Some Applications of the Residue Theorem Supplementary Lecture Notes MATH 322, Complex Analysis Winter 2005

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1 Some Applications of the Residue Theorem Supplementary Lecture Notes MATH 3, Complex Analysis Winter 5 Pawe l Hitczenko Department of Mathematics Drexel University Philadelphia, PA 94, U.S.A. phitczenko@math.drexel.edu I would like to thank Frederick Akalin for pointing out a couple of typos.

2 Introduction These notes supplement a freely downloadable book Complex Analysis by George Cain (henceforth referred to as Cain s notes, that I served as a primary text for an undergraduate level course in complex analysis. Throughout these notes I will make occasional references to results stated in these notes. The aim of my notes is to provide a few examples of applications of the residue theorem. The main goal is to illustrate how this theorem can be used to evaluate various types of integrals of real valued functions of real variable. Following Sec.. of Cain s notes, let us recall that if C is a simple, closed contour and f is analytic within the region bounded by C except for finitely many points z, z,..., z k then C f(zdz = πi k Res z=zj f(z, j= where Res z=a f(z is the residue of f at a. Evaluation of Real-Valued Integrals.. Definite integrals involving trigonometric functions We begin by briefly discussing integrals of the form π F (sin at, cos btdt. ( Our method is easily adaptable for integrals over a different range, for example between and π or between ±π. Given the form of an integrand in ( one can reasonably hope that the integral results from the usual parameterization of the unit circle z = e it, t π. So, let s try z = e it. Then (see Sec. 3.3 of Cain s notes, cos bt = eibt + e ibt = zb + /z b, sin at = eiat e iat i = za /z a. i Moreover, dz = ie it dt, so that Putting all of this into ( yields π F (sin at, cos btdt = dt = dz iz. C ( z a /z a F, zb + /z b dz i iz, where C is the unit circle. This integral is well within what contour integrals are about and we might be able to evaluate it with the aid of the residue theorem.

3 It is a good moment to look at an example. We will show that π cos 3t 5 4 cos t dt = π. ( Following our program, upon making all these substitutions, the integral in ( becomes (z 3 + /z 3 / dz = z 6 + C 5 4(z + /z/ iz i C z 3 (z 4z 4 dz = z 6 + i C z 3 (z 5z + dz = z 6 + i z 3 (z (z dz. The integrand has singularities at z =, z = /, and z =, but since the last one is outside the unit circle we only need to worry about the first two. Furthermore, it is clear that z = is a pole of order 3 and that z = / is a simple pole. One way of seeing it, is to notice that within a small circle around z = (say with radius /4 the function C z 6 + (z (z is analytic and so its Laurent series will have all coefficients corresponding to the negative powers of z zero. Moreover, since its value at z = is 6 + ( ( =, the Laurent expansion of our integrand is z 6 + z 3 (z (z = z 3 ( + a z +... = z 3 + a z +..., which implies that z = is a pole of order 3. By a similar argument (using a small circle centered at z = / we see that z = / is a simple pole. Hence, the value of integral in ( is ( z πi (Res 6 + z= z 3 (z (z + Res z=/ ( z 6 + z 3. (z (z The residue at a simple pole z = / is easy to compute by following a discussion preceding the second example in Sec.. in Cain s notes: z 6 + z 3 (z (z = z 6 + z 3 (z 5z + = z 6 + z 5 5z 4 + z 3 3

4 is of the form p(z/q(z with p(/ = 6 + and q(/ =. Now, q (z = z 4 z 3 + 6z, so that q (/ = / 4 / 3 + 6/ = 3/ 3. Hence, the residue at z = / is p(/ q (/ = + 3 (6 6 = The residue at a pole of degree 3, z =, can be obtained in various ways. First, we can take a one step further a method we used to determine the degree of that pole: since on a small circle around, z 6 + (z (z = z 6 (z (z + (z (z. (3 is analytic, the residue of our function will be the coefficient corresponding to z of Taylor expansion of the function given in (3. The first term will not contribute as its smallest non-zero coefficient is in front of z 6 so we need to worry about the second term only. Expand each of the terms /(z and /(z into its Taylor series, and multiply out. As long as z < we get (z (z = z z/ = ( + z + z +... z z ( = ( + A z + z + 4 z + 4z +... = ( ( + A z z = + a z + 8 z +... so that the residue at z = is /8 (from the calculations we see that A = + /, but since our interest is the coefficient in front of z we chose to leave A unspecified. The same result can be obtained by computing the second derivative (see Sec.. of Cain s notes of z 6 +! z3 z 3 (z (z, and evaluating at z =. Alternatively, one can open Maple session and type: residue((z^6+/z^3/(*z-/(z-,z=; to get the same answer again. Combining all of this we get that the integral in ( is z 6 + i z 3 (z (z dz = ( i (πi = π = π 4 4, C as required. 4

5 . Evaluation of improper integrals involving rational functions Recall that improper integral is defined as a limit R lim R f(xdx f(xdx, provided that this limit exists. When the function f(x is even (i.e. f(x = f( x, for x R one has R f(xdx = R R f(xdx, and the above integral can be thought of as an integral over a part of a contour C R consisting of a line segment along the real axis between R and R. The general idea is to close the contour (often by using one of the semi-circles with radius R centered at the origin, evaluate the resulting integral by means of residue theorem, and show that the integral over the added part of C R asymptotically vanishes as R. As an example we will show that dx (x + = π 4. (4 Consider a function f(z = /(z +. This function is not analytic at z = i (and that is the only singularity of f(z, so its integral over any contour encircling i can be evaluated by residue theorem. Consider C R consisting of the line segment along the real axis between R x R and the upper semi-circle A R := {z = Re it, t π}. By the residue theorem C R dz (z + = πires z=i The integral on the left can be written as R dz (z + + R A R ( (z + dz (z +. Parameterization of the line segment is γ(t = t + i, so that the first integral is just R dx R R (x + = dx (x +. Hence, R ( dx (x + = πires z=i (z + A R. dz (z +. (5 5

6 Since (z + = (z i (z + i, and /(z + i is analytic on the upper half-plane, z = i is a pole of order. Thus (see Sec.. of Cain s notes, the residue is ( ( d (z i dz (z i (z + i = (z + i 3 = (i 3 = 4i 3 = 4i z=i which implies that the first term on the right-hand side of (5 is πi 4i = π 4. Thus the evaluation of (4 will be complete once we show that lim R A R But this is straightforward; for z A R we have z=i dz (z =. (6 + z + z = R, so that for R > (z + (R. Using our favorite inequality g(zdz M length(c, (7 C where g(z M for z C, and observing that length(a R = πr we obtain dz (z + πr (R, A R as R. This proves (6 and thus also (4..3 Improper integrals involving trigonometric and rational functions. Integrals like one we just considered may be spiced up to allow us to handle an apparently more complicated integrals with very little extra effort. We will illustrate it by showing that cos 3x π (x dx = + e 3. 6

7 We keep the same function /(x +, just to illustrate the main difference. This time we consider the function e 3iz f(z, where f(z is, as before /(z +. By following the same route we are led to R R e 3xi (x + dx = πires z=i(f(ze 3zi A R f(ze 3zi dz = π e 3 A R f(ze 3zi dz. At this point it only remains to use the fact that the real parts of both sides must the same, write e 3xi = cos 3x + i sin 3x, and observe that the real part of the left-hand side is exactly the integral we are seeking. All we need to do now is to show that the real part of the integral over A R vanishes as R. But, since for a complex number w, Re(w w we have (A Re f(ze dz 3iz f(ze 3iz dz, R A R and the same bound as in (7 can be established since on A R we have e 3iz = e 3i(x+iy = e 3y e 3xi = e 3y, since A R is in the upper half-plane so that y. Remark. (i This approach generally works for many integrals of the form f(z cos azdz and f(z sin azdz, where f(z is a rational function (ratio of two polynomials, where degree of a polynomial in the denominator is larger than that of a polynomial in the numerator (although in some cases working out the bound on the integral over A R may be more tricky than in the above example. The following inequality, known as Jordan s inequality, is often helpful (see Sec..4 for an illustration as well as a proof π e R sin t dt < π R. (8 (ii The integrals involving sine rather than cosine are generally handled by comparing the imaginary parts. The example we considered would give sin 3x (x dx =, + but that is trivially true since the integrand is an odd function, and, clearly, the improper integral sin 3x (x + dx converges. For a more meaningful examples see Exercises 4-6 at the end. 7

8 .4 One more example of the same type. Here we will show that sin x x dx = π. (9 This integral is quite useful (e.g. in Fourier analysis and probability and has an interesting twist, namely a choice of a contour (that aspect is, by the way, one more thing to keep in mind; clever choice of a contour may make wonders. Based on our knowledge form the previous section we should consider f(z = e iz /z and try to integrate along our usual contour C R considered in Sections. and.3. The only problem is that our function f(z has a singularity on the contour, namely at z =. To avoid that problem we will make a small detour around z =. Specifically, consider pick a (small ρ > and consider a contour consisting of the arc A R that we considered in the last two sections followed by a line segment along a real axis between R and ρ, followed by an upper semi-circle centered at with radius ρ and, finally, a line segment along the positive part of the real exit from ρ to R (draw a picture to see what happens. We will call this contour B R,ρ and we denote the line segments by L and L +, respectively and the small semi-circle by A ρ. Since our function is analytic with B R,ρ its integral along this contour is. That is AR e iz z dz = + ρ R e ix x Aρ dx + e iz z dz + R Since ρ e ix R R x dx = e ix ρ x dx, by combining the second and the fourth integral we obtain R ρ e ix e ix dx + x AR e iz z dz + Since the integrand in the leftmost integral is we obtain R i ρ or upon dividing by i R ρ i eix e ix ix = i sin x x, Aρ e iz z ρ dz =. sin x x AR dx = e iz z Aρ dz e iz dz =, z sin x x dx = i ( A R e iz z Aρ dz + e iz z dz. e ix x dx, The plan now is to show that the integral over A R vanishes as R and that e lim ρ Aρ iz dz = πi. ( z 8

9 To justify this last statement use the usual parameterization of A ρ : z = ρe it, t π. Then dz = iρe it dt and notice (look at your picture that the integral over A ρ is supposed to be clockwise (i.e. in the negative direction. Hence, Aρ e iz z dz = π e iρeit iρe it dt = i ρe it Thus to show ( it suffices to show that π lim ρ To this end let us look at π π e iρeit dt π = e iρeit dt e iρeit dt = π. π π π dt = e iρeit dt. ( e iρeit dt. We will want to use once again inequality (7. Since the length of the curve is π we only need to show that But we have e iρeit, as ρ. ( e iρeit = e iρ(cos t+i sin t = e ρ sin t e iρ cos t = e ρ sin t e iρ cos t e ρ sin t + e ρ sin t e ρ sin t e iρ cos t + e ρ sin t e iρ cos t + e ρ sin t. For t π, sin t so that e ρ sin t and thus the second absolute value is actually equal to e ρ sin t which is less than ρ sin t, since for any real u, u e u (just draw the graphs of these two functions. We will now bound e iρ cos t. For any real number v we have e iv = (cos v + sin v = cos v + sin v cos v + = ( cos v. We will show that cos v v If we knew that, then ( would follow since we would have ( e iρeit e iρ cos t + e ρ sin t ρ( cos t + sin t ρ. But the proof of ( is an easy exercise in calculus: let h(v := v + cos v. 9

10 Than ( is equivalent to h(v for v R. Since h( =, it suffices to show that h(v has a global minimum at v =. But h (v = v sin v so that h ( =, and h (v = cos v. That means that v = is a critical point and h(v is convex. Thus, it has to be the mininimum of h(v. All that remains to show is that e lim R AR iz dz =. (3 z This is the place where Jordan s inequality (8 comes into picture. Going once again through the routine of changing variables to z = Re it, t π, we obtain e AR iz z dz π = i π e ireit dt e ir(cos t+i sin t dt π = π e ir cos t e R sin t dt e R sin t dt π R, where the last step follows from Jordan s inequality (8. It is now clear that (3 is true. It remains to prove (8. To this end, first note that π e R sin t dt = π/ e R sin t dt. That s because the graph of sin t and (thus also of e R sin t is symmetric about the vertical line at π/. Now, since the function sin t is concave between t π, its graph is above the graph of a line segment joining (, and (π/, ; in other words sin t t π. Hence π/ π/ e R sin t dt e R π t dt = π ( e R π R R which proves Jordan s inequality. 3 Summation of series. As an example we will show that n = π 6, (4 n=

11 but as we will see the approach is fairly universal. Let for a natural number N C N be a positively oriented square with vertices at (± ± i(n + / and consider the integral cos πz z dz. (5 sin πz C N Since sin πz = for z =, ±, ±,... the integrand has simple poles at ±, ±,... and a pole of degree three at. The residues at the simple poles are (z k cos πz cos πk π(z k lim z k z = sin πz k lim z k π sin(π(z k = πk. The residue at the triple pole z = is π/3. To see that you can either ask Maple to find it for you by typing: residue(cos(pi*z/sin(pi*z/z^,z=; in your Maple session or compute the second derivative (see Sec.. of Cain s notes again of cos πz z3! z sin πz = z cos πz sin πz, and evaluate at z =, or else use the series expansions for the sine and cosine, and figure out from there, what s the coefficient in front of /z in the Laurent series for cos πz z sin πz around z =. Applying residue theorem, and collecting the poles within the contour C N we get C N ( cos πz z dz = πi π N sin πz 3 + k= N πk + k= ( πk = πi π N 3 + k= πk. The next step is to show that as N the integral on the left converges to. Once this is accomplished, we could pass to the limit on the right hand side as well, obtaining lim N ( π N k= k = π 3, which is equivalent to (4. In order to show that the integral in (5 converges to as N, we will bound cos πz z sin πz dz C N using our indispensable inequality (7.

12 First, observe that each side of C N has length N + so that the length of the contour C N is bounded by 8N + 4 = O(N. On the other hand, for z C N, z N + N, so that z N, so that it would be (more than enough to show that cos πz/ sin πz is bounded on C N by a constant independent of N. By Exercise 9 from Cain s notes sin z = sin x + sinh y, cos z = cos x + sinh y, where z = x + iy. Hence cos πz = cos πx + sinh πy sin πz sin πx + sinh πy = cos πx sin πx + sinh πy + sinh πy sin πx + sinh πy cos πx sin πx + sinh πy +. On the vertical lines of the contour C N x = ±(N + / so that cos(πx = and sin(πx = ±. Hence the first term above is. We will show that on the horizontal lines the absolute value of sinh πy is exponentially large, thus making the first term exponentially small since sine and cosine are bounded functions. For t > we have sinh t = et e t et. Similarly, for t < so that, in either case sinh t = et e t e t = e t, sinh t e t. Since on the upper horizontal line y = N + / we obtain sinh πy eπ(n+/, as N. All of this implies that there exists a positive constant K such that for all N and all for z C N cos πz sin πz K. Hence, we conclude that C N cos πz z sin πz dz K 8N + 4 N,

13 which converges to as N. This establishes (4. The above argument is fairly universal and applies generally to the sums of the form f(n. n= Sums from (or to infinity are then often handled by using a symmetry of f(n or similarly simple observations. The crux of the argument is the following observation: Proposition 3. Under mild assumptions on f(z, n= f(n = π ( Res f(z where the sum extends over the poles of f(z. cos πz sin πz In our example we just took f(z = /z. Variations of the above argument allow us to handle other sums. For example, we have Proposition 3. Under mild assumptions on f(z, n=, ( n f(n = π ( f(z Res, sin πz where the sum extends over the poles of f(z. 3

14 4 Exercises.. Evaluate. Evaluate 3. Evaluate 4. Evaluate 5. Evaluate 6. Evaluate 7. Evaluate 8. Evaluate 9. Show that π π π n= x x 4 + 5x + 4 dx. Answer: π/4. x (x + (x dx. Answer: π/5. + x + dx x 4 +. Answer: π. sin x x + 4x + 5 dx. Answer: π sin. e x sin x x dx. Answer: π sin. e x sin πx x + x + 5 dx. Answer: πe π. dt + sin t. Answer: π. cos 3x 5 4 cos x dx. Answer: 3 8 π. n 4 = π4 9.. Use a suggestion of Proposition 3. to show that ( n+ n= n = π. 4