Man will occasionally stumble over the truth, but most of the time he will pick himself up and continue on.

Transcription

1 hapter 3 The Residue Theorem Man will occasionally stumble over the truth, but most of the time he will pick himself up and continue on. - Winston hurchill 3. The Residue Theorem We will find that many integrals on closed contours may be evaluated in terms of the residues of a function. We first define residues and then prove the Residue Theorem. 626

2 Result 3.. Residues. Let f(z) be single-valued an analytic in a deleted neighborhood of z. Then f(z) has the Laurent series expansion f(z) = n= a n (z z ) n, The residue of f(z) at z = z is the coefficient of the z z term: Res(f(z), z ) = a. The residue at a branch point or non-isolated singularity is undefined as the Laurent series does not exist. If f(z) has a pole of order n at z = z then we can use the Residue Formula: ( d n [ Res(f(z), z ) = lim (z z ) n f(z) ] ). z z (n )! dz n See Exercise 3.4 for a proof of the Residue Formula. Example 3.. In Example we showed that f(z) = z/ sin z has first order poles at z = nπ, n Z \ {}. Now 627

3 B Figure 3.: Deform the contour to lie in the deleted disk. we find the residues at these isolated singularities. ( z ) ( Res sin z, z = nπ z ) = lim (z nπ) z nπ sin z z nπ = nπ lim z nπ = nπ lim z nπ = nπ ( ) n = ( ) n nπ sin z cosz Residue Theorem. We can evaluate many integrals in terms of the residues of a function. Suppose f(z) has only one singularity, (at z = z ), inside the simple, closed, positively oriented contour. f(z) has a convergent Laurent series in some deleted disk about z. We deform to lie in the disk. See Figure 3.. We now evaluate f(z) dz by deforming the contour and using the Laurent series expansion of the function. 628

4 f(z) dz = = = B f(z) dz B n= n= n = a ı2π a n (z z ) n dz a n [ (z z ) n+ n + ] r e ı(θ+2π) r e ıθ + a [log(z z )] r eı(θ+2π) r e ıθ f(z) dz = ı2π Res(f(z), z ) Now assume that f(z) has n singularities at {z,...,z n }. We deform to n contours,..., n which enclose the singularities and lie in deleted disks about the singularities in which f(z) has convergent Laurent series. See Figure 3.2. We evaluate f(z) dz by deforming the contour. f(z) dz = n k= k f(z) dz = ı2π n Res(f(z), z k ) k= Now instead let f(z) be analytic outside and on except for isolated singularities at {ζ n } in the domain outside and perhaps an isolated singularity at infinity. Let a be any point in the interior of. To evaluate f(z) dz we make the change of variables ζ = /(z a). This maps the contour to. (Note that is negatively oriented.) All the points outside are mapped to points inside and vice versa. We can then evaluate the integral in terms of the singularities inside. 629

5 2 3 Figure 3.2: Deform the contour n contours which enclose the n singularities. ( ) f(z) dz = f ζ + a ζ dζ ( ) 2 = z 2f z + a dz = ı2π ( ( ) Res z 2f z + a n, ) ( ( ) ) + ı2π Res ζ n a z 2f z + a,. 63

6 a Figure 3.3: The change of variables ζ = /(z a). Result 3..2 Residue Theorem. If f(z) is analytic in a compact, closed, connected domain D except for isolated singularities at {z n } in the interior of D then f(z) dz = f(z) dz = ı2π Res(f(z), z n ). D k k n Here the set of contours { k } make up the positively oriented boundary D of the domain D. If the boundary of the domain is a single contour then the formula simplifies. f(z) dz = ı2π Res(f(z), z n ) n If instead f(z) is analytic outside and on except for isolated singularities at {ζ n } in the domain outside and perhaps an isolated singularity at infinity then f(z) dz = ı2π ( ( ) ) ( ( ) ) Res z 2f z + a, + ı2π Res ζ n n a z 2f z + a,. Here a is a any point in the interior of. 63

7 Example 3..2 onsider ı2π sin z z(z ) dz where is the positively oriented circle of radius 2 centered at the origin. Since the integrand is single-valued with only isolated singularities, the Residue Theorem applies. The value of the integral is the sum of the residues from singularities inside the contour. The only places that the integrand could have singularities are z = and z =. Since sin z lim z z = lim z cosz =, there is a removable singularity at the point z =. There is no residue at this point. Now we consider the point z =. Since sin(z)/z is analytic and nonzero at z =, that point is a first order pole of the integrand. The residue there is ( ) sin z Res z(z ), z = sin z = lim(z ) z z(z ) = sin(). There is only one singular point with a residue inside the path of integration. The residue at this point is sin(). Thus the value of the integral is sin z dz = sin() ı2π z(z ) Example 3..3 Evaluate the integral cotz cothz z 3 where is the unit circle about the origin in the positive direction. The integrand is cotz cothz cosz cosh z = z 3 z 3 sin z sinh z 632 dz

8 sin z has zeros at nπ. sinh z has zeros at ınπ. Thus the only pole inside the contour of integration is at z =. Since sin z and sinh z both have simple zeros at z =, sin z = z + O(z 3 ), sinh z = z + O(z 3 ) the integrand has a pole of order 5 at the origin. The residue at z = is lim z 4! ( d 4 z dz 4 ) 5cotz cothz = lim z 3 z 4! = 4! lim z d 4 ( z 2 cotz cothz ) dz 4 ( 24 cot(z) coth(z)csc(z) 2 32z coth(z)csc(z) 4 6z cos(2z) coth(z)csc(z) z 2 cot(z) coth(z)csc(z) 4 + 2z 2 cos(3z) coth(z)csc(z) cot(z) coth(z)csch(z) csc(z) 2 csch(z) 2 48z cot(z)csc(z) 2 csch(z) 2 48z coth(z)csc(z) 2 csch(z) z 2 cot(z) coth(z)csc(z) 2 csch(z) 2 + 6z 2 csc(z) 4 csch(z) 2 + 8z 2 cos(2z)csc(z) 4 csch(z) 2 32z cot(z)csch(z) 4 6z cosh(2z) cot(z)csch(z) z 2 cot(z) coth(z)csch(z) 4 + 6z 2 csc(z) 2 csch(z) 4 ) + 8z 2 cosh(2z)csc(z) 2 csch(z) 4 + 2z 2 cosh(3z) cot(z)csch(z) 5 = ( 56 ) 4! 5 =

9 Since taking the fourth derivative of z 2 cotz cothz really sucks, we would like a more elegant way of finding the residue. We expand the functions in the integrand in Taylor series about the origin. ( )( ) z2 + z4 + z2 + z cosz cosh z z 3 sin z sinh z = z ( 3 z z3 + z5 )( z + z3 + z5 + ) z4 6 = + z ( 3 z 2 + z ( 6 + ) ) = z4 + 6 z 5 z4 + 9 = ( ) ) ( z4 z z4 9 + ) = z 5 ( 7 45 z4 + = z z + Thus we see that the residue is 7. Now we can evaluate the integral. 45 cotz cothz dz = ı 4 z 3 45 π 3.2 auchy Principal Value for Real Integrals 3.2. The auchy Principal Value First we recap improper integrals. If f(x) has a singularity at x (a...b) then b a f(x) dx lim ǫ + x ǫ a f(x) dx + lim δ + b x +δ f(x) dx. 634

10 For integrals on (... ), f(x) dx b lim f(x) dx. a, b a Example 3.2. dx is divergent. We show this with the definition of improper integrals. x ǫ dx = lim x ǫ + dx + lim x δ + δ x dx = lim [ln x ] ǫ ǫ + + lim [ln x ] δ + δ = lim ln ǫ lim ln δ ǫ + δ + The integral diverges because ǫ and δ approach zero independently. Since /x is an odd function, it appears that the area under the curve is zero. onsider what would happen if ǫ and δ were not independent. If they approached zero symmetrically, δ = ǫ, then the value of the integral would be zero. lim ǫ + ( ǫ + ǫ ) x dx = lim ǫ +(lnǫ ln ǫ) = We could make the integral have any value we pleased by choosing δ = cǫ. ( ǫ lim + ǫ + cǫ ) x dx = lim ǫ +(lnǫ ln(cǫ)) = ln c We have seen it is reasonable that x dx has some meaning, and if we could evaluate the integral, the most reasonable value would be zero. The auchy principal value provides us with a way of evaluating such integrals. If f(x) is continuous on (a, b) except at the point x (a, b) This may remind you of conditionally convergent series. You can rearrange the terms to make the series sum to any number. 635

11 then the auchy principal value of the integral is defined b ( x ǫ b ) f(x) dx = lim f(x) dx + f(x) dx. a ǫ + a x +ǫ The auchy principal value is obtained by approaching the singularity symmetrically. The principal value of the integral may exist when the integral diverges. If the integral exists, it is equal to the principal value of the integral. The auchy principal value of dx is defined x ( ǫ dx lim x ǫ + x dx + ǫ = lim ǫ + ( [log x ] ǫ [log x ] ǫ = lim (log ǫ log ǫ ) ǫ + =. ) x dx ) (Another notation for the principal value of an integral is PV f(x) dx.) Since the limits of integration approach zero symmetrically, the two halves of the integral cancel. If the limits of integration approached zero independently, (the definition of the integral), then the two halves would both diverge. Example x dx is divergent. We show this with the definition of improper integrals. x 2 + x x 2 + dx = b lim a, b a [ = lim a, b = 2 lim ln a, b x x 2 + dx ] b 2 ln(x2 + ) ( ) b 2 + a 2 + a 636

12 The integral diverges because a and b approach infinity independently. Now consider what would happen if a and b were not independent. If they approached zero symmetrically, a = b, then the value of the integral would be zero. 2 lim b ln ( ) b 2 + = b 2 + We could make the integral have any value we pleased by choosing a = cb. We can assign a meaning to divergent integrals of the form auchy principal value of the integral is defined f(x) dx with the auchy principal value. The a f(x) dx = lim f(x) dx. a a The auchy principal value is obtained by approaching infinity symmetrically. The auchy principal value of x dx is defined x 2 + x dx = lim x 2 + a = lim a =. a x a x 2 + dx [ 2 ln ( x 2 + ) ] a a 637

13 Result 3.2. auchy Principal Value. If f(x) is continuous on (a, b) except at the point x (a, b) then the integral of f(x) is defined b a f(x) dx = lim ǫ + x ǫ a f(x) dx + lim δ + The auchy principal value of the integral is defined b a f(x) dx = lim ǫ + ( x ǫ a f(x) dx + b b x +ǫ x +δ f(x) dx. ) f(x) dx. If f(x) is continuous on (, ) then the integral of f(x) is defined f(x) dx = lim a, b The auchy principal value of the integral is defined b a f(x) dx. a f(x) dx = lim f(x) dx. a a The principal value of the integral may exist when the integral diverges. If the integral exists, it is equal to the principal value of the integral. Example learly x dx diverges, however the auchy principal value exists. x dx = lim a 638 [ ] x 2 2 a a =

14 In general, if f(x) is an odd function with no singularities on the finite real axis then f(x) dx =. 3.3 auchy Principal Value for ontour Integrals Example 3.3. onsider the integral r z dz, where r is the positively oriented circle of radius r and center at the origin. From the residue theorem, we know that the integral is { r z dz = for r <, ı2π for r >. When r =, the integral diverges, as there is a first order pole on the path of integration. However, the principal value of the integral exists. 2π ǫ dz = lim r z ǫ + ǫ e ıθ ıeıθ dθ [ = lim log(e ıθ ) ] 2π ǫ ǫ + ǫ 639

15 We choose the branch of the logarithm with a branch cut on the positive real axis and arg log z (, 2π). ( ( = lim log e ı(2π ǫ) ) log (e ıǫ ) ) ǫ + ( (( = lim log iǫ + O(ǫ 2 ) ) ) log (( + iǫ + O(ǫ 2 ) ) )) ǫ + ( ( = lim log iǫ + O(ǫ 2 ) ) log ( iǫ + O(ǫ 2 ) )) ǫ + ( ( = lim Log ǫ + O(ǫ 2 ) ) + ı arg ( ıǫ + O(ǫ 2 ) ) Log ( ǫ + O(ǫ 2 ) ) ı arg ( ıǫ + O(ǫ 2 ) )) ǫ + = ı 3π 2 ıπ 2 = ıπ Thus we obtain r z dz = for r <, ıπ for r =, ı2π for r >. In the above example we evaluated the contour integral by parameterizing the contour. This approach is only feasible when the integrand is simple. We would like to use the residue theorem to more easily evaluate the principal value of the integral. But before we do that, we will need a preliminary result. Result 3.3. Let f(z) have a first order pole at z = z and let (z z )f(z) be analytic in some neighborhood of z. Let the contour ǫ be a circular arc from z + ǫe ıα to z + ǫe ıβ. (We assume that β > α and β α < 2π.) lim ǫ + ǫ f(z) dz = ı(β α) Res(f(z), z ) The contour is shown in Figure 3.4. (See Exercise 3.9 for a proof of this result.) 64

16 ε β α z ε Figure 3.4: The ǫ ontour p ε Figure 3.5: The indented contour. Example onsider z dz where is the unit circle. Let p be the circular arc of radius that starts and ends a distance of ǫ from z =. Let ǫ be the positive, circular arc of radius ǫ with center at z = that joins the endpoints of p. Let i, be the union of p and ǫ. ( p stands for Principal value ontour; i stands for Indented ontour.) i is an indented contour that avoids the first order pole at z =. Figure 3.5 shows the three contours. 64

17 Note that the principal value of the integral is dz = lim z ǫ + p z dz. We can calculate the integral along i with the residue theorem. i z dz = ı2π We can calculate the integral along ǫ using Result Note that as ǫ +, the contour becomes a semi-circle, a circular arc of π radians. ( ) lim dz = ıπ Res ǫ + ǫ z z, = ıπ Now we can write the principal value of the integral along in terms of the two known integrals. z dz = i z dz ǫ z dz = ı2π ıπ = ıπ In the previous example, we formed an indented contour that included the first order pole. You can show that if we had indented the contour to exclude the pole, we would obtain the same result. (See Exercise 3..) We can extend the residue theorem to principal values of integrals. (See Exercise 3..) 642

18 Result Residue Theorem for Principal Values. Let f(z) be analytic inside and on a simple, closed, positive contour, except for isolated singularities at z,...,z m inside the contour and first order poles at ζ,...,ζ n on the contour. Further, let the contour be at the locations of these first order poles. (i.e., the contour does not have a corner at any of the first order poles.) Then the principal value of the integral of f(z) along is f(z) dz = ı2π m Res(f(z), z j ) + ıπ j= n Res(f(z), ζ j ). j= 3.4 Integrals on the Real Axis Example 3.4. We wish to evaluate the integral x 2 + dx. We can evaluate this integral directly using calculus. x 2 + dx = [arctanx] = π Now we will evaluate the integral using contour integration. Let R be the semicircular arc from R to R in the upper half plane. Let be the union of R and the interval [ R, R]. We can evaluate the integral along with the residue theorem. The integrand has first order poles at z = ±ı. For 643

19 R >, we have ( ) dz = ı2π Res z 2 + z 2 +, ı = ı2π ı2 = π. Now we examine the integral along R. We use the maximum modulus integral bound to show that the value of the integral vanishes as R. R z 2 + dz πr max z R z 2 + = πr R 2 as R. Now we are prepared to evaluate the original real integral. z 2 + dz = π R R x 2 + dx + R z 2 + dz = π We take the limit as R. x 2 + dx = π We would get the same result by closing the path of integration in the lower half plane. Note that in this case the closed contour would be in the negative direction. 644

20 If you are really observant, you may have noticed that we did something a little funny in evaluating The definition of this improper integral is x 2 + dx. dx = lim x 2 + a + a b dx+ = lim x 2 + b + x 2 + dx. In the above example we instead computed R lim R + R x 2 + dx. Note that for some integrands, the former and latter are not the same. onsider the integral of x. x 2 + x dx = lim x 2 + R a + = lim a + a x x 2 + ( 2 log a2 + dx + lim ) b + b + lim b + x x 2 + dx ( 2 log b2 + Note that the limits do not exist and hence the integral diverges. We get a different result if the limits of integration approach infinity symmetrically. R ( ) x lim dx = lim R + x 2 + R + 2 (log R2 + log R 2 + ) = (Note that the integrand is an odd function, so the integral from R to R is zero.) We call this the principal value of the integral and denote it by writing PV in front of the integral sign or putting a dash through the integral. PV f(x) dx f(x) dx 645 lim R + R R f(x) dx )

21 The principal value of an integral may exist when the integral diverges. If the integral does converge, then it is equal to its principal value. We can use the method of Example 3.4. to evaluate the principal value of integrals of functions that vanish fast enough at infinity. Result 3.4. Let f(z) be analytic except for isolated singularities, with only first order poles on the real axis. Let R be the semi-circle from R to R in the upper half plane. If ( ) lim R max f(z) = R z R then f(x) dx = ı2π m Res (f(z), z k ) + ıπ k= n Res(f(z), x k ) where z,...z m are the singularities of f(z) in the upper half plane and x,...,x n are the first order poles on the real axis. Now let R be the semi-circle from R to R in the lower half plane. If then lim R f(x) dx = ı2π ( R max z R f(z) ) = m Res (f(z), z k ) ıπ k= k= n Res(f(z), x k ) where z,...z m are the singularities of f(z) in the lower half plane and x,...,x n are the first order poles on the real axis. k= 646

22 This result is proved in Exercise 3.3. Of course we can use this result to evaluate the integrals of the form where f(x) is an even function. 3.5 Fourier Integrals f(z) dz, In order to do Fourier transforms, which are useful in solving differential equations, it is necessary to be able to calculate Fourier integrals. Fourier integrals have the form e ıωx f(x) dx. We evaluate these integrals by closing the path of integration in the lower or upper half plane and using techniques of contour integration. onsider the integral Since 2θ/π sin θ for θ π/2, π/2 e R sin θ dθ. e R sin θ e R2θ/π for θ π/2 π/2 e R sin θ dθ = π/2 e R2θ/π dθ [ π ] π/2 2R e R2θ/π = π 2R (e R ) π 2R as R 647

23 We can use this to prove the following Result (See Exercise 3.7.) Result 3.5. Jordan s Lemma. π e R sin θ dθ < π R. Suppose that f(z) vanishes as z. If ω is a (positive/negative) real number and R is a semi-circle of radius R in the (upper/lower) half plane then the integral R f(z) e ıωz dz vanishes as R. We can use Jordan s Lemma and the Residue Theorem to evaluate many Fourier integrals. onsider f(x) eıωx dx, where ω is a positive real number. Let f(z) be analytic except for isolated singularities, with only first order poles on the real axis. Let be the contour from R to R on the real axis and then back to R along a semi-circle in the upper half plane. If R is large enough so that encloses all the singularities of f(z) in the upper half plane then m n f(z) e ıωz dz = ı2π Res(f(z) e ıωz, z k ) + ıπ Res(f(z) e ıωz, x k ) k= where z,...z m are the singularities of f(z) in the upper half plane and x,...,x n are the first order poles on the real axis. If f(z) vanishes as z then the integral on R vanishes as R by Jordan s Lemma. m n f(x) e ıωx dx = ı2π Res(f(z) e ıωz, z k ) + ıπ Res(f(z) e ıωz, x k ) k= For negative ω we close the path of integration in the lower half plane. Note that the contour is then in the negative direction. k= k= 648

24 Result Fourier Integrals. Let f(z) be analytic except for isolated singularities, with only first order poles on the real axis. Suppose that f(z) vanishes as z. If ω is a positive real number then f(x) e ıωx dx = ı2π m Res(f(z) e ıωz, z k ) + ıπ k= n Res(f(z) e ıωz, x k ) k= where z,...z m are the singularities of f(z) in the upper half plane and x,...,x n are the first order poles on the real axis. If ω is a negative real number then f(x) e ıωx dx = ı2π m Res(f(z) e ıωz, z k ) ıπ k= n Res(f(z) e ıωz, x k ) k= where z,...z m are the singularities of f(z) in the lower half plane and x,...,x n are the first order poles on the real axis. 3.6 Fourier osine and Sine Integrals Fourier cosine and sine integrals have the form, f(x) cos(ωx) dx and f(x) sin(ωx) dx. If f(x) is even/odd then we can evaluate the cosine/sine integral with the method we developed for Fourier integrals. 649

25 Let f(z) be analytic except for isolated singularities, with only first order poles on the real axis. Suppose that f(x) is an even function and that f(z) vanishes as z. We consider real ω >. f(x) cos(ωx) dx = 2 f(x) cos(ωx) dx Since f(x) sin(ωx) is an odd function, 2 f(x) sin(ωx) dx =. Thus f(x) cos(ωx) dx = 2 f(x) e ıωx dx Now we apply Result f(x) cos(ωx) dx = ıπ m Res(f(z) e ıωz, z k ) + ıπ 2 k= n Res(f(z) e ıωz, x k ) k= where z,...z m are the singularities of f(z) in the upper half plane and x,...,x n are the first order poles on the real axis. If f(x) is an odd function, we note that f(x) cos(ωx) is an odd function to obtain the analogous result for Fourier sine integrals. 65

26 Result 3.6. Fourier osine and Sine Integrals. Let f(z) be analytic except for isolated singularities, with only first order poles on the real axis. Suppose that f(x) is an even function and that f(z) vanishes as z. We consider real ω >. f(x) cos(ωx) dx = ıπ m Res(f(z) e ıωz, z k ) + ıπ 2 k= n Res(f(z) e ıωz, x k ) k= where z,...z m are the singularities of f(z) in the upper half plane and x,...,x n are the first order poles on the real axis. If f(x) is an odd function then, f(x) sin(ωx) dx = π µ Res(f(z) e ıωz, ζ k ) + π 2 k= n Res(f(z) e ıωz, x k ) k= where ζ,...ζ µ are the singularities of f(z) in the lower half plane and x,...,x n are the first order poles on the real axis. Now suppose that f(x) is neither even nor odd. We can evaluate integrals of the form: f(x) cos(ωx) dx and f(x) sin(ωx) dx by writing them in terms of Fourier integrals f(x) cos(ωx) dx = 2 f(x) sin(ωx) dx = ı 2 f(x) e ıωx dx + 2 f(x) e ıωx dx + ı 2 f(x) e ıωx dx f(x) e ıωx dx 65

27 R ε Figure 3.6: 3.7 ontour Integration and Branch uts Example 3.7. onsider x a dx, < a <, x + where x a denotes exp( a ln(x)). We choose the branch of the function f(z) = z a z + z >, < arg z < 2π with a branch cut on the positive real axis. Let ǫ and R denote the circular arcs of radius ǫ and R where ǫ < < R. ǫ is negatively oriented; R is positively oriented. onsider the closed contour that is traced by a point moving from ǫ to R above the branch cut, next around R, then below the cut to ǫ, and finally around ǫ. (See Figure 3.6.) We write f(z) in polar coordinates. f(z) = exp( a log z) z + = exp( a(log r + iθ)) r e ıθ + 652

28 We evaluate the function above, (z = r e ı ), and below, (z = r e ı2π ), the branch cut. f(r e ı ) = f(r e ı2π ) = exp[ a(log r + i)] r + exp[ a(log r + ı2π)] r + = r a r + = r a e ı2aπ. r + We use the residue theorem to evaluate the integral along. f(z) dz = ı2π Res(f(z), ) R r a R r + dr + r a e ı2aπ f(z) dz dr + f(z) dz = ı2π Res(f(z), ) R r + ǫ The residue is ǫ ǫ Res(f(z), ) = exp( a log( )) = exp( a(log + ıπ)) = e ıaπ. We bound the integrals along ǫ and R with the maximum modulus integral bound. ǫ a ǫ a f(z) dz 2πǫ = 2π ǫ ǫ ǫ R a R a f(z) dz 2πR = 2π R R R Since < a <, the values of the integrals tend to zero as ǫ and R. Thus we have ıaπ r a e dr = ı2π r + e ı2aπ x a x + dx = π sin aπ 653

29 Result 3.7. Integrals from Zero to Infinity. Let f(z) be a single-valued analytic function with only isolated singularities and no singularities on the positive, real axis, [, ). Let a Z. If the integrals exist then, f(x) log x dx = 2 x a f(x) log x dx = f(x) dx = x a f(x) dx = n Res (f(z) log z, z k ), k= ı2π e ı2πa n Res (z a f(z), z k ), k= n Res ( f(z) log 2 ) n z, z k + ıπ Res (f(z) log z, z k ), k= ı2π e ı2πa n Res (z a f(z) log z, z k ) k= x a f(x) log m x dx = m a m ( ı2π e ı2πa k= + π2 a sin 2 (πa) n Res (z a f(z), z k ), k= ) n Res (z a f(z), z k ), where z,...,z n are the singularities of f(z) and there is a branch cut on the positive real axis with < arg(z) < 2π. k= 654

30 3.8 Exploiting Symmetry We have already used symmetry of the integrand to evaluate certain integrals. For f(x) an even function we were able to evaluate f(x) dx by extending the range of integration from to. For x α f(x) dx we put a branch cut on the positive real axis and noted that the value of the integrand below the branch cut is a constant multiple of the value of the function above the branch cut. This enabled us to evaluate the real integral with contour integration. In this section we will use other kinds of symmetry to evaluate integrals. We will discover that periodicity of the integrand will produce this symmetry Wedge ontours We note that z n = r n e ınθ is periodic in θ with period 2π/n. The real and imaginary parts of z n are odd periodic in θ with period π/n. This observation suggests that certain integrals on the positive real axis may be evaluated by closing the path of integration with a wedge contour. Example 3.8. onsider + x n dx 655

31 where n N, n 2. We can evaluate this integral using Result x dx = n n k= n = k= n = k= n ( log z Res lim z e ıπ(+2k)/n lim z e ıπ(+2k)/n eıπ(+2k)/n + zn, ( ) ıπ( + 2k)/n = n e ıπ(+2k)(n )/n k= ) ( ) (z e ıπ(+2k)/n ) log z + z n ( ) log z + (z e ıπ(+2k)/n )/z nz n ıπ n = ( + 2k) e ı2πk/n n 2 e ıπ(n )/n = = = ı2π eıπ/n n 2 ı2π eıπ/n n 2 k= n k e ı2πk/n k= π n sin(π/n) n e ı2π/n This is a bit grungy. To find a spiffier way to evaluate the integral we note that if we write the integrand as a function of r and θ, it is periodic in θ with period 2π/n. + z n = + r n e ınθ The integrand along the rays θ = 2π/n, 4π/n, 6π/n,... has the same value as the integrand on the real axis. onsider the contour that is the boundary of the wedge < r < R, < θ < 2π/n. There is one singularity inside the 656

32 contour. We evaluate the residue there. Res We evaluate the integral along with the residue theorem. ( ) z e ıπ/n eıπ/n + zn, = lim z e ıπ/n + z n = lim z e ıπ/n nz n = eıπ/n n ı2π eıπ/n dz = + zn n Let R be the circular arc. The integral along R vanishes as R. R + z dz n 2πR n max z R + z n 2πR n R n as R We parametrize the contour to evaluate the desired integral. + x dx + n + x n eı2π/n dx = ı2π eıπ/n dx = + xn n( e ı2π/n ) + x n dx = π n sin(π/n) ı2π eıπ/n n 657

33 3.8.2 Box ontours Recall that e z = e x+ıy is periodic in y with period 2π. This implies that the hyperbolic trigonometric functions cosh z, sinh z and tanhz are periodic in y with period 2π and odd periodic in y with period π. We can exploit this property to evaluate certain integrals on the real axis by closing the path of integration with a box contour. Example onsider the integral [ ( ( ıπ cosh x dx = ı log tanh 4 + x 2))] = ı log() ı log( ) = π. We will evaluate this integral using contour integration. Note that cosh(x + ıπ) = ex+ıπ + e x ıπ 2 = cosh(x). onsider the box contour that is the boundary of the region R < x < R, < y < π. The only singularity of the integrand inside the contour is a first order pole at z = ıπ/2. We evaluate the integral along with the residue theorem. cosh z ( dz = ı2π Res cosh z, ıπ 2 = ı2π lim z ıπ/2 = ı2π lim z ıπ/2 = 2π z ıπ/2 cosh z sinh z ) 658

34 The integrals along the sides of the box vanish as R. ±R+ıπ ±R cosh z dz π max z [±R...±R+ıπ] cosh z π max 2 y [...π] e ±R+ıy + e R ıy 2 = e R e R π sinh R as R The value of the integrand on the top of the box is the negative of its value on the bottom. We take the limit as R. cosh x dx + cosh x dx = π dx = 2π cosh x 3.9 Definite Integrals Involving Sine and osine Example 3.9. For real-valued a, evaluate the integral: f(a) = 2π dθ + a sin θ. What is the value of the integral for complex-valued a. Real-Valued a. For < a <, the integrand is bounded, hence the integral exists. For a =, the integrand has a second order pole on the path of integration. For a > the integrand has two first order poles on the path of integration. The integral is divergent for these two cases. Thus we see that the integral exists for < a <. 659

35 For a =, the value of the integral is 2π. Now consider a. We make the change of variables z = e ıθ. The real integral from θ = to θ = 2π becomes a contour integral along the unit circle, z =. We write the sine, cosine and the differential in terms of z. sin θ = z z ı2, cosθ = z + z, dz = ı e ıθ dθ, dθ = dz 2 ız We write f(a) as an integral along, the positively oriented unit circle z =. /(ız) f(a) = + a(z z )/(2ı) dz = 2/a z 2 + (ı2/a)z dz We factor the denominator of the integrand. 2/a f(a) = (z z )(z z 2 ) dz ( ) ( ) + a 2 a 2 z = ı, z 2 = ı a a Because a <, the second root is outside the unit circle. z 2 = + a 2 a >. Since z z 2 =, z <. Thus the pole at z is inside the contour and the pole at z 2 is outside. We evaluate the contour integral with the residue theorem. 2/a f(a) = z 2 + (ı2/a)z dz = ı2π 2/a z z 2 = ı2π ı a 2 66

36 f(a) = 2π a 2 omplex-valued a. We note that the integral converges except for real-valued a satisfying a. On any closed subset of \ {a R a } the integral is uniformly convergent. Thus except for the values {a R a }, we can differentiate the integral with respect to a. f(a) is analytic in the complex plane except for the set of points on the real axis: a (... ] and a [... ). The value of the analytic function f(a) on the real axis for the interval (...) is 2π f(a) =. a 2 By analytic continuation we see that the value of f(a) in the complex plane is the branch of the function f(a) = 2π ( a 2 ) /2 where f(a) is positive, real-valued for a (...) and there are branch cuts on the real axis on the intervals: (... ] and [... ). Result 3.9. For evaluating integrals of the form a+2π a F(sin θ, cosθ) dθ it may be useful to make the change of variables z = e ıθ. This gives us a contour integral along the unit circle about the origin. We can write the sine, cosine and differential in terms of z. sin θ = z z ı2, cosθ = z + z, dθ = dz 2 ız 66

37 3. Infinite Sums The function g(z) = π cot(πz) has simple poles at z = n Z. The residues at these points are all unity. Res(π cot(πz), n) = lim z n π(z n) cos(πz) sin(πz) = lim z n π cos(πz) π(z n) sin(πz) π cos(πz) = Let n be the square contour with corners at z = (n + /2)(± ± ı). Recall that First we bound the modulus of cot(z). cosz = cosxcosh y ı sin x sinh y and sin z = sinxcosh y + ı cosxsinh y. cot(z) = cosxcosh y ı sin x sinh y sin x cosh y + ı cosxsinh y cos = 2 x cosh 2 y + sin 2 x sinh 2 y sin 2 x cosh 2 y + cos 2 x sinh 2 y cosh 2 y sinh 2 y = coth(y) The hyperbolic cotangent, coth(y), has a simple pole at y = and tends to ± as y ±. Along the top and bottom of n, (z = x ± ı(n + /2)), we bound the modulus of g(z) = π cot(πz). π cot(πz) π coth(π(n + /2)) 662

38 Along the left and right sides of n, (z = ±(n + /2) + ıy), the modulus of the function is bounded by a constant. g(±(n + /2) + ıy) = + /2)) cosh(πy) ı sin(π(n + /2)) sinh(πy) πcos(π(n sin(π(n + /2)) cosh(πy) + ı cos(π(n + /2)) sinh(πy) = ıπ tanh(πy) π Thus the modulus of π cot(πz) can be bounded by a constant M on n. Let f(z) be analytic except for isolated singularities. onsider the integral, n π cot(πz)f(z) dz. We use the maximum modulus integral bound. π cot(πz)f(z) dz n (8n + 4)M max z n f(z) Note that if then lim zf(z) =, z lim π cot(πz)f(z) dz =. n n This implies that the sum of all residues of π cot(πz)f(z) is zero. Suppose further that f(z) is analytic at z = n Z. The residues of π cot(πz)f(z) at z = n are f(n). This means n= f(n) = ( sum of the residues of π cot(πz)f(z) at the poles of f(z) ). 663

39 Result 3.. If lim zf(z) =, z then the sum of all the residues of π cot(πz)f(z) is zero. If in addition f(z) is analytic at z = n Z then f(n) = ( sum of the residues of π cot(πz)f(z) at the poles of f(z) ). n= Example 3.. onsider the sum n= (n + a) 2, a Z. By Result 3.. with f(z) = /(z + a) 2 we have n= ( (n + a) = Res π cot(πz) 2 n= d = π lim z a dz cot(πz) (z + a) ) 2, a = π π sin2 (πz) π cos 2 (πz) sin 2. (πz) (n + a) 2 = π 2 sin 2 (πa) Example 3..2 Derive π/4 = /3 + /5 /7 + /9. 664

40 onsider the integral I n = ı2π n dw w(w z) sinw where n is the square with corners at w = (n + /2)(± ± ı)π, n Z +. With the substitution w = x + ıy, sin w 2 = sin 2 x + sinh 2 y, we see that / sin w on n. Thus I n as n. We use the residue theorem and take the limit n. = [ n= ( ) n nπ(nπ z) + ] ( )n + nπ(nπ + z) z sin z z 2 sin z = z 2z ( ) n n 2 π 2 z 2 n= = z [ ] ( ) n nπ z ( )n nπ + z n= We substitute z = π/2 into the above expression to obtain π/4 = /3 + /5 /7 + /9 665