Complex varibles:contour integration examples. cos(ax) x

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1 1 Problem 1: sinx/x omplex varibles:ontour integration examples Integration of sin x/x from to is an interesting problem 1.1 Method 1 In the first method let us consider e iax x dx = cos(ax) dx+i x sin(ax) dx x So when Z is real the integral is the imaginary part of the above integral. So if we select a contour as shown below, then on the real axis, Z = x and if we can find the values of the other integrals we can find the answer as the imaginary part of the real line integral. Thus, we replace the integral with ǫ z dz = z ǫ dz + z dz + ǫ z dz + R z dz As ǫ tends to zero we find that the first and the third part give us the real line integral e iax x dx As for R integral, using Jordan s lemma, it goes to zero for a > 0. With regard to ǫ we have for z dz z = 0 is a simple pole. Thus, one of the theorems says ǫ ǫ z dz = iφ 1 = iπ where 1 is the residue at the pole. φ = π because we go in the W direction. Thus we have z dz = e iax x dx iπ = 0 since no pole is enclosed. Thus, Matching real to real etc., we have e iax x dx = iπ sinax x dx = π 1

2 R -R R z=0 Figure 1: ontour for problem Method 2 In the second method, we will include the pole within the contour. Thus, z dz = ǫ And the result follows. 1. Method z dz = z dz + ǫ z dz + ǫ sin(ax) dx = x z z dz + z ǫ dz +iπ = 2πi z dz = πi e iax e iax dx 2ix z dz + R z dz dz = 2πi(ResF(z = 0)) dz The logic here is this: The first integral (what we want) is exactly equal to the second representation. We make it into a complex integral (with real limits) such that the complex integral will equal the two prior integrals when z = x. Hence the complex integral (with real limits) must be a branch of any contour we choose. Next, since the complex integral is a sum then it has to be split into its components and then each integral has a singularity at z = 0. We must then calculate 2

3 the penalty for integrating through the singularity. The first term is dz If this is integrated over the same contour as in figure 2, dz = The R integral is zero by Jordan lemma. The second term is dz = dz = If this is integrated over the same contour dz = ǫ dz + w R dz + ǫ dz + w dz = 0 iπ dz + 2i dz = π 2 dz = 0 ǫ dz + w R dz + The R integral is infinity because for a > 0 the Jordan lemma holds for the upper half of the complex plane. So we take the bottom contour and include the pole. dz = dz + ǫ w dz = 2πiRes(F(z = 0)) where the negative on the residue is because the direction of the contour is clockwise. dz = π dz + 2 = ( 1)2πiRes(F(z = 0)) Thus So dz = +π 2 π dz = +π 2 dz = π = sin(ax) dx x

4 1.4 Method 4 At times, a point z seems like a singularity, but when the limits are computed, then it is not a singularity. If an integral has to go through a singularity then there is no choice but to go through it using an ǫ contour and see if this is a integrable singularity by taking ǫ to zero. There is no avoiding it by deforming the contour. Avoiding the singularity by starting with an indented contour can be done only when the seeming singularity is not a singularity, i.e., the function is analytic in the domain. With a singularity it is not. However, in this case, there is no singularity in sin(ax)/x to begin with at x = 0. So when we break it into sin(ax) dx = x e iax e iax dx 2ix dz the complex integral is continuous at z = 0 and analytic all the way along the real axis. First let us observe that around the origin lim ǫ 0 ǫ dz, withz = ǫ iθ, dz = iǫ iθ dθ we get e iaǫ(cosθ+isinθ) e iaǫ(cosθ+isinθ) lim ǫ 0 ǫ 2iǫ iθ iǫ iθ dθ = 0. If f is continuous in a domain D and integrals round closed contours are zero, then the function is analytic in D. Thus, the function is not singular and has no integral contributions from z = 0. So nothing new can come integrating through the z = 0 point. Hence, we choose an indented contour to begin with as shown in figure below (figure 2). The value of the integral must be the same passing through the origin or on the indented contour. R z=0 R Figure 2: ontour for problem method 4. First Term: dz We have two terms in the integral above. We take the first term. We now choose to deform the already indented path. If we continuously deform the path now to + this function is analytic throughout and follows Jordan s Lemma. Given a > 0 and 1/z goes to zero uniformly as z goes to infinity. Thus, dz = 0. 4

5 Second Term: dz This term because of th = ia(x + iy) = iax + ay goes to infinity as y becomes large positive in the upper half plane. This function is non-analytic at z = in the upper half. (If the value of a function becomes infinity then it has a singularity at that point). So we deform it the contour downward as shown in the figure. So for the second integral we have the downward R z=0 R R contour Figure : ontour for problem method 4a. dz If this is integrated over the new downward contour R dz = ) e iaz + ǫ w ( dz + + to 0 0 to R dz ++ We will hold onto the end points, i.e., + and, and continuously deform the contour downward, into R at. The value of this R integral by Jordan s Lemma is zero. The two upward and downward integrals sum to zero, since the function is analytic and uniquely defined. The integral limits are opposite. Thus, we are left with the integral round the singularity in the clockwise direction. This gives a residue Thus as before. ǫ w 2 The motivation problem The following integral needs to be worked out dz = 2πi 1 2i = π dz = π v(x) = iωf 2πB e ikx k 4 kp 4 dk. (1) The evaluation of the above improper real integral will be done using complex analysis. We now extend the function v(k) over the complex k plane. The motivation in doing so being that 5

6 in the complex plane using auchy integral theorem and Jordan s lemma, infinite integrals of the form in equation (1), can be easily computed. This solution methodology is more elegant and powerful. auchy integral formula states that for a function f(z) analytic except at a finite number of points (z 1, z 2,...z N ) lying inside a closed contour 0, the following relation holds N f(z)dz = i2π Residue [f(z)]. (2) z=zn 0 n=1 Jordan s lemma states that if in the above contour integral, f(z) = g(z) with a > 0 and 0 consists of a contour from R to R on the real line followed by a semicircle centered at the origin and lying in the upperhalf plane (lower half plane if a < 0), then under certain conditions on g(z), the contribution from the semi-circular part to the contour integral approaces zero as R. Thus, in such case we have = = i2π Residue[f(z)], z=zn n I where I is an index set such that z n, n I is a pole of f(z) lying strictly over the real line. We shall use auchy s integral formula and Jordan s lemma extensively. We return to our problem of inverting v(k) for x > 0 through equation (1). The symmetry of the problem suggests identical results for x < 0. The form of the integral in equation (1) satisfies the conditions required by Jordan s lemma, with g(k) having singularities at ±k p & ±ik p. The pole ik p lies inside the semi-circle prescribed in Jordan s lemma and hence gives a contribution to the integral. This is not so with the pole ik p as it lies outside the contour. The poles ±k p lie on the contour and we need to deal with these separately. The contour and the pole locations are indicated in figure 4 for both x > 0 and x < 0. ikp ikp -Kp K p -Kp K p -ik p -ik p X>0 x<0 Figure 4: Schematic showing the contours along with the pole positions for the inversion integral in equation 1. Let us now displace the poles located on the real axis (±k p ) by a small distance ǫ > 0 along the imaginary axis. In so doing, the auchy integral formula will be applicable as we will be able to ascertain whether or not the poles lie inside the contour. This small perturbation of the poles from the real axis may be carried in many ways as described below. We then find the limit as 6

7 ǫ 0 in each case and judge the physical relevance of the solution obtained. Figure 5 depicts pictorially all the possible pole shifts. The poles can be shifted in the following ways: 1. Both poles ±k p are shifted upward. In that case, for x > 0 three poles lie inside the contour namely ik p, kp+iǫ, kp+iǫ. This is depicted in case 1 of figure 5. Thus, by auchy integral formula we expect v(x) for x > 0 to be of the form Ae kpx + Be ikpx ǫx + e ikpx ǫx. In the limit ǫ 0, the third term in the above represents a negative travelling wave in the region x > 0. Recall, the forcing is at x = 0. Thus, the incoming wave solution violates the causality condition and hence is physically unacceptable. 2. Both poles ±k p are shifted downward. In that case, for x < 0 three poles lie inside the contournamely ik p, kp iǫ, kp iǫ. Thisisdepictedincase2offigure5. Thus, byauchy integral formula we expect v(x) for x < 0 to be of the form Ae kpx +Be ikpx+ǫx +e ikpx+ǫx. In the limit ǫ 0, the second term in the above represents a positive travelling wave in the region x < 0. As the force is at x = 0, this solution violates the causality condition and hence is physically unacceptable.. The pole k p is shifted to k p iǫ and the pole k p is shifted to k p + iǫ. For x > 0, the interior poles are ik p & k p +iǫ. This is depicted in case of figure 5. Thus, by auchy integral formula we expect v(x) for x > 0 to be of the form Ae kpx + Be ikpx ǫx. In the limit ǫ 0, the third term in the above represents negative travelling wave in the region x > 0. As the force is at x = 0, this solution violates the causalty condition and hence is physically unacceptable. 4. The pole k p is shifted to k p + iǫ and the pole k p is shifted to k p iǫ. For x > 0, the interior poles are ik p & k p + iǫ. This is depicted in case 4 of figure 5. Thus, by auchy integral formula we expect v(x) for x > 0 to be of the form Ae kpx +Be ikpx ǫx. The first term represents an evanescent wave, the second term in the limit ǫ 0 represent a positive traveling wave. This solution is physically acceptable. Similarly for x < 0, the interior poles are ik p & k p iǫ. Thus, by auchy integral formula we expect v(x) for x < 0 to be of the form Ae kpx +Be ikpx+ǫx. The first term represents an evanescent wave, the second term with ǫ 0 represent a negative travelling wave. In the limit ǫ 0, for all x, the solution is decomposed into a travelling and an evanescent wave contribution. Hence, out of the four ways of perturbing the poles, this is the only one which gives physically meaningful results. In dealing with integrals where the poles lie on the contour in complex analysis, the notion of principal value of the integral is commonly used. The principal value interpretation of the integral too fails to give a physically meaningful results in this case. Generalizing this idea, we observe that for x > 0 we need to account for the pole on the positive real axis and ignore the ones on the negative real axis. Similarly, for x < 0 we need to account for the pole on the negative real axis and ignore the ones on the positive real axis. The equivalent approach would be, instead of taking the integral from to on the real axis, we deform the contour slightly upward in the left half plane and slightly downward in the right half plane (the principle of deformation of contours aloows this). Thus, for x > 0 when we close the contour in the upper half, the pole on the positive real axis will contribute to the integral and that on the negative real axis would be left out. Similarly, for x < 0 when we close the contour in the lower half, the pole on the negative real axis will contribute to the integral and that on the positive real axis would be left out. This scheme is depicted in figure 6. 7

8 ase 1 Both poles displaced upwards ase 2 Both Poles Displaced downwards ase Right Pole Shifted Downwards Left Pole Shifted Upwards ase 4 Right Pole Shifted upwards Left Pole Shifted downwards Figure 5: Different Strategies to shift poles by a small amount For x>0 For x<0 Figure 6: Redefining the contour integral path for computing the inverse Fourier transform. For applying the auchy residue theorem with x > 0 (x < 0), the semicircular contour in the upper (lower) half plane is chosen. 8

9 2.1 Structural Response We return to the evaluation of equation (1). In light of the preceeding discussion, for x > 0 the residue contributions corresponding to the poles at k p & ik p need to be considered. The residues corresponding to k p and ik p are eikpx 4k p and ie kpx, respectively. Thus, for x > 0 we have 4kp v(x) = iωf [ ] e ik px 2πB 2πi 4kp + ie kpx 4kp = ωf [ 4kpB e ikpx +ie kpx]. () For x < 0 by taking the residue contribution from poles at k p & ik p we arrive at a similar result, v(x) = iωf [ ] e ik px 2πB 2πi 4kp + iekpx 4kp = ωf [ 4kpB e ikpx +ie kpx]. (4) The above relations indicate that the velocity response of the line-driven plate can be decomposed in terms of free wavenumbers (from equation??). The line excitation has no spatial extent and hence enforces only the frequency. And the free wavenumbers at this frequency are generated in the plate. Jordan s Lemma was easily invoked to determine the structural response. The theorem required the function under consideration to be single-valued and analytic except at the poles. For the present case of uncoupled analysis the function e ikx /(k 4 k 4 p) satisfied these conditions. 9