lim when the limit on the right exists, the improper integral is said to converge to that limit.

Transcription

1 hapter 7 Applications of esidues - evaluation of definite and improper integrals occurring in real analysis and applied math - finding inverse Laplace transform by the methods of summing residues. 6. Evaluation of Improper Integrals In calculus lim f( x) dx f( x) dx when the limit on the right exis, the improper integral is said to converge to that limit. If f(x) is continuous for all x, its improper integral over the < x < is defined by lim lim f ( x ) dx f ( x ) dx + f ( x ) dx () When both of the limits here exi, integral () converges to their sum. There is another value that is assigned to integral (). i.e., The auchy principal value (P.V.) of integral (). lim - PV.. f( x) dx f( x) dx (3) provided this single limit exis

2 If integral () converges its auchy principal value (3) exis. If is not, however, always true that integral () converges when its auchy P.V. exis. Example. (ex8, sec. 6) f( x) x x x xdx lim xdx + lim xdx lim + lim lim( ) + ( ) lim neither of these limits exi On the other hand, P.V. xdx x xdx ( ) lim lim lim lim 3 But suppose that f( x) (- < x< ) is an even function, f( x) f( x) for each x f( x) dx ( ) f x dx f ( x) dx ( ) ( ) lim f x dx lim f x dx () (3) converge if exi to P.V. Moreover, f( x) dx ( ) f x dx f( x) dx lim + lim lim ( ) f x dx 4

3 when f( x) is even f( x) dx f( x) dx PV.. f( x) dx To evaluate improper integral of px ( ) f( x) qx ( ) p, q are polynomials with no factors in common. and q(x) has no real eros. See example 5 Example x 6 x + dx f( ) 6 has isolated singularities at 6th roots of. + and is analytic everywhere else. Those roots are k k exp i ( + ) k,,,3, 4,5 6 6 i i5 6 6 e i, e lie in the upper half plane when > - f( x) dx+ f( ) d i( B + B + B ) esidues of f( ) at ( k,,) k - 6

4 k are simple poles, k k e s Z K k k B ( k,,) B 6i B i ( B + B+ B) 6i 3 B 6i f( x) dx f( ) d f( x) dx f( ) d - 3 when for on f( ) f( ) d as 6 x x x lim dx or P.V x + 3 dx dx - x + 3 x + 6 8

5 6. Improper Integrals Involving sines and cosines To f ( x) sin ax dx or f ( x) cos ax dx ( a > ) evaluate Previous method does not apply since sin a, cos a as y sinhay (See p.7) However, we note that iax f( x) cos ax dx+ i f( x) sin ax dx f( x) e dx ia iax ay e e e is bounded in y - 9 Ex. cos3x? ( ) dx 3 ( x + ) e An even function Let f( ) ( + ) i3 And note that f( ) e above the real axis except at is analytic everywhere on and i. i3x e ( ) ( x + ) i3 dx ib f e d, ( > )

6 Take real part B e s f( ) e i3 i i3 e φ( ) ( + i) f( ) pole of order ( i) ( i) e cos3 ( x + ) e i3 + 3 i ( + i) ( i) + ie x 3 e ( ) dx i 3 f e d f e e ( ) i3 3y but ( ), i3 i3 3 ie B φ '( ) i e + ( ) 3 ( ) + i + i 3 3 i3 i3 e f( ) e d ( ) as f e d ( ) i i It is sometimes necessary to use a result based on Jordan s inequality to evaluate f( x)cos ax dx,... sinθ e dθ < (>) θ y pf: sin θ when θ y sinθ for > θ y θ sinθ e e, θ θ sinθ So e dθ e dθ e < since y sin θ is symmetric w. r.t. θ sinθ e dθ < ( > ) θ

7 Suppose f is analytic at all points i y and above e θ Let ( θ ) denote iθ e ( θ ), > ia iθ iθ iθ f( ) e d f( e ) exp ( iae ) ie dθ iθ If f ( e ) M, and M as Since we also have exp( iae ) e e iθ asinθ iacosθ ia asinθ M f( ) e d M e d as θ < a ia lim f( ) e d (a>) Jordan's Lemma 3 Example. Sol: x sin xdx x + x + f( ) + + ( )( ) Z + i i is a simple pole of f( ) e with residue i i( + i) e ( + i) e ( + i) e (cos isin) B ( i) ( i) i + ( i+ )(cos isin) e [ cos+ sin + i(cos sin) ] e Z 4

8 when > ix xe dx i ib f( e ) d x + x+ x sin x dx i Im( i B ) Im f( ) e d x + x + i i But Im f( ) e d f( ) e d ( ) f( ) ( ) i y and e e 5 i f( ) e d does not tend to ero as ( ) But from Jordan s Lemma f( ) as i x sin x dx lim f( ) e d P.V. Im( i B ) - x + x+ (sin+ cos) e 6

9 6. Definite Integrals Involving Sines and osines To evaluate F(sin θ,cos θ) dθ iθ θ : consider an unit circle : e, θ, d + sin θ cos θ dθ i i + d F(, ) parametric form i i dθ Example : (-< a<) + asinθ a a, trivial suppose a 7 a ( ) + asinθ + i i dθ / a d + asinθ i + a poles, i + a a i 4 ± + 4 a a i± a a > + a > a ± i i a ± a i ( ) a a < Z 8

10 but, < φ( ) / a f( ), where φ( ) - / a B φ( ) i a / a d i B + ( i ) - a a Indented Paths If a function f( ) has a simple pole at a point x on the real axis, with a Laureut series representation in a punctured disk < - x < and with residue B. If ρ denotes the upper half of a circle where ρ<, and negative oriented, then lim f( ) d B i ρ ρ lim f( ) d i B ρ - x ρ, y ρ ρ X x

11 pf: f( ) can be written as B f g B ( ) ( ) +, x ( < - x < ) continuous in - x <, not sure if it is analytic. d f ( ) d g( ) d B + ρ ρ ρ x since g( ) is continuous, there is a positive M such that g ( ) Mfor all - x ρ gd ( ) Mρ as ρ iθ ρ x ρ x ρe ( ρ) iθ - ρ : x + ρe ( θ ) iθ d d ρie dθ i lim f( ) d B i ρ ρ Ex. sin x To show dx x onsider a simple closed contour ρ <, L L : ρ x L : x ρ From auchy-goursat theorem i i i i e e e e d + d d d L + + L ρ i i i i e e e e or d + d d d Since : ( r ) L L ρ i L re r ρ i L : re r ( ρ r ) ρ L

12 i i ir ir e e e e d d dr ( dr) ρ r ρ ( r) L L ir ir e e sin r dr i dr ρ r ρ r i i sin r e e i dr d d ρ r ρ i e i ( i) Now !! 3 i i i < <!! 3! has a simple pole at, with residue i e lim d i ρ ρ also M as i sinr From (9) in sec. 6, lim ed dr Jordan s Lemma r Integrating Along a Branch ut (P.8, complex exponent) a onsider x, where x>, and < a<. a Let it denote the principal vabue of x a aln x ln Log i.e., x e alog x a x e is multivalued shall evaluate dx ( < a < ) x + important in gamma function udy -a x ( has an infinite discontinuity at x. x + The integral converges when <a< since the x x x as x tends to infinity) integrand behaves like - a near, and like -a- 4

13 Begin by ρ : ρ ρ < < : consider the branch a f( ) ( >, < arg < ) + of the multiple-valued function a, with branch cut arg + Since it is piecewise continuous on ρ and, then integrals f( ) d and f( ) d exi ρ [ a r+ iθ ] exp( alog ) exp (ln ) iθ Write f( ), ( re ) iθ + re + and use θ and θ along the upper and lower "edges", respectively. ρ 5 Then [ + ] a i exp a(ln r i r for re, f ( ) r+ r+ a i exp [ a(ln r+ i ] r e for re, f( ) r+ r+ from residue thm, ia a a ia r r e dr + f ( ) d dr + f ( ) d ρ r+ ρ r+ ρ i e s f( ) [ f( ) is not defined on the branch cut involved, but...see Ex.9] a Note that φ( ) exp( alog ) ( >, < arg < ) is analytic at - and [ ] ia φ( ) exp a(ln + i ) e 6

14 This shows that - is a simple pole of the function f( ) and e s f( ) e ia a ia ia r f( ) d+ ( ) ( ) f d ie + e dr ρ ρ r + a ρ a but f( ) d ρ ρ as ρ ρ ρ ρ a and f( ) d as a -a ia r ie i also lim dr ρ ρ ia ia ia r+ e e e -a x dx + sina Argument Principle and ouche s Theorem A function f is said to be meromorphic in a domain D if it is analytic throughout D - except possibly for poles. Suppose f is meromorphic inside a positively oriented simple close contour, and analytic and nonero on. The image Γ of under the transformation w f(), is a closed contour, not necessarily simple, in the w plane. y v w w x u Γ 8

15 As a point traverses in the positive direction, its image w traverses Γ in a particular direction that determines the orientation of Γ. Note, since f has no eros on, the contour Γ does not pass through the origin in the w plane. Let w and w be points on Γ, where w is fixed and φ arg w. Let arg w vary continuously, arting with w, as w begins at w and traverses Γ once. When w returns to the arting point w, arg w assumes a particular value of arg w, φ. Thus the change in arg w as w describes Γ once is φ - φ. 9 This change is independent of ω. φ φ is in fact the change in arg f( ) as describes once, arting at. arg f( ) φ φ. a multiple of The integer arg f ( ) represents the number of times the point ω winds around the origin in the ω plane. called winding number of Γ w.r.t. the origin ω. Positive: Negative: winding number when Γ does not enclose ω 3

16 The winding number can be determined from the number of eros and poles of f interior to. Number of poles eros Argument principle are finite (Ex 5, sec. 57) (Ex 4) Thm. f is meromorphic inside a simple closed, positively oriented ontour, and is analytic and nonero on. If, counting multiplicities, Z is the number of eros and P is the number of poles inside, then arg f( ) Z P. 3 Pf. Let be ( t) (a t b) [ ] [ ] f [ t ] f '( ) b f ' ( t) '( t) d dt f( ) a ( ) The image of t ( ) can be expressed as iφ () t (t) e ( f( ) ) ω ρ ρ iφ () t Thus f ( t) ( t) e ( t b) and d d iφ () t f '[ ( t) ] '() t f [ () t ] () t e dt dt ρ iφ() t iφ() t ρ'( te ) + iρ( te ) φ'( t) f '( ) bρ '( t) b d dt + i φ '( t) dt f( ) ρ( t) a a b b ln ρ( t) + iφ( t) a a 3

17 But ρ( b) ρ( a) and φ( b) φ( a) arg f( ) f '( ) d i arg f ( ) f( ) Another way to evaluate f '( ) d f( ) f '( ) Since is analytic inside and on except at f( ) points inside at which eros and poles of f occur. If f( ) has a ero of order m at, then m ( ) ( ) ( ) m m Hence f '( ) m ( ) g( ) + ( ) g'( ) or f g f '( ) m g'( ) + f( ) g( ) analytic, nonero at. 33 g '( ) Since is analytic at, it has Taylor series representation g ( ) about that point. f '( ) has a simple pole at, with residue m. f( ) On the other hand, if f has a pole of order mp at, φ( ) f( ) analytic and nonero at m. p ( ) mpφ( ) φ '( ) f '( ) + mp+ mp ( ) ( ) f '( ) mp φ '( ) + f( ) ( ) φ( ) has a simple pole at, with residue - m p. 34

18 Applying the residue theorem, then f '( ) f '( ) d i es i ( Z P) f( ) f( ) poles, eros arg f( ) Z P Example. f( ), pole of order. i Let e θ,( θ ) arg( ) by theorem. ω e θ iθ or ( ) 35 ouche s theorem Thm. Let two functions f( ) and g( ) be analytic inside and on a simple closed contour, and suppose f( ) > g( ) at each point on. Then f( ) and f( ) + g( ) have the same number of eros, counting multiplicities, inside. Pf. Since f( ) > g( ) on f( ) + g( ) f( ) g( ) > on. g ( ) arg [ f( ) + g( ) ] arg f( ) + f( ) g( ) arg f( ) + arg + f( ) 36

19 g( ) Let F( ) + f( ) g ( ) then F( ) <, f( ) Therefore under the transformation w F( ), the image of lies in w <, which does not enclose w. g ( ) arg + f( ) arg f( ) + g( ) arg f( ) [ ] They have the same number of eros. (No poles) 37 Ex. To determine the number of roots inside, f g write ( ) 4 and ( ) 3 7 ( ) 4 4, and ( ) when. f g since f ( ) has three eros, inside, has three roots inside. 38

20 66. Inverse Laplace Transforms Suppose that a function F of complex variable s is analytic throughout the finite s plane except for a finite number of isolated singularities. y Let L denote a vertical line segment from s γ i to s γ + i, SN where γ is positive and large enough that singularities of F all lie to the S S left of the segment. define f( t) lim e F( s) ds ( t ) i > L provided the limit exis γ + i or f( t) P.V. e F( s) ds ( t > ) i γ i Bromwich integral r r i L 39 It can be shown that f( t) is the inverse Laplace transform of Fs ( ). That is, if Fs ( ) is [ ] Fs ( ) e ftdt ( ) then f( t) L F( s) Let s (n,,...n) denote the singularities of F( s). Let n denote the large of their moduli and consider a semicircle iθ 3 s γ + e ( θ ) where > + γ 4

21 Note that for each s, s γ s + γ + γ < n Hence all singularities are inside the semicircle and e F() s ds i e s e F() s L n n n s sn - e F( s) ds N Suppose that Fs ( ) M, where M as, 3 () iθ iθ iθ exp( + ) ( + e ) 3 γt () t cos θ e F s ds γ t te F γ ie dθ e F s ds e M e dθ < e t sinφ γ t e M t r + i dφ < ( φ θ ) t Jordan s inequality 4 lim e F( s) ds N f() t e s e F( s) n can be exteded to the case with infinite singular points f( t) e s e F( s) ( t > ) n s sn s sn 4

22 67. Example Exercise Suppose that F( s) has a pole of order m at s, b b b Fs ( ) a( s s) ( bm ) m ( ) ( ) n m n n s s s s s s Then e ( )... s s b bm s e F s e b + t+ + t! ( m )! when s α + iβ ( β ) and Fs ( ) Fs ( ), s α iβ is also a pole of order m. Moreover, e s e F( s) + e s e F( s) s s s s b b e e b + t t! ( m )! When t is real αt iβt m m e m 43 If s is a simple pole sefs e sfs e ( ) e ( ) s s s s αt iβt and e s e F( s) e s e F( s) e e e e sf( s) s s + s s s s Ex. s Fs () ( s + a ) singularities s ia and s ia φ() s s Fs ( ), φ( s) ( s ia ) ( s+ ia) s s s is a pole of order. F( s) ( s + a ) ( x y + a ixy) s s Furthermore Fs ( ) Fs ( ). Fs ( ) ( s + a ) ( x y + a + ixy) 44

23 e s e F( s) + e s e F( s) e e ( b + b t) iat s s s s φ '( ai) But Fs () φ() s φ( ai) + ( s ai) +... ( s ia) ( s ia)! φ( ai) φ'( ia) ( < s ia < a) ( s ia) s ia i φ( ia), φ'( ia) 4a ( s + ia) s ( s + ia) φ '( s) 4 i ( s+ ia) b, b 4a iat i e s e F( s) e e ( t) tsin at s s n 4a a f( t) tsin at ( t > ) a provided Fs ( ) satisfies the boundedness condition (in sec. 66) p iθ 3 Let s γ + e ( θ ) where γ >, > a+ γ iθ s γ + e γ + iθ s γ + e γ γ > a Since s + a s a ( γ ) a > s γ + Fs () as. s + a ( γ ) a 46