Residue Theory Basic Residue Theory The Residue Theorem

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1 7 Residue theory is bsiclly theory for computing integrls by looking t certin terms in the Lurent series of the integrted functions bout pproprite points on the complex plne We will develop the bsic theorem by pplying the Cuchy integrl theorem nd the Cuchy integrl formuls long with Lurent series expnsions of functions bout the singulr points We will then pply it to compute mny, mny integrls tht cnnot be esily evluted otherwise Most of these integrls will be over subintervls of the rel line 7 Bsic The Residue Theorem Suppose f is function tht, except for isolted singulrities, is single-vlued nd nlytic on some simply-connected region R Our initil interest is in evluting the integrl C f (z) dz where C is circle centered t point z t which f my hve pole or essentil singulrity We will ssume the rdius of C is smll enough tht no other singulrity of f is on or enclosed by this circle As usul, we lso ssume C is oriented counterclockwise In the region right round z, we cn express f (z) s Lurent series f (z) = k= k (z z ) k, nd, s noted erlier somewhere, this series converges uniformly in region contining C So f (z) dz = C k (z z ) k dz = k (z z ) C k dz C k= k= But we ve seen C (z z ) k dz for k =, ±, ±, ±3, 3//4

2 Chpter & Pge: 7 before You cn compute it using the Cuchy integrl theorem, the Cuchy integrl formuls, or even (s you did wy bck in exercise 44 on pge 4 7) by direct computtion fter prmeterizing C However you do it, you get, for ny integer k, { if k = (z z ) k dz = C iπ if k = So, the bove integrl of f reduces to { } if k = f (z) dz = k C iπ if k = k= = iπ This shows tht the coefficient in the Lurent series of function f bout point z completely determines the vlue of the integrl of f over sufficiently smll circle centered t z This quntity,, is clled the residue of f t z, nd is denoted by,z or Res z ( f ) or In fct, it seems tht every uthor hs their own nottion We will use Res z ( f ) If, insted of integrting round the smll circle C, we were computing f (z) dz C where C is ny simple, counterclockwise oriented loop in R tht touched no point of singulrity of f but did enclose points z, z, z, t which f could hve singulrities, then, consequence of Cuchy s integrl theorem (nmely, theorem 55 on pge 5 7) tells us tht f (z) dz = f (z) dz C k C k where ech C k is counterclockwise oriented circle centered t z k smll enough tht no other point of singulrity for f is on or enclosed by this circle Combined with the clcultions done just bove, we get the following: Theorem 7 (Residue Theorem) Let f be single-vlued function on region R, nd let C be simple loop oriented counterclockwise Assume C encircles finite set of points {z, z, z,} t which f might not be nlytic Assume, further, tht f is nlytic t every other point on or enclosed by C Then f (z) dz = iπ Res zk ( f ) (7) C k In prctice, most people just write eqution (7) s f (z) dz = iπ [ sum of the enclosed residues C The residue theorem cn be viewed s generliztion of the Cuchy integrl theorem nd the Cuchy integrl formuls In fct, mny of the pplictions you see of the residue theorem cn be done nerly s esily using theorem 55 (the consequence of the Cuchy integrl theorem used bove) long with the Cuchy integrl formuls

3 Bsic Chpter & Pge: 7 3 Computing Residues Remember, wht we re now clling the residue of function f t z in the Lurent series expnsion of f, is simply the vlue of f (z) = k= k (z z ) k, right round z, nd, for this to be nonzero, f must hve either pole or essentil singulrity t z So the discussion bout such singulrities in section 63 pplies both for identifying where function my hve residues nd for computing the residues The bsic pproch to computing the residue t z is be to simply find the bove Lurent series Then Res z ( f ) = This my be necessry if f hs n essentil singulrity t z If f hs pole of finite order, sy, of order M, then we cn use formul (68) on pge 6 4 for, d Res z ( f ) = = M [ (z z (M )! dz M ) M f (z) (7) z=z If the pole is simple (ie, M = ), this simplifies to Res z ( f ) = = (z z ) f (z) z=z (73) Often, we my notice tht f (z) = g(z) z z for some function g which is nlytic nd nonzero t z In this cse, we clerly hve simple pole, nd formul (73), bove, clerly reduces to Res z ( f ) = g(z ) This will mke computing residues very esy in mny cses More generlly, from our erlier discussion of poles, we know tht if f (z) = g(z) (z z ) M (74) for some function g which is nlytic nd nonzero t z, then f hs pole of order M t z In this cse, formul (7) reduces to Res z ( f ) = (M )! g(m ) (z ) (75) Keep in mind tht, if g(z ) =, then the pole of f (z) = g(z) (z z ) M hs order less thn M, nd little more work will be needed to determine the precise order of the pole nd the corresponding residue version: 3//4

4 Chpter & Pge: Simple Applictions The min ppliction of the residue theorem is to compute integrls we could not compute (or don t wnt to compute) using more elementry mens We will consider some of the common cses involving single-vlued functions not hving poles on the curves of integrtion Lter, we will dd poles nd del with multi-vlued functions Integrls of the form π f (sin θ, cosθ) dθ Suppose we hve n integrl over (, π) of some formul involving sin(θ) nd cos(θ), sy, π dθ + cosθ We cn convert this to n integrl bout the unit circle by using the substitution z = e iθ Note tht z does go round the unit circle in the counterclockwise direction s θ goes from to π Under this substitution, we hve So, dz = d [ e iθ = ie iθ dθ = iz dθ dθ = iz dz = iz dz For the sines nd cosines, we hve nd cosθ = eiθ + e iθ sin θ = eiθ e iθ i = z + / z = z / z i = z + z = z z i These substitutions convert the originl integrl to n integrl of some function of z over the unit circle, which cn then be evluted by finding the enclosed residues nd pplying the residue theorem! Exmple 7: Let s evlute π dθ + cosθ Letting C denote the unit circle nd pplying the substitution z = e iθ, s described bove, we get π dθ + cosθ = iz dz C + [ z + z z = C z iz + [ z + z dz i = 4z + [ z dz = i + z + 4z + dz C C

5 Simple Applictions Chpter & Pge: 7 5 Letting f (z) = z + 4z + nd pplying the residue theorem, the bove becomes π dθ + cosθ = i iπ [ sum of the residues of f (z) in the unit circle = 4π [ sum of the residues of f (z) in the unit circle To find the necessry residues we must find where the denomintor of f (z) vnishes, z + 4z + = Using the qudrtic formul, these points re found to be So f (z) = z ± = 4 ± 4 4 = ± 3 ( z [ + 3 )( z [ 3 ) Now, since 3 7, z + = while z = = 3 7 = 37 Clerly, + 3 is the only singulr point of f (z) enclosed by the unit circle, nd the singulrity there is simple pole We cn rewrite f (z) s f (z) = g(z) z [ + 3 where g(z) = z [ 3 Thus, Res z+ [ f = g(z + ) = g( + 3) = [ [ = Plugging this bck into the lst formul obtined for our integrl, we get π dθ + cosθ = 4π [ sum of the residues of f (z) in the unit circle = 4π Res z+ [ f = 4π 3 = π 3 version: 3//4

6 Chpter & Pge: 7 6 Y Y Γ + R R I R R X R I R R C R X Γ R () (b) Figure 7: Closed curves for integrting on the X xis when () Γ R = Γ + R = I R + nd when (b) Γ R = Γ R = I R +C R Integrls of the form Now let s consider evluting ssuming tht: f (x) dx f (x) dx, Except for finite number of poles nd/or essentil singulrities, f is single-vlued nlytic function on the entire complex plne None of these singulrities re on the X xis 3 One of the following holds: () z f (z) s z (b) f (z) = g(z)e iαz where α > nd g(z) s z (c) f (z) = g(z)e iαz where α > nd g(z) s z Under these ssumptions, we cn evlute the integrl by constructing, for ech R >, suitble closed loop Γ R contining the intervl ( R, R) The integrl over Γ R is computed vi the residue theorem, nd R is llowed to go to The conditions listed under the third ssumption bove ensure tht the integrl over tht prt of Γ R which is not the intervl ( R, R) vnishes s R, leving us with the integrl over (, ) The exct choice for the closed loop Γ R depends on which of the three conditions under ssumption 3 is known to hold If either (3) or (3b) holds, we tke Γ R = Γ + R = I R + where I R is the subintervl ( R, R) of the X xis nd is the semicirle in the upper hlf plne of rdius R nd centered t (see figure 7) In this cse, with Γ + R oriented

7 Simple Applictions Chpter & Pge: 7 7 counterclockwise, the direction of trvel on the intervl I R is from x = R to x = R (the norml positive direction of trvel on the X xis) nd, so, f (z) dz = I R R R f (x) dx Since we re letting R nd there re only finitely mny singulrities, we cn lwys ssume tht we ve tken R lrge enough for Γ + R to enclose ll the singulrities of f in the upper hlf plne (UHP) Then iπ [ sum of the residues of f in the UHP = f (z) dz Γ + R = f (z) dz + I R f (z) dz Thus, R R f (x) dx = I R f (z) dz = iπ [ sum of the residues of f in the UHP f (z) dz (76) To del with the integrl over, first note tht this curve is prmeterized by z = z(θ) = Re iθ where θ π So dz = d[re iθ = i Re iθ dθ nd f (z) dz = π f ( Re iθ) i Re iθ dθ Now ssume ssumption (3) holds, nd let M(R) = the mximum of z f (z) when z = R By definition, then, Re iθ f ( Re iθ) M(R) Moreover, it is esily verified tht ssumption (3) (tht z f (z) s z ) implies tht M(R) s R So R f (z) dz = R π f ( Re iθ) i Re iθ dθ π f ( Re iθ) i Re iθ dθ R π M(R) dθ = M(R)π = R R version: 3//4

8 Chpter & Pge: 7 8 Thus, Under ssumption (3b), for some α > Now, R f (z) dz = f (z) dz = (77) g(z)e iαz dz e iαz = e iα(x+iy) = e iαx e αy = e αy, which goes to zero very quickly s y So it is certinly resonble to expect eqution (77) to hold under ssumption (3b) And it does but rigorous verifiction requires little more spce thn is pproprite here Anyone interested cn find the detils in the proof of lemm 7 on pge 7 4 So, fter letting R, formul (76) for the integrl of f (x) on ( R, R) becomes f (x) dx = iπ [ sum of the residues of f in the UHP (78) On the other hnd, if it is ssumption (3c) tht holds, then f (z) dz = g(z)e iαz dz for some α > In this cse, though, e iαz = e iα(x+iy) = e iαx e αy = e αy, which rpidly blows up s y gets lrge So it is not resonble to expect eqution (77) to hold here Insted, tke Γ R = Γ R = I R + C R where C R is the semicirle in the lower hlf plne of rdius R nd centered t (see figure 7b) (Observe tht, this time, the direction of trvel on I R is opposite to the norml positive direction of trvel on the X xis) Agin, since we re letting R nd there re only finitely mny singulrities, we cn ssume tht we ve tken R lrge enough for Γ R to enclose ll the singulrities of f in the lower hlf plne (LHP) Then iπ [ sum of the residues of f in the LHP = f (z) dz Thus, R R f (x) dx = I R f (z) dz Γ R = f (z) dz + I R = f (z) dz + I R = iπ [ sum of the residues of f in the UHP + C R C R f (z) dz f (z) dz f (z) dz

9 Simple Applictions Chpter & Pge: 7 9 But, s before, it cn be shown tht f (z) fst enough s z (on the lower hlf plne) to ensure tht f (z) dz = R C R So, fter letting R, the bove formul for the integrl of f (x) on ( R, R) becomes! Exmple 7: f (x) dx = iπ [ sum of the residues of f in the LHP (79) Let us evlute the Fourier integrl Here we hve e iπ x + x dx f (z) = g(z)e iπz with g(z) = + z Clerly, g(z) s z nd π > ; so condition (3b) on pge 7 6 holds Thus, we will be pplying eqution (78), which requires the residues of f from the upper hlf plne By inspection, we see tht f (z) = eiπz + z = e iπz (z + i)(z i), which tells us tht the only singulrities of f (z) re t z = i nd z = i Only i, though, is in the upper hlf plne, so we re only interested in the residue t i Since we cn write f (z) s f (z) = h(z) z i with (nd h(i) = ), we know f (z) hs simple pole t i nd Thus, pplying eqution (78) Res i [ f = h(i) = eiπi i + i h(z) = eiπz z + i = e π i e iπ x + x dx = iπ[ sum of the residues of f in the UHP = iπ [ Res i [ f [ e π = iπ = πe π i Rther thn memorize tht condition (3b) implies tht (78) is used, keep in mind the derivtion nd the fct tht you wnt the integrl over one of the semicircles to vnish s R Write out the exponentil in terms of x nd y nd see if this exponentil is vnishing s y + or s y Then rederive the residue-bsed formul for the integrl of interest using the semicircle in the upper hlf plne if the exponentil vnishes s y + nd the semicircle in the lower hlf plne if the exponentil vnishes s y For our exmple So we re using the upper hlf plne e iπz = e iπ(x+iy) = e iπ e π y s y + version: 3//4

10 Chpter & Pge: 7 Stndrd Simple Tricks Using Rel nd Imginry Prts It is often helpful to observe tht one integrl of interest my be the rel or imginry prt of nother integrl tht my, possibly, be esier to evlute Do remember tht b [ b Re[ f (x) dx = Re f (x) dx nd b [ b Im[ f (x) dx = Im f (x) dx (If this isn t obvious, spend minute to (re)derive it) In this regrd, it is especilly useful to observe tht cos(x) = Re [ e i X nd sin(x) = Im [ e i X! Exmple 73: Consider cos(π x) + x dx Using the bove observtions nd our nswer from the previous exercise, cos(π x) [ e iπ x dx = Re dx + x + x [ e = Re iπ x + x dx = Re [ πe π = πe π Clever Choice of Curve nd Function The min trick to pplying residue theory in computing integrls of rel interest (ie, integrls tht ctully do rise in pplictions) s well s other integrls you my encounter (eg, other integrls in ssigned homework nd tests) is to mke clever choices for the functions nd the curves so tht you relly cn extrct the vlue of desired integrl from the integrl over the closed curve used Some suggestions, such s were given on pge 7 6 for computing certin integrls on (, ), cn be given In generl, though, choosing the right curves nd functions is cross between n rt nd skill tht one just hs to develop A good exmple of using both clever choices of functions nd curves, nd in using rel nd imginry prts, is given in computing the Fresnel integrls (which rise in optics)! Exmple 74: Consider the Fresnel integrls cos ( x ) dx nd sin ( x ) dx Rther thn use cos ( x ) nd sin ( x ) directly, it is clever to use e i x nd the fct tht e i x = cos ( x ) + i sin ( x ) So, if we cn evlute e i x dx,

11 Simple Applictions Chpter & Pge: 7 Y Re iπ/4 l R Γ R A R π/ 4 I R R X Figure 7: The closed curve for computing the Fresnel integrls then we cn get the two integrls we wnt vi nd Now, to compute cos ( x ) dx = sin ( x ) dx = Re [ e i x dx = Re [ Im [ e i x dx = Im [ e i x dx, e i x dx e i x dx we nturlly choose the function f (z) = e iz The clever choice for the closed curve is sketched in figure 7 It is Γ R = I R + A R l R where, for ny choice of R >, I R = the stright line on the rel line from z = to z = R = the intervl on from x = to x = R, nd A R = the circulr rc centered t strting t z = R nd going to z = Re iπ/4, l R = the stright line from z = to z = Re iπ/4 Notice tht the chosen function, f (z) = e iz, is nlytic on the entire complex plne So there re no residues, nd, for ech R >, we hve = dz = dz + dz dz So, Γ R e iz R x= = e iz I R R x= e i x dx = Letting R, this becomes e i x dx = R e i x dx + l R e iz l R e iz A R e iz dz A R e iz dz A R e iz l R e iz dz l R e iz dz dz R e iz dz A R (7) version: 3//4

12 Chpter & Pge: 7 Now, l R is prmeterized by z = z(r) = re iπ/4 with r going from to R Using this, nd R e iz dz = R l R R iz = ir e iπ/ = ir i = r, dz = d [ re iπ/4 = e iπ/4 dr e r e iπ/4 dr = e iπ/4 e r dr The vlue of the lst integrl is π/ This cn be found by chep tricks not involving residues (see the ppendix on the integrl of the bsic Gussin strting on pge 7 6) Thus, e iz dz = e iπ/4 π (7) R l R The curve A R for the other integrl in eqution (7) is prmeterized by Using this, z = z(θ) = Re iθ with θ going from to π / 4 iz = i R e iθ = i R [cos(θ) + i sin(θ) = i R cos(θ) R sin(θ), dz = d [ Re iθ = i Re iθ dθ nd e iz dz = R A R R π/4 e i R cos(θ) R sin(θ) i Re iθ dθ Note tht π/4 = R π/4 e i R cos(θ) e R sin(θ) i Re iθ dθ π/4 e i R cos(θ) e R sin(θ) i Re iθ dθ e i R cos(θ) e R sin(θ) i Re iθ dθ = π/4 e R sin(θ) R dθ For ech θ in (, π / 4, the exponentil in the lst integrl goes to zero much fster thn R increses So this integrl clerly vnishes s R (see exercise 7 on pge 7 6 for rigorous verifiction) Thus, R e iz dz = A R (7)

13 Simple Applictions Chpter & Pge: 7 3 get Gthering together wht we ve derived (equtions (7),(7) nd (7)), we finlly e i x dx = e iz dz e iz dz R l R R A R nd Consequently, = e iπ/4 π + [ ( ) ( ) π π π = cos + i sin 4 4 [ = + i π = cos ( x ) [ dx = Re e i x dx sin ( x ) [ dx = Im e i x dx π + i = = π π π? Exercise 7: Consider the lst exmple : Why ws the curve Γ R = I R + A R l R, s illustrted in figure 7, such clever choice of curves? (Consider wht hppened to the function e iz on I R ) b: Why would the curve Γ R = I R + illustrted in figure 7 not be clever choice for computing the Fresnel integrls? In the lst exmple, s in previous exmple, the closed curve constructed so the residue theorem cn be pplied included piece of circle of rdius R, sy, the A R in our lst exmple Our formul for the desired integrl then includes term of the form R A R f (z) dz, nd it is importnt tht this it be zero (or some other computble finite number) Otherwise, either this generl pproch fll prt, or we cn show tht the desired integrl does not converge Keep in mind tht the length of A R increses s R increses; so the vnishing of the integrnd is not enough to ensure the vnishing of the bove it To tke into ccount the incresing length of the curve, it is often good ide to prmeterize it using ngulr mesurement, z = z(θ) = Re iθ with θ ited to some fixed intervl (α,β) Then, s we ve seen in exmples, β f (z) dz = f ( Re iθ) i Re iθ dθ A R α β α f ( Re iθ ) i Re iθ dθ = β α f ( Re iθ ) R dθ version: 3//4

14 Chpter & Pge: 7 4 With luck, you will be ble to look t the inside of the lst integrl nd tell whether the desired it is zero or whether the it does not exist Of course, you my sk, why not just compute the it by bringing the it inside the integrl, β ( ) β ( f Re iθ R dθ = ) f Re iθ R dθ? R α α R Becuse this lst eqution is not lwys vlid There re choices for f such tht R β α f ( Re iθ) R dθ = For exmple, f (z) = e z with (α,β) = (, π / ) β α R f ( Re iθ) R dθ! Appendices to this Section Two issues re ddressed here One is how to rigorously verify tht certin integrls over rcs of rdii R vnish s R goes to infinity The other is how to evlute e s ds These ppendices re included for the ske of completeness You should be cquinted with the generl results, but don t worry bout reproducing the sort of nlysis given here The Vnishing of Certin Integrls s R Often, when using residues to compute integrls with exponentils, it is necessry to verify tht certin integrls over rcs of rdii R vnish s R goes to infinity Sometimes this is esy to show; sometimes it is not To illustrte how we might verify the cses involving complex exponentils, we will rigorously verify the following lemm It is the lemm needed to rigorously verify eqution (77) on pge 7 8 Lemm 7 Let α >, nd ssume g is ny resonbly smooth function on the complex plne stisfying Then where g(z) s z g(z)e iαz dz = R is the upper hlf of the circle of rdius R bout the origin PROOF: Keep in mind tht is getting longer s R gets lrger So the simple fct tht the integrnd gets smller s R gets lrger is not enough to ensure tht the bove it is zero To help tke into ccount the incresing length of the curve nd to help convert the integrl to something little more esily to del with, we mke use of the fct tht this curve cn be prmeterized by z = z(θ) = Re iθ = R [cos(θ) + i sin(θ) where θ π For convenience, let M( R) = mximum vlue of g(z(θ)) when θ π it suffices to ssume g(z) is continuous on the region where z > R for some finite rel vlue R

15 Simple Applictions Chpter & Pge: 7 5 Y y = θ y = sin(θ) π π Θ Figure 73: The grphs of y = sin(θ) nd y = θ / on the intervl [, π / It isn t hrd to show tht the fct tht g(z) s z implies tht M(R) when R Also note tht tht, using the bove prmetriztion, dz = i Re iθ dθ nd e iαz(θ) = e iαr[cos(θ)+i sin(θ) = e iαr cos(θ) αr sin(θ) = e iαr cos(θ) e αr sin(θ) = e αr sin(θ) So, g(z)e iαz dz = π π g(z(θ))e iαz(θ) i Re iθ dθ g(z(θ))e iαz(θ) i Re iθ dθ π M(R)e αr sin(θ) R dθ Pulling out M(R) nd using the fct tht sin(θ) is symmetric bout θ = π /, this becomes π/ g(z)e iαz dz M(R) e αr sin(θ) R dθ (73) Now, it is esy to see nd esy to confirm tht sin(θ) > θ (see figure 73) It then follows tht for θ π (74) e αr sin(θ) < e αrθ/ for θ π (75) Plugging this into inequlity (73) nd computing the resulting integrl yields π/ g(z)e iαz dz M(R) e αrθ/ R dθ = 4M(R) α [ e αrπ/4 < 4M(R) α version: 3//4

16 Chpter & Pge: 7 6 Thus, R which, of course, mens tht g(z)e iαz dz < R 4M(R) α g(z)e iαz dz =, R = 4 α =, s ced? Exercise 7: Using the ides behind inequlities (74) nd (75), show tht π/4 e R sin(θ) R dθ π/4 e Rθ R dθ = R s R Integrl of the Bsic Gussin While there is no simple formul for the indefinite integrl of e x, the vlue of the definite integrl is esily computed vi clever trick To begin, observe tht, by symmetry, where I = e s ds = e s ds e s ds = I e x dx = e y dy The clever trick is bsed on the observtion tht I, the product of I with itself, cn be expressed s double integrl over the entire XY plne, ( ) ( ) I = e x dx e y dy ( = = ( e x dx e x e y dx ) e y dy ) dy = e (x +y ) dx dy This double integrl is esily computed using polr coordintes (r,θ) where x = r cos(θ) nd y = r sin(θ) Recll tht x + y = r nd dx dy = r dr dθ

17 Integrls Over Brnch Cuts of Multi-Vlued Functions Chpter & Pge: 7 7 So, converting to polr coordintes nd using elementry integrtion techniques, I = Tking the squre root gives Becuse e s π e r r dr dθ = I = ± π π dθ = π is positive function, its integrl must be positive Thus, e s ds = I = π nd e s ds = I = π 73 Integrls Over Brnch Cuts of Multi-Vlued Functions If you need to compute β α f (x) dx when f (z) is multi-vlued function, then clever choice for the closed curve my include using the intervl (α,β) s brnch cut for f, nd letting it serve s two prts of the curve enclosing the residues of f Just wht I men will be lot clerer if we do one exmple But first, go bck nd re-red the discussion of multi-vlued functions strting on pge 4 5 Also, re-red the bit bout Squre Roots nd Such strting on pge 4 8! Exmple 75: Nturlly, we will let Consider evluting x + x dx f (z) = z / + z This function hs singulrities t z = ±i (where the denomintor is zero) Also, becuse of the z / fctor, f (z) is multi-vlued with brnch point t z = We will cleverly tke the cut line to be the positive X xis, nd define z / to be given by z / = z e iθ/ where θ is the polr ngle (rgument) of z stisfying < θ < π version: 3//4

18 Chpter & Pge: 7 8 Y i C R Γ εr C ε ε I + εr I εr Γ εr Γ εr R X i Figure 74: The curves for exmple 75 The region enclosed by Γ εr is shded, nd the rrows indicte the direction of trvel long Γ εr This defines the brnch of f tht we will use This lso mens tht we will hve to be creful bout the vlue of this function on the positive X xis since, for ech point x on this intervl, while Thus, while z / z x Im z> = θ + x e iθ/ = x e i = x z x z / = x e iθ/ θ π = x e iπ = x Im z< z x Im z> z x Im z< f (z) = z x Im z> f (z) = z x Im z> z / + z = z / x + x x = + z + x The closed curve we will use with the residue theorem is constructed from the curves in figure 74 For ech pir of positive vlues ε nd R stisfying ε < < R, we let Γ εr be the closed curve given by where, s indicted in figure 74, Γ εr = I + εr + C R + I εr C ε nd C R = circle of rdius R bout, oriented counterclockwise, C ε = circle of rdius ε bout, oriented counterclockwise, I + εr I εr = stright line on the X xis from x = ε to x = R, = stright line on the X xis from x = R to x = ε

19 Integrls Over Brnch Cuts of Multi-Vlued Functions Chpter & Pge: 7 9 The circle C ε isoltes the brnch point from the region encircled by Γ εr This is good ide becuse bd things cn hppen ner brnch points Lter, we will check tht we cn (or cnnot) let ε The curves I + εr nd I εr re ech, in fct, simply the subintervl (ε, R) on the X xis oriented in opposing directions They re treted, however, s two distinct pieces of Γ εr, with I + εr viewed s lower boundry to the region bove it, nd I εr viewed s the upper boundry to the region below it Indeed, it my be best to first view them s lying smll distnce bove nd below the X xis, exctly s sketched in figure 74, with this distnce shrunk to zero by the end of the computtions Consequently, the region encircled by Γ εr is the shded region in figure 74 Applying the residue theorem, we hve iπ [sum of the resides of f in the shded region = f (z) dz Γ = f (z) dz + f (z) dz + C R I + εr I εr f (z) dz C ε f (z) dz where nd I εr Thus, f (z) dz = I + εr ε R f (z) dz = R ε [ f (z) z x Im z> dx = R ε x + x dx [ R [ x f (z) dx = dx = z x Im z< ε + x iπ [sum of the resides of f in the shded region R x = ε + x dx + f (z) dz f (z) dz, C ε nd, so, R x dx = iπ [sum of the resides of f in the shded region ε + x + f (z) dz f (z) dz C ε C R (Notice tht, becuse of the multi-vlueness of f, f (z) dz nd I + εr C R I εr f (z) dz R ε x + x dx (76) did not cncel ech other out even though I + εr nd I εr re the sme curve oriented in opposite directions Tht is wht will mke these computtions work) Nturlly, we wnt ε nd R Now, if z = ε nd ε <, then z / f (z) = + z ε / ε version: 3//4

20 Chpter & Pge: 7 So, using the prmetriztion z = z(θ) = εe iθ, π f (z) dz = f (z(θ)) iεe iθ dθ C ε π f (z(θ)) ε dθ π ε / By very similr computtions, we get (since R > ), C R π f (z) dz 3/ ε ε dθ = ε π s ε ε R / R R dθ = R 3/ π s R R Thus, fter letting ε nd R, eqution (76) reduces to x dx = iπ [sum of the resides of f in the shded region (77) + x Clerly, the only singulrities of re t z = ±i, nd f (z) = z / + z f (z) = g(z) z i nd f (z) = h(z) z ( i) where g(z) = z / z + i So these singulrities re simple poles, nd nd h(z) = z / z i x dx = iπ [sum of the resides of f in the shded region + x = iπ [ Res i ( f ) + Res i ( f ) = iπ[g(i) + h( i) [ i / = iπ + ( i) / i + i i i [ e iπ/4 = π ei3π/4 = = π

21 Integrting Through Poles nd the Cuchy Principl Vlue Chpter & Pge: 7 74 Integrting Through Poles nd the Cuchy Principl Vlue The Cuchy Principl Vlue of n Integrl The Cuchy principl vlue of n integrl is ctully redefinition of the integrl so tht certin types of singulrities re ignored through the process of tking symmetric its To be precise: Let f be function on n intervl (, b) nd ssume f is well behved (sy, is nlytic) t every point in the intervl except one, x We then define The Cuchy principl vlue of b b f (x) dx = CPV = ε + [ x ε f (x) d x f (x) dx + (my nottion) b x +ε f (x) dx By the wy, in Arfken, Weber nd Hrris you ll find b CPV b P f (x) dx = ε + b f (x) dx nd f (x) dx denoting the bove Cuchy principl vlue If f is continuous, or even just hs jump discontinuity, t x, then [ x ε b f (x) dx + f (x) dx = = x b f (x) dx + f (x) dx b x f (x) dx This simply mens tht the Cuchy principl vlue reduces to the integrl when the integrl is well defined However, when the singulrity is simple pole then you cn esily verify tht the Cuchy principl vlue removes the effect of the singulrity by cnceling out infinities 3! Exmple 76: Consider x dx nd CPV The function being integrted hs singulrity t x =, nd 3 x +ε x dx 4 3 x dx = 3 x dx + 4 x dx = ln x 3 + ln x 4 = ln 3 + ln 4 () 3 In prctice, mke sure you cn justify this cnceling out of infinities Otherwise, your use of the Cuchy principl vlue is probbly frudulent version: 3//4

22 Chpter & Pge: 7 = + ln 4 3, which, for very good resons, is considered to be undefined On the other hnd, 4 [ ε 4 CPV dx = x ε + x dx + x dx 3 3 = [ln ε ln 3 + ln 4 ln ε ε + = ε + [ ln 3 + ln 4 = ln 4 3 The Cuchy principl prt should not just be thought of s wy to get round technicl difficulties with integrls tht blow up, nd its indiscriminte use cn led to dngerously misleding results For exmple, if the force on some object t position x is given by / x, then one could nively rgue tht the work needed to move tht object from x = 3 to x = 4 is just work = CPV 4 3 x dx = ln 4 3 but just see if you relly cn push tht object pst x = The Cuchy principl vlue cn be useful, but its use must be justified by something more thn desire to cncel out infinities Also, sometimes those infinities just don t cncel out ε,? Exercise 73: Let n be ny positive integer, nd show tht CPV x n dx = { if n is odd if n is even By the wy, if the integrnd hs severl singulrities on the intervl (, b), then the Cuchy principl vlue is computed by tking the bove described symmetric it t ech singulrity Also, if (, b) = (, ), then we hve the improper integrl version of the Cuchy principl vlue given by CPV Agin, its use must be justified f (x) dx = R R + R f (x) dx Linerity of the Cuchy Principl Vlue It should be briefly noted tht, like the regulr integrl, the Cuchy principl vlue is liner To be precise, you cn esily show using the linerity of integrls nd its tht, if t lest two of the following exist s finite numbers, CPV b f (x) dx, CPV then they ll exist nd, moreover, b g(x) d x nd CPV b [ f (x) + g(x) dx, CPV b [α f (x) + βg(x) dx = α CPV b b f (x) dx + β CPV g(x) dx

23 Integrting Through Poles nd the Cuchy Principl Vlue Chpter & Pge: 7 3 for ny pir of numbers α nd β In prticulr, if the principl vlues exist, we hve CPV b [u(x) + iv(x) dx = CPV From this it lmost immeditely follows tht nd We will be using this lter [ b Re CPV [ b Im CPV f (x) dx f (x) dx b = CPV = CPV u(x) dx + i CPV b b Re[ f (x) dx b Im[ f (x) dx v(x) dx Integrting Through Singulrities Sometimes the function you re integrting hs singulrity t point on the curve over which you must integrte the function When this hppens, there re t lest three things you cn do to del with this integrl: Use the Cuchy principl vlue Modify the function slightly to move the singulrity off the curve by distnce of ε nd then see wht hppens s ε 3 Look very closely t your problem nd decide if the fct tht this integrl doesn t relly exist is telling you something bout the physics of the problem If you do either the first or the second of the bove, then mke sure you cn justify your choice by something other thn I don t like infinities This mens tht you often should do the third thing in the bove list to help justify the your choice Simple Poles nd the Cuchy Principl Vlue The discussion strting on pge 7 6 concerning integrls of the form f (x) dx, cn be esily dpted to tke into ccount the possibility of f hving simple poles on the X xis Wht we will discover is tht we don t ctully end up with the bove integrl, but with the Cuchy principl vlue of tht integrl So let s consider evluting the bove integrl ssuming tht: Except for finite number of poles nd/or essentil singulrities, f is single-vlued nlytic function on the entire complex plne Only one of these singulrities is on the X xis, nd tht is simple pole t x 3 One of the following holds: () z f (z) s z version: 3//4

24 Chpter & Pge: 7 4 Y Y C + ε R x R X R x R X () (b) C ε Figure 75: Isolting pole on the X xis (Compre with figure 7 on pge 7 6) (b) f (z) = g(z)e iαz where α > nd g(z) s z The computtions here differ from those on pge 7 6 only in tht we initilly isolte the pole t x by circle of rdius ε centered t x (oriented counterclockwise) with, unsurprisingly, C ε + nd Cε denoting the upper nd lower hlves (see figures 75 nd 75b) This lso mens tht, insted of using the entire intervl ( R, R) in the closed curve for the residue theorem, we use the subintervls ( R, x ε) nd ( R, x ε) long with either C ε + or Cε (possibly re-oriented) For these clcultions, C ε + will be used, s in figure 75 Tke R lrge enough nd ε smll enough tht the closed curve figure 75 encircles ll the singulrities of f in the upper hlf plne (UHP) The residue theorem then tells us tht iπ [ sum of the residues of f in the UHP = x ε R f (x) dx C + ε f (z) dz + By the sme rguments s given before, the integrl over letting R, the lst eqution becomes iπ [ sum of the residues of f in the UHP = x ε f (x) dx R x +ε f (x) dx + f (z) dz shrinks to s R So, fter C + ε f (z) dz + x +ε f (x) dx Hence, x ε f (x) dx + Letting ε this becomes CPV x +ε f (x) dx = iπ [ sum of the residues of f in the UHP + C + ε f (z) dz f (x) dx = iπ [ sum of the residues of f in the UHP + ε + C ε + f (z) dz (78)

25 Integrting Through Poles nd the Cuchy Principl Vlue Chpter & Pge: 7 5 On C + ε we cn use the Lurent series for f round x, long with the prmetriztion z = z(θ) = x + εe iθ with θ π Since f is ssumed to hve simple pole t x, the Lurent series will be of the form f (z(θ)) = c k z k = k= c k ε k e ikθ = c ε e iθ + k= c k ε k e ikθ k= Remember c = Res x ( f ) So ε + C ε + π f (z) dz = ε + [ π = c i + ε + = ε + = ε + = ε + [ [ [ = c iπ + [ c ε e iθ + c k ε k e ikθ iεe }{{ iθ dθ} k= dz c k iε k+ e i(k+)θ dθ ic π ic π + ic π + k= = iπ Res x ( f ) k= dθ + π ic k ε k+ e i(k+)θ dθ k= π ic k ε k+ e i(k+)θ dθ k= k= Plugging this result bck into eqution 78 gives us c k ε k+ ( e i(k+)π ) k + c k k+ ( e i(k+)π ) k + CPV f (x) dx = iπ [ sum of the residues of f in the UHP + iπ Res x ( f ) (79) This lst eqution ws derived using the curves in figure 75 It turns out tht you get exctly the sme result using the curves in figure 75b You should verify this yourself If there is more thn one simple pole on the X xis, then the residue of ech contributes, nd we hve the following result: Lemm 73 Let f be function which, except for finite number of poles nd/or essentil singulrities, is single-vlued nlytic function on the entire complex plne Assume, further, tht the singulrities on the X xis re ll simple poles, nd tht either z f (z) s z version: 3//4

26 Chpter & Pge: 7 6 or f (z) = g(z)e iαz where α > nd g(z) s z Then CPV f (x) dx = iπ [ sum of the residues of f in the UHP + iπ [ sum of the residues of f in the X xis (7) Deriving the corresponding results when we use the residues in the lower hlf plne (LHP) will be left to you? Exercise 74: Assume the following concerning some function f on : nd Except for finite number of poles nd/or essentil singulrities, f is single-vlued nlytic function on the entire complex plne All the singulrities on the X xis re simple poles 3 f (z) = g(z)e iαz where α > nd g(z) s z Then show tht CPV f (x) dx = iπ [ sum of the residues of f in the LHP iπ [ sum of the residues of f in the X xis (7) Wht if the singulrities on the X xis re not simple poles sy, poles of higher order or essentil singulrities? The computtions done bove cn still be ttempted, but, in computing f (z) dz ε + C + ε there will generlly be terms involving / ε tht blow up So, in generl, we cnnot del with singulrities worse thn simple poles on the X xis, t lest not using the Cuchy principl vlue Let s do n exmple where the Cuchy principl vlue is of vlue! Exmple 77: Consider sin(x) x dx The only possible point where we my hve pole in the integrnd is t x = z = In fct, though, you cn esily show tht sin(z) z z =

27 Integrting Through Poles nd the Cuchy Principl Vlue Chpter & Pge: 7 7 So, in fct, sin(z) / z is nlytic t z =, the integrnd in the bove integrl is continuous t x =, nd we hve sin(x) sin(x) dx = CPV dx x x This becomes something other thn stupid, obvious observtion fter lso noting tht, for x, [ sin(x) e i x = Im x x nd tht f (z) = eiz z stisfies the conditions for f in lemm 73 Applying tht lemm (nd noting tht the only residue of f is t z = ), we hve ( ) e CPV i x e x dx = iπ Res i x ( = iπ Res ) e i = iπ = iπ x So, sin(x) x dx = CPV = CPV [ = Im CPV sin(x) dx x [ e i x Im dx x e i x x dx = Im[iπ = π Moving Singulrities To be written, somedy Bsiclly, this is the ide of replcing with either b y + b g(x) x x dx g(x) b dx or x (x + y ) y + g(x) x (x y ) dx Agin, there must be some justifiction for this Be wrned tht, in generl, b CPV g(x) x x dx = b y + b = y + g(x) x (x + y ) dx g(x) b dx = CPV x (x y ) g(x) x x dx version: 3//4