residue theorem greatly simplifies a

number of integrals and series that arisein theoretical physics. Prof. Townsendclaims, however, that Richard Feynmanchallenged colleagues to find an integralthey could do via contour integrationthat he couldn’t do using some othermethod. He never lost. But I still reallylike the residue theorem!

The residue theorem greatly simplifies anumber of integrals and series that arisein theoretical physics. Prof. Townsendclaims, however, that Richard Feynmanchallenged colleagues to find an integralthey could do via contour integrationthat he couldn’t do using some othermethod. He never lost. But I still reallylike the residue theorem!

you can readily verify by returning to Eq. (2) and noting that we now integrateonly through an angular range of π.

5. Fourier representation of δ(x)

We now seek to establish that

δ(x) = 1

2π

∫ ∞

−∞e i kx dk (4)


5. FOURIER REPRESENTATION OF δ(x)

where δ(x) is the Dirac delta “function.” First a plausibility argument. If we useEuler’s theorem to express the complex exponential as e i kx = coskx + i sinkx,then we see immediately that the imaginary part must vanish because sinkx isan odd function of k and we are integrating over the (symmetric) infinite inter-val. All the cosines peak at the origin and then proceed to oscillate at differentspatial frequencies as we move away. These oscillations tend to wipe one an-other out at any nonzero value of x, but they all add up at x = 0 to produce anenormous (infinite) spike. With only a tiny bit of pixie dust to supply the factorof 1/2π, we can see how the integral might just produce δ(x).

Of course, whenever you see a delta function you feel an Pavlovian urge to inte-grate. I know I do. So consider

I =∫ x2

x1

(1

2π

∫ ∞

−∞e i kx dk

)d x = 1

2π

∫ ∞

−∞dk

∫ x2

x1

e i kx d x

where I have once again blithely interchanged the order of integration. If Eq. (4)holds, then I = 0 if x1 and x2 don’t bracket 0, and I = 1 if x2 > 0 and x1 < 0. Weevaluate I in two steps. The first integral is child’s play:∫ x2

x1

e i kx d x = e i kx

i k

∣∣∣∣∣x2

x1

= e i kx2

i k= e i kx2 −e i kx1

i k

Re(k)

Im(k)

R

Figure 2: Contour to evaluateEq. (5).

Now we have to evaluate integrals of the form

J (x) =∫ ∞

−∞e i kx

i kdk (5)

in which the integrand diverges right in themiddle of the path of integration. If we ex-pand the exponential in a series e i kx = 1 +i kx + (i kx)2

2! + ·· · , we can readily see that thecoefficient of the a−1 term is just 1/i . Haveno fear. There are two possibilities to worryabout: x > 0 and x < 0. If x > 0, then wecan create a closed contour on the complex kplane using a semicircular arc of radius k = Rin the upper half plane, as illustrated in Fig. 2,and take the limit as R →∞. All along this arc, the imaginary part of the com-plex variable k is positive and enormous, so the semicircular integral contributesnothing in the limit as R →∞. By the residue theorem, then, J (k) is equal to 2πitimes the residue of the single pole at k = 0, except that this pole lies right onthe contour of integration. By the argument leading to Eq. (2), we get half theresidue. Hence,

J (x) = 1

i(πi ) =π, x > 0


6. LAPLACE TRANSFORMS

If x < 0, then we must close in the lower half plane and we now traverse theclosed contour in the negative (clockwise) direction. Now we get minus half thepole, so

J (x) = 1

i(−πi ) =−π, x < 0

Putting this all together, we have

I = 1

2π[J (x2)− J (x1)] =

1 x1 < 0 < x2

−1 x2 < 0 < x1

0 otherwise

This confirms Eq. (4).

Exercise 1 One definition of the Fourier transform of a function f (t ) (assumedto die off appropriately as |t |→∞) is

f̃ (ω) ≡∫ ∞

−∞f (t )e iωt d t (6)

Given f̃ (ω), show that the original function f (t ) may be recovered by computingthe inverse transform, given by

f (t ) = 1

2π

∫ ∞

−∞f̃ (ω)e−iωt dω (7)

Caution: take care not to confuse dummy variables of integration with “real”variables.

6. Laplace Transforms

Some dynamics problems may be solved using Laplace transforms to convertdifferential equations to algebraic ones. The Laplace transform of a functiong (t ) is defined by

G (s) ≡∫ ∞

0g (t )e−st d t ≡L

{g (t )

}(8)

where we require that g (t ) = 0 for t < 0. (You can always shift the origin of time ifyour function starts up at some distinct value of t less than 0.) Note that as s getslarge, only the portions of g (t ) at small t contribute to G (s). You can readily con-firm that the Laplace transform of g (t ) = e−αt is G (s) = 1

α+s , and that the Laplacetransform of g (t ) = sinωt is G (s) = ω

ω2+s2 . (You’ll need to do two integrations byparts for this one.)



Re(k)

Im(k)

b

Figure 3: Contour for evaluating the inverse Laplace transform.

Lots and lots of Laplace transforms are tabulated in books; engineers especiallylove them. Typically, they solve a problem by finding the Laplace transform ofthe function they really want and then have to go backwards to find that func-tion. One approach is to look in the book of transforms until you find the onethat matches your transform and then copy down the corresponding functionthat produces this transform. This is sort of like figuring out the integral of afunction f by taking the derivative of all sorts of functions until you find a deriva-tive that matches f . Eventually, you’d like to learn how to integrate!

So, our problem is, given a function G (s) that is defined for s ≥ 0 and whichdoesn’t blow up as s →∞, how do we find the function g (t ) for which G (s) is theLaplace transform?

From the previous section, we know that δ(x) = 12π

∫ ∞−∞ e i kx d x. If we could

somehow turn the e−st into a delta function with argument t − t ′, then we couldpick off the value of g (t0). So, let’s multiply G (s) by e st ′ and integrate:∫

G (s)e st ′ d s =∫ ∫ ∞

0g (t )e−st d t e st ′ d s =

∫ ∫ ∞

0g (t )e−s(t−t ′) d t d s (9)

I’ve been a bit coy about the limits of integration for s. If we want to use theFourier representation of the delta function, then we have to change s into i ssomehow. Suppose we integrate along the line s = b + i y , where b is a real con-stant that puts the line, shown in Fig. 3, to the right of any poles in G (s). Then wecould close the contour using the arc at infinity shown in the dotted line withoutchanging the value of the integral, since the e st ′ebt ′ term will cause it to vanish



for large negative values of s. Then∫ b+i∞

b−i∞G (s)e st ′ d s =

∫ b+i∞

b−i∞

∫ ∞

0g (t )e s(t ′−t ) d t d s (10)

Now we blithely interchange the order of integration on the right (acceptable,provided that g (t ) is well behaved at infinity) to get∫ b+i∞

b−i∞G (s)e st ′ d s =

∫ ∞

0g (t )

∫ b+i∞

b−i∞e s(t ′−t ) d s d t

=∫ ∞

0g (t )eb(t ′−t )

[∫ ∞

−∞e i y(t ′−t )i d y

]d t (11)

The term in brackets is just 2πiδ(t ′− t ), which makes the integral over t straight-forward. We get ∫ b+i∞

b−i∞G (s)e st ′ d s = 2πi g (t ′) (12)

Changing t ′ to t and tidying up, we have the Laplace transform inversion for-mula,

g (t ) = 1

2πi

∫ b+i∞

b−i∞G (s)e st d s (13)

where the value of b must be chosen to be greater than the real coordinate ofany poles in G (s). Eq. (13) is sometimes called the Bromwich integral. By theresidue theorem, therefore, the inverse transform is just the sum of the residuesof G (s)e st .

Example 1: Bromwich integral

Given the Laplace transform G (s) = s +a

(s +a)2 +k2 , find g (t ).

The poles ofs +a

(s +a)2 +k2 e st occur where s + a = ±i k, so let s = −a ±i k + z and rewrite the integrand in terms of the new variable z:

G (s)e st = ±i k + z

(±i k + z)2 +k2 e(±i k−a+z)t = ±i k + z

±2i kz + z2 e(±i k−a+z)t (14)

We can now take the limit as z → 0 and look for the term proportionalto z−1; we get

a±−1 =

1

2e±i kt−at (15)

Summing the residues, we find

g (t ) = 1

2e−at

(e i kt +e−i kt

)= e−at coskt (16)

You can confirm that L{e−at coskt

}is indeed s+a

(s+a)2+k2 .


7. EXERCISES AND PROBLEMS

7. Exercises and Problems

Exercise 2 Show that the Laplace transform of the derivative of a function g (t )is

L

{d g

d t

}= sG (s)− g (0) (17)

where G (s) = L{

g (t )}. In other words, the Laplace transform converts deriva-

tives to powers of s.

Exercise 3 The convolution of two functions is defined by

f (t ) =∫ t

0g1(t − t ′)g2(t ′)d t ′ (18)

Show that L{

f (t )}=G1(s)G2(s).

Exercise 4 Calculate the Laplace transform of sinωt .


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