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1 EVALUATIOM OF INTEGRALS USING ONTOUR INTEGRATION I our lectures o itegral solutios to differetial equatios usig Laplace kerels,we ecoutered itegrals of the type- y( x) f ( t)exp( xt) t + where tγ+iτ ad is a closed cotour withi the complex plae. To evaluate this type of itegrals uder coditios where the curve partially lies at ifiity, oe makes use of complex variable methods ad i particular the auchy Itegral Theorem. We preset here this evaluatio method usig several specific examples. Our a startig poit is the auchy Itegral Theorem- F ( z) 0 whe F( z) is aalytic everywhere withi ad auchy s Itegral Formula- d G( z0)! i G( z) + ( z z0),where G(z) is aalytic everywhere i but the itegrad has a (+) order pole at zz 0 provided is a positive iteger. osider first the itegral- H ( a, b, c) ( az + bz + c) a[ z z][ z z] which has simple poles at- b b z + i ac b ad z i ac b a a where we are assumig a, b, ad c are real ad that ac>b. We cosider itegratio aroud the closed lie itegral defied as the ifiite radius semicircle ad the lie zx alog the real z axis from mius to plus ifiity as show-

2 Here oly the first order pole at zz lies withi the closed cotour, so that the auchy Itegral Formula reads- (/( z a z z ) / a i z a( z z) ac b From this last result ad the fact that the itegrad /(az +bz+c)vaishes alog the semicircle, leads to the result- H ( a, b, c) + ax dx + bx + c ( / a) ac b The simplest special case occurs for ac ad b0. There H(,0,). Look ext at the itegral- K( a, ) ( z + a )

3 about the same semicircle cotour show above. Here we have th order poles at zia ad z-ia provided agai that is a positive iteger. Agai applyig the auchy Itegral Formula, oe fids that- dx i d (/( z + ia) K a x x a (, ) + 0 ( ) ( )! z ia Thus for a ad we have 4 x 0 ( z dx + x + ) i! d ( ) z + i z i 4 Aother defiite itegral which ca be solved by complex variable methods is- K( a) exp( iz) ( z + a ) [cos( z) + isi( z)] ( z ia)( z + ia) where is agai the same semicircle cotour show above. Here we have simple poles at ±ia ad eed oly cosider the oe lyig alog the positive Im(z) axis. Oe fids that- so that- i exp( iz) K( a) exp( a) 0! ( z ia) + z ia a + cos( x) dx exp( a) x + a a ad + si( x) dx 0 x + a The vaishig of the secod itegral ca also be see directly by otig the odd symmetry of si(x). Next cosider a itegral ivolvig a multi-valued fuctio requirig the use of a brach cut i its itegratio. Specifically we cosider-

4 α z L where 0 < α < z + This time z α- is multivalued ad so requires a brach cut to avoid o-uiqess. The cotour chose looks as follows- Now accordig to the auchy theorem the value of the closed cotour is just i(-) α-. Furthermore if we let R become ifiite ad ε go to zero, the oly ozero cotributios remai the cotributios alog the two straight lies zx exp(i0) ad zx exp(i). Oe has- or- 0 α x α α [ xexp(i)] i( ) dx + dx x + x x 0 + x α α x i ( ) dx x 0 + x exp[( α )i ] si( α ) You will otice that this result is also equal to the product Γ(α) Γ(-α) of the Gamma fuctio so that oe has-

5 dx du Γ( / ) where x sih( u) x 0 ( + x) x x 0 cosh( u) Aother example of cotour itegratio is that used to prove the idetity- si( ) x P dx x x Here we look at the related cotour itegral- exp( iz) z where is the cotour show- cos( z) + isi( z) z Sice there are o poles(or brach poits)withi the cotour show, oe has, accordig to auchy s Theorem

6 Rexp( ix) / 0 dx + i ε x θ i θ / exp[ iε exp( iθ )] exp[ ir exp( iθ )] ε exp( iθ ) dθ R exp( iθ ) dθ + ε exp( y) i dy y R y If oe ow lets R become ifiite ad ε approach zero, the theta itegrals vaish ad we are left with- x 0 [cos( x) + isi( x)] dx x i y 0 exp( y) dy y i Γ(/ ) ( + i) from which we have our result. Aother example which ca be icely treated by cotour itegral methods is the itegral- M + x dx cosh( x) This time oe otices that the complex fuctio /cosh(z) has simple poles alog the imagiary axis at zi(+)/ ad that the fuctio vaishes at z±. This suggest the use of a rectagular cotour with sides alog the zx axis, alog the lie zx+i, plus the two remaiig sides at z± +iy. This cotour ecloses the simple pole at zi/ ad by the auchy Itegral Formula leads to- dx x cosh( x) + idy y 0 cosh( + iy) dx + x cosh( x + i ) + 0 idy i cosh( ) i y + iy sih( ) Now. otig that the two itegrals ivolvig y vaish ad that the third itegral just equals the first, we fid our desired result that M. Fially, it should be poited out that oe ca also use cotour itegratios to obtai iverse Laplace trasforms via the operatio- F( t) L [ f ( s)] γ + i f ( s)exp( st) ds i γ i

7 where the closed cotour curve closes to the left of the lie sγ+iτ. To demostrate we cosider f(s)/s so that oe has a sigle th order pole at s0. It the follows from the auchy Itegral that- F( t) i d exp( st) i ( )! ds s 0 t ( )! A secod example of ivertig a Laplace trasform we take- L f ( s) γ + i γ i i exp( k s s) exp( st) This trasform will be recogized as that obtaied whe dealig with timedepedet coductio ito a semi-ifiite bar whose iitial temperature is zero ad which has its ed at x0 maitaied at costat temperature. Here oe is dealig with a brach poit at s0 so oe cosiders a cotour 3 cosistig of the lie sγ+i τ coected to a circle at abs(s), two lies o the top ad bottom of the egative Re(s) axis which serves as a brach cut, plus a small circle of radius ε about the origi. The cotour is show here- Sice there are o poles withi the cotour 3, the value of the itegral about the etire closed cotour must be zero. Furthermore the cotributio aroud the outer circle vaishes as R goes to ifiity ad that about the ier circle is just i as ε is allowed to go to zero. Oe is left with the remaiig two cotributios alog the

8 lies sx exp(i) ad sx exp(-i). After some maipulatio (as show i class) this results i- k k F ( t) erf ( ) erfc( ) t t Typically oe does ot use the cotour itegratio approach for simpler fuctios f(s) sice these will be foud i tables of Laplace trasforms. However there are may more complex forms for which cotour itegratio remais the oly aveue of approach. For istace, I remember about te years ago oe of my studets ( Whitey Joes, Ijectio oolig withi a Micro-Heat Exchager, MS thesis, Uiversity of Florida, 999 )was strugglig with the iversio of Laplace trasforms such as- a f ( s) + s b s cosh( d cosh( c s) s s) + K sih( d s)coth( ce ) for which o iverses ca be foud i eve the most extesive trasform tables. September 008

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