Lightning Overview of Contour Integration Wednesday, ÕÉ October óþÕÕ Physics ÕÕÕ

e prettiest math I know, and extremely useful.

Õ. Overview is handout summarizes key results in the theory of functions of a complex variable, leading to the inverse Laplace transform. I make no pretense at rigor, but I’ll do my best to ensure that I don’t say anything false. e main points are: Õ. e Cauchy-Riemann conditions, which must be satised by a dišerentiable function f z of a complex variable z. ó. f z dz þ if f z is continuous and dišerentiable everywhere inside the closed path.( ) ì. If∮ f (z) blows= up inside( ) a closed contour in the complex plane, then the value of f z dz is óπi times the sum of the residues of all the poles inside the contour. ( ) ∞ ikx ¦. e Fourier representation of the delta function is −∞ e dk óπδ x . ¢. ∮e( inverse) Laplace transform equation. ∫ = ( ) ó. Cauchy-Riemann For the derivative of a function of a complex variable to exist, the value of the derivative must be independent of the manner in which it is taken. If z x iy is a complex variable, which we may wish to represent on the x y plane, and f z u x, y iυ x, y , then the d f = + value of should be the same at a point z whether we approach the point along a line parallel dz ( ) = ( ) + ( ) to the x axis or a line parallel to the y axis. erefore,

d f ∂u ∂υ i (along x) dz ∂x ∂y d f = ∂u+ ∂υ ∂υ ∂u i i (along y) dz ∂ iy ∂ iy ∂y ∂y = + = − Comparing real and imaginary parts,( we) obtain( the) Cauchy-Riemann equations:

∂u ∂υ ∂υ ∂u and (Õ) ∂x ∂y ∂x ∂y = = − ì. Integrals around closed paths Stokes’s theorem holds that F dl F dA, indicating that the integral of the curl of a vector function over an enclosed area is equal to the line integral of the function ∮ ⋅ = ∯ (∇ × ) ⋅

Physics ÕÕÕÕ Peter N. Saeta ¦. RESIDUES over the boundary to the area. e right-hand rule is used to orient the contour with respect to the outward normal to the area. F ∂Fy ∂Fx e z component of the curl is z ∂x ∂y . By Stokes’s theorem, the integral of this quantity over a closed area in the x y plane must be equal to the line integral around the area, (∇ × ) = − ∂Fy ∂Fx dx dy Fx dx Fy dy ∂x ∂y Œ − ‘ = ( + ) Now consider theU integral of a function of a complexc variable around a closed contour on the complex plane,

f z dz u iυ dx i dy u dx υ dy i u dy υ dx

( ) = ( + )( + ) = ( − ) + ( + ) We canc now use Stokes’sc theorem to show thatc each of the integralsc on the right vanishes. For the rst integral, let Fx u and Fy υ. en

= = − ∂υ ∂u u dx υ dy dx dy ∂x ∂y c ( − ) = U Œ− − ‘ but the integrand vanishes by the Cauchy-Riemann equation. Similarly for the second inte- gral, let Fx υ and Fy u. en

= = ∂u ∂υ υ dx u dy dx dy þ ∂x ∂y c ( + ) = U Œ − ‘ = erefore, the integral of f z around a closed contour vanishes, provided that f z is everywhere dišerentiable inside the contour. ( ) ( ) ¦. Residues

Suppose that f z blows up at a point zþ inside a closed contour. Although I will not justify this, suppose further that in the neighborhood of zþ the function may be expanded in a Laurent series,( which) is a generalization of Taylor’s series:

∞ n f z an z zþ n=−∞ ( ) = Q ( − ) Notice that unlike Taylor’s series, the Laurent series includes both negative and positive powers. Since the integral around any closed contour that doesn’t contain a singularity van- ishes, we are free to deform the contour to include a detour that comes within є of zþ, then goes around a circle of radius є centered on zþ and then heads right back out from where it came, as illustrated in Fig. Õ. e contributions along the path in and then out precisely cancel, and we are leŸ with the contribution around the small circle.

Peter N. Saetaó Physics ÕÕÕ ¢. FOURIER REPRESENTATION OF δ X

iθ ( ) Let ζ z zþ єe , where є is a constant. en y ∞ ∞ = − = n n f z dz anζ dζ anζ dζ n=−∞ n=−∞ є wherec ( I) have= blithelyc Œ Q interchanged‘ = theQ orderc of inte- gration and summation. See the math department for the details. We now must evaluate ζn dζ, but unless x + n ζ n Õ n Õ, ζ dζ n+Õ . is is a single-valued function of ζ, which means that when we evaluate∮ at ζ єeiθ for = − = θ óπ ∫and θ þ, we get the same value. Hence, the = integral vanishes. Figure Õ: Deforming a contour to =When n =Õ we have to proceed more carefully: remove a pole from the interior of −Õ a region. Only the a−Õζ term in dζ= − iθ óπ óπ the Laurent expansion of f z con- ln єe lnє iθ þ óπi (ó) ζ þ tributes to the integral around the cir- ( ) So, thec only term= in( the) inniteT = ( series+ that)S contributes= is cle of innitesimal radius є; the coef- cient a−Õ is called the residue of the the one with n Õ, and it contributes óπi times the co- pole. e›cient a−Õ. is coe›cient is called the residue. us, we have the residue= − theorem: f z dz óπi residues (ì) where the sum of residues meansc the( sum) of= the residuesQ of all poles lying within the contour. If the contour runs through a pole, then we get half the residue, as you can readily verify by returning to Eq. (ó) and noting that we now integrate only through an angular range of π.

¢. Fourier representation of δ(x) We now seek to establish that Õ ∞ δ x eikx dk (¦) óπ −∞ where δ x is the Dirac delta “function.”( ) = FirstS a plausibility argument. If we use Euler’s theo- rem to express the complex exponential as eikx cos kx i sin kx, then we see immediately that the( imaginary) part must vanish because sin kx is an odd function of k and we are in- tegrating over the innite interval. All the cosines= peak+ at the origin and then proceed to oscillate at dišerent spatial frequencies as we move away. ese oscillations tend to wipe one another out at any nonzero value of x, but they all add up at x þ to produce an enormous (innite) spike. With only a tiny bit of pixie dust to supply the factor of Õ óπ, 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 integrate. I know I do. So consider ( ) xó Õ ∞ Õ ∞ xó I eikx dk dx dk eikx dx xÕ óπ −∞ óπ −∞ xÕ = S ‹ S  = S S Physics ÕÕÕì Peter N. Saeta ¢. FOURIER REPRESENTATION OF δ X where I have once again blithely interchanged the order of integration. If Eq. (¦) holds, then( ) I þ if xÕ and xó don’t bracket þ, and I Õ if xó þ and xÕ þ. We evaluate I in two steps. e rst integral is child’s play: = = > < xó xó eikx eikxó eikxó eikxÕ eikx dx xÕ ik ik ik xÕ − = W = = Now we have to evaluateS integrals of the form Im(k) ∞ eikx J x dk (¢) −∞ ik in which the integrand( ) = divergesS right in the middle of the path of integration. If we expand the exponential in R ( )ó a series eikx Õ ikx ikx , we can readily see that ó! Re(k) the coe›cient of the a−Õ term is just Õ i. Have no fear. ere are two= possibilities+ + to+⋯ worry about: x þ and x þ. If x þ, then we can create a closed~ contour on the complex k plane using a semicircular arc of radius> k R in< the upper> half plane, as illustrated in Fig. ó, and take Figure ó: Contour to evaluate Eq. (¢). the limit as R . All along this arc, the imaginary part= of the complex variable k is positive and enormous, so the semicircular integral contributes nothing in the→ limit∞ as R . By the residue theorem, then, J k is equal to óπi times the residue of the single pole at k þ, except that this pole lies right on the contour of integration. By the argument leading to→ Eq.∞ (ó), we get half the residue. Hence,( ) = Õ J x πi π, x þ i If x þ, then we must close in the( ) lower= ( half) = plane and we> now traverse the closed contour in the negative (clockwise) direction. Now we get minus half the pole, so < Õ J x πi π, x þ i Putting this all together, we have( ) = (− ) = − <

Õ xÕ þ xó Õ I J xó J xÕ ⎧ Õ xó þ xÕ óπ ⎪ < < ⎪þ otherwise = [ ( ) − ( )] = ⎨− < < ⎪ is conrms Eq. (¦). ⎩⎪ Exercise Õ One denition of the Fourier transform of a function f t (assumed to die oš appropriately as t ) is ∞ ( ) f˜ ω f t eiωt dt (ä) S S → ∞ −∞ ( ) ≡ S ( ) Peter N. Saeta¦ Physics ÕÕÕ ä. LAPLACE TRANSFORMS

Given f˜ ω , show that the original function f t may be recovered by computing the inverse transform, given by ( ) Õ ∞( ) f t f˜ ω e−iωt dω (ß) óπ −∞ Caution: take care not to confuse( dummy) = S variables( ) of integration with “real” variables.

ä. Laplace Transforms Some dynamics problems may be solved using Laplace transforms to convert dišerential equations to algebraic ones. e Laplace transform of a function g t is dened by

∞ G s g t e−st dt g t ( ) (˜) þ where we require that g t (þ) for≡ St þ.( (You) can≡ alwaysL { ( shiŸ)} the origin of time if your function starts up at some distinct value of t less than þ.) Note that as s gets large, only the portions of g t at small( ) =t contribute< to G s . You can readily conrm that the Laplace −αt Õ transform of g t e is G s α+s , and that the Laplace transform of g t sin ωt is ω ( ) ( ) G s ωó+só . (You’ll need to do two integrations by parts for this one.) Lots and lots( ) of= Laplace transforms( ) = are tabulated in books; engineers especially( ) = love them.( ) = Typically, they solve a problem by nding the Laplace transform of the function they really want and then have to go backwards to nd that function. One approach is to look in the book of transforms until you nd the one that matches your transform and then copy down the corresponding function that produces this transform. is is sort of like guring out the integral of a function f by taking the derivative of all sorts of functions until you nd a derivative that matches f . Eventually, you’d like to learn how to integrate! So, our problem is, given a function G s that is dened for s þ and which doesn’t blow up as s , how do we nd the function g t for which G s is the Laplace transform? ( ) ≥ Õ ∞ ikx From→ the∞ previous section, we know that( δ) x óπ −∞(e) dx. If we could somehow turn the e−st into a delta function with argument t t′, then we could pick oš the value of st′ ( ) = g tþ . So, let’s multiply G s by e and integrate: ∫ − ′ ∞ ′ ∞ ′ ( ) G s est ds ( ) g t e−st dt est ds g t e−s(t−t ) dt ds (É) þ þ I’ve beenS a( bit) coy about= SS the limits( ) of integration= forSSs. If we( want) to use the Fourier representation of the delta function, then we have to change s into is somehow. Suppose we integrate along the line s b iy, where b is a real constant that puts the line, shown in Fig. ì, to the right of any poles in G s . en we could close the contour using the arc at innity ′ ′ shown in the dotted line= without+ changing the value of the integral, since the est ebt term will cause it to vanish for large( negative) values of s. en

b+i∞ ′ b+i∞ ∞ ′ G s est ds g t es(t −t) dt ds (Õþ) b−i∞ b−i∞ þ S ( ) = S S ( ) Physics ÕÕÕ¢ Peter N. Saeta ä. LAPLACE TRANSFORMS

Im(k)

b Re(k)

Figure ì: Contour for evaluating the inverse Laplace transform.

Now we blithely interchange the order of integration on the right (acceptable, provided that g t is well behaved at innity) to get

( ) b+i∞ ′ ∞ b+i∞ ′ G s est ds g t es(t −t) ds dt b−i∞ þ b−i∞ ∞ ′ ∞ ′ S ( ) = S g(t)eSb(t −t) eiy(t −t)i dy dt (ÕÕ) þ −∞ = S ( ) S  e term in brackets is just óπiδ t′ t , which makes the integral over t straightforward. We get (b+i∞− ) ′ G s est ds óπig t′ (Õó) b−i∞ Changing t′ to t and tidying up,S we have( the) Laplace= transform( ) inversion formula,

Õ b+i∞ g t G s est ds (Õì) óπi b−i∞ ( ) = S ( ) where the value of b must be chosen to be greater than the real coordinate of any poles in G s . Eq. (Õì) is sometimes called the Bromwich integral. By the residue theorem, therefore, the inverse transform is just the sum of the residues of G s est. ( ) ( ) Example Õ: Bromwich integral s a Given the Laplace transform G s , nd g t . s a ó kó + ( ) = ( ) ( + ) + Peter N. Saetaä Physics ÕÕÕ ß. EXERCISES AND PROBLEMS s a e poles of est occur where s a ik, so let s a ik z and s a ó kó + rewrite the integrand in terms of the new variable+ = z±: = − ± + ( + ) + ik z ik z G s est e(±ik−a+z)t e(±ik−a+z)t (Õ¦) ik z ó kó óikz zó ± + ± + ( ) = = We can now take the( limit± + as)z + þ and look for the± term+ proportional to z−Õ; we get → ± Õ ±ikt−at a− e (Õ¢) Õ ó Summing the residues, we nd = Õ g t e−at eikt e−ikt e−at cos kt (Õä) ó

( ) =−at ‰ + Ž = s+a You can conrm that e cos kt is indeed (s+a)ó+kó . L ™ ž

ß. Exercises and Problems Exercise ó Show that the Laplace transform of the derivative of a function g t is

dg ( ) sG s g þ (Õß) dt L › = ( ) − ( ) where G s g t . In other words, the Laplace transform converts derivatives to pow- ers of s. ( ) = L { ( )}

Exercise ì e convolution of two functions is dened by

t ′ ′ ′ f t gÕ t t gó t dt (՘) þ ( ) = ( − ) ( ) Show that f t GÕ s Gó s . S

L { ( )} = ( ) ( ) Exercise ¦ Calculate the Laplace transform of sin ωt.

Physics ÕÕÕß Peter N. Saeta