INTEGRATION. 1 Integrals of Complex Valued functions of a REAL variable

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1 INTEGRATION NOTE: These notes re supposed to supplement Chpter 4 of the online textbook. 1 Integrls of Complex Vlued funtions of REAL vrible If I is n intervl in R (for exmple I = [, b] or I = (, b)) nd h : I C writing h = u + iv where u, v : I C, we n extend ll lulus 1 onepts to h by simply sying H stisfies property P if nd only if u nd v stisfy P. Speifilly: The funtion h is ontinuous on I if nd only if u, v re ontinuous on I. The funtion h is differentible on I if nd only if u, v re differentible on I. In this se we DEFINE for t I. h (t) = u (t) + iv (t) The funtion h is integrble on [, b] if nd only if u, v re integrble on [, b]. In this se we define = u(t) dt + i v(t) dt. The expeted properties hold, even if some re not s obvious s one might think. The fundmentl theorem of lulus still holds, if H = U + iv : [, b] C nd H = h = u + iv, then u = U, v = V nd = u(t) dt + i v(t) dt = U(b) U() + i(v (b) V ()) = (U(b) + iv (b)) (U() + iv ()) = H(b) H(). The formul for hnge of vribles still holds: If φ is rel vlued differentible funtion defined on [, b] nd h is defined on the rnge of φ, then h(φ(s)) φ (s) ds = φ(b) φ().

2 1 INTEGRALS OF COMPLEX VALUED FUNCTIONS OF A REAL VARIABLE2 A more fmilir version of this formul is the following: In the integrl we n mke the substitution t = φ(s), dt = φ (s) ds to get = φ 1 (d) φ 1 () h(φ(s))φ (s) ds. It should be ler why this is lso vlid if h = u + iv. It should be quite ler lso tht the integrl of sum or differene is the sum or differene of the integrl. It is bit more omplited to verify tht omplex onstnt pulls out of the integrl. Sy α = + ib,, b rel nd h = u + iv s before. Then α = = = ( + ib)(u(t) + iv(t)) dt = (u(t) bv(t)) dt + i u(t) dt b ( = ( + ib) u(t) dt + i v(t) dt + i [(u(t) bv(t)) + i(v(t) + bu(t))] dt (v(t) + bu(t)) dt ( v(t) dt ) v(t) dt + b = α. u(t) dt ) For rel vlued ontinuous funtions h it is true tht h(t) dt. (1) The proof is rther immedite; one hs h(t) h(t) h(t) for ll t; integrls of rel vlued fntions preserve the order so tht h(t) dt h(t) dt, whih is preisely (1). Things get hrder in the omplex relm. If h = u + iv, then (1) sttes tht ( 2 ( 2 b b u(t) dt) + v(t) dt) u(t)2 + v(t) 2 dt. As usul, rewriting omplex vlued funtion in terms of its rel nd imginry prts is the hrdest wy to go. Here is proof (1) if h is omplex vlued. I

3 1 INTEGRALS OF COMPLEX VALUED FUNCTIONS OF A REAL VARIABLE3 think it is quite n ingenious nd plesnt proof. We n write, s for every omplex number, where r = = re iθ, nd θ [, 2π) is the rgument. Then = r = e iθ = e iθ. The lst equlity is due to the ft tht onstnt n go in or out of the integrl, s proved bove. We will use now ( ) b Re = beuse bsolute vlues re rel; we lso use tht Re w w for ll omplex numbers w, tht the rel prt of omplex integrl is the integrl of the rel prt of the integrnd, nd tht integrls of rel vlued funtions preserve the order. We left off bove in = e iθ ; tking rel prts of both sides, using the mentioned fts, = Re = Re e iθ = e iθ h(t) dt = h(t) dt proving (1); the lst equlity is due to e iθ = 1. Re ( e iθ h(t) ) dt Finlly, here is useful differentition formul. Suppose h(t) = e αt where α = + ib,, b R. Then h (t) = αe αt ; tht is, just s in the rel se. Here is the proof: d dt eαt = αe αt, (2) d dt eαt = d dt et+ibt = d { e t os bt + ie t sin bt } dt = e t os bt be t sin bt + i sin bt + i ( e t sin bt + be t os bt ) = ( + ib) ( e t os bt + ie t sin bt ) = αe αt.

4 2 INTEGRATION OVER PATHS 4 2 Integrtion over Pths Definition 1 Assume : [, b] C is pieewise smooth pth ( defintion is given in the doument titled Conordne With the Online Text ); ssume thus tht we hve points t,..., t n, = t < t 1 < < t n = b suh tht (t) is ontinuously differentible (i.e., differentible with ontinuous derivtive) in eh open intervl (t i 1, t i ) nd (t) hs limit s t pprohes the endpoints of the intervl, from the right t t i=1, from the left t t i. Assume lso tht f is funtion defined t ll points of the urve desribed by so tht f((t)) mkes sense for ll t [, b]. Finlly ssume tht f((t)) is ontinuous for t b. We then define f(z) dz by n i=1 ti t i 1 f((t)) (t) dt. I wnt to emphsize: In this ourse, ny time you see the nottion f(z) dz, or vrint thereof (the urve will not lwys be denoted by, nor will the funtion lwys be f, or the vrible of integrtion z) the following onditions must hold, even if not expliitly stted: 1. is pieewise smooth pth. 2. The domin of f ontins the rnge of ; tht is f is defined t ll points desribed by. 3. The omposition of f with is ontinuous. We will dopt nother onvention, to keep the nottion from beoming too umbersome. Suppose, t,..., t n re s bove. We define or write n ti f((t)) (t) dt = f((t)) (t) dt t i 1 i=1 so tht we n lwys write f((t)) (t) dt. Of ourse, if we tully hve to ompute the integrl by the definition nd is not smooth, just pieewise smooth, we will need to brek up the intervl into portions on whih is smooth. One thing to keep in mind is tht the min use of pths will be to integrte over them. This llows us to mke definition tht, like ll definitions, mens extly wht it sys;

5 2 INTEGRATION OVER PATHS 5 Definition 2 Two pths 1 : [, b] C nd 2 : [, d] C will be sid to be equivlent nd we will even buse nottion to the point of writing 1 = 2 if (nd only if): 1. They desribe the sme set of points; tht is, both hve the sme rnge, desribe the sme urve. 2. If we denote this rnge by C, then if f is ny funtion whose domin ontins C nd is ontinuous on C (see below), then f(z) dz. 1 2 To sy tht f is ontinuous on C mens tht if {z n } is ny sequene of points on C onverging to point w C, then {f(z n )} onverges to f(w). But it is equivlent to sying tht f((t)) is ontinuous s funtion of t for one nd then lso for every pth desribing C. Here re few exmples. i. Suppose 1 : [, 2π] C is given by 2 : [, 1] C is given by 3 : [ π, pi] C is given by 1 (t) = e it = os t + i sin t, 2 (t) = e 2πit = os(2πt) + i sin(2πt), 3 (t) = e it = os t + i sin t, nd 4 by 4 (t) = ie it = sin t + i os t. As funtions no two of 1, 2, 3, 4 re the sme; they re different funtions. But in the sense of the definition bove, 1 = 2 = 3 4. The geometri objet desribed by ll four of them is the unit irle C = {z C : z = 1} entered t the origin. Thus the first ondition for equivlene is stisfied for ll of them. It is up to the seond ondition. Deomposing everything into rel nd imginry prts is bsilly no help t ll. For exmple, suppose f u + iv is defined on C. Then (plese, hek this!) 1 = 2π 2π f(os t, sin t)( sin t + i os t) dt ( u(os t, sin t) v(os t, sin t) os t) dt + i 2π (u(os t, sin t) os t v(os t, sin t) sin t) dt

6 2 INTEGRATION OVER PATHS 6 The more one deomposes, the worse it gets. The best wy of writing these integrls s ordinry Riemnn integrls is to sty in the omplex relm. Using the eqution (2) to ompute i (t) for i = 1, 2, 3, 4, we n write 2π i f(e it )e it dt, (3) 1 1 2πi f(e 2πit )e 2πit dt, (4) 2 π i f(e it )e it dt, (5) 3 π 2π f(ie it )e it dt. (6) 4 To see tht pth 1 = 2 (in the sense explined bove!) we n mke the substitution t = 2πs, dt = 2π ds in the first integrl (7); one gets the seond integrl (4). To see tht the first integrl equls the third n be bit more triky. One n use tht the integrnd is periodi of period 2π nd the integrl of periodi funtion over ny intervl of length equl to the period is the sme. Or one n write i 2π f(e it )e it dt, = i π f(e it )e it dt, +i 2π π f(e it )e it dt nd in the seond integrl of the right hnd side we mke the substitution t = s + 2π, so s = t 2π, dt = ds. We use tht e 2πi = 1. Then i 2π f(e it )e it dt = i = i π π f(e it )e it dt + i f(e it )e it dt + i 2π 2π π 2π π f(e i(s+2π) )e i(s+2π) dt f(e is )e is dt = i π π f(e it )e it dt. Coming to the fourth integrl, to see it might not be equl to the other three, we n tke onrete f nd evlute. For exmple, tke f(z) = Re z. Then Re (e it ) = os t, Re (ie it ) = sin t so tht 2π 2π i (os t)e it dt = i (os 2 t + i os t sin t) dt = i(π + i) = iπ, 1 while 4 2π (sin t)e it dt = 2π Sine iπ iπ, the pths re not equivlent. (sin t os t i sin 2 t) dt = ( iπ) = iπ. Rule of thumb (mostly works): If you hve two prmetriztions desribing the sme urve nd if the the urve is gone through the sme number of times

7 2 INTEGRATION OVER PATHS 7 nd in the sme diretion in both ses, then the pths desribing tht urve re equivlent. The most ommon equivlene of pths is by substitution: If : [, b] C is pth, if φ : [, d] [, b] is ontinuous, inresing, differentible in the open intervl (, d) nd φ() = φ(d) = b, then the pth 1 : [, d] C defined by 1 (t) = (φ(t)) is equivlent to 1. This is beuse 1(t) = (φ(t))φ (t) so tht if in f( 1 (t)) 1(t) dt = f((φ(t))) (φ(t))φ (t) dt 1 we mke the hnge of vribles (substitution) s = φ(t), ds = φ (t) dt, we get f((φ(t))) (φ(t))φ (t) dt = f((s)) (s) ds = f(z) dz. One onsequene is: Every pth is equivlent to one prmeterized by the intervl [, 1]. In ft, onsider pth : [.b] C. Define φ : [, 1] [, b] by φ(s) = + s(b ), whih is nie inresing differentible funtion stisfying φ() =, φ(1) = b.