A Gaussian integral with a purely imaginary argument

Transcription

1 Physics 15 Winter 18 A Gaussian integral with a purely imaginary argument The Gaussian integral, e ax dx =, Where ea >, (1) is a well known result. tudents first learn how to evaluate this integral in the case where a is a real, positive constant. It is not difficult to show that eq. (1) is valid for complex values of a in the case of ea >. Writing a = a +ia I, where a >, it follows that e ax = e a x e ia Ix. The presence of the e a x term guarantees that the integral given in eq. (1) converges, due to the exponential suppression of the integrand as x. In this note, I wish to evaluate the integral in eq. (1) in the case of an exponential function with a purely imaginary argument; i.e., a =. To treat this case, we shall first consider the following integral that is integrated over a closed contour in the complex plane, e iaz dz, where a > is a real constant, () and the closed contour is exhibited below. Imz 45 ez ince there are no singularities in the region of the complex plane enclosed by (and no singularities on the contour itself), we can use auchy s theorem to conclude that e iaz dz =. We can evaluate this integral in another way by considering three separate contributions, e iaz dz = e iax dx+ e iaz dz + e iaz dz =, (3) where z = x+iy, indicates the integral over the arc portion of the contour and indicates the integral over the diagonal portion of the contour (in the direction indicated by the arrows). 1

2 Along, we have x = y, where x = rcos(π/4) and y = rsin(π/4). That is, z = re iπ/4 and dz = e iπ/4 dr. Hence, e iaz dz = e iπ/4 e ia(reiπ/4 ) dr = e iπ/4 e ar dr, (4) after using e iπ/ = i in the argument of the exponent. Along, z = e iθ, with θ π/4. Then dz = ie iθ dθ and π/4 e iaz dz = e ia (cosθ+isinθ) ie iθ dθ. (5) We shall now demonstrate that lim e iaz dz =. (6) First, we employ the generalization of the triangle inequality, a+b a + b, to integrals, b b f(x)dx f(x) dx. Applying this inequality to eq. (5), it follows that π/4 e a sinθ dθ. hanging variables to α = θ, it follows that a a π/ e a sinα dα. (7) To make further progress, we employ the following result, known as Jordan s inequality, which is established in Appendix A, Using this result, eq. (7) yields π/ e a sinα dα α π sinα α, for α 1 π. (8) π/ e aα/π dα π ( ) 1 e a. Taking the limit and recalling that a >, we end up with the result quoted in eq. (6). In light of eq. (6), eqs. (3) and (4) yields e iax dx = lim e iaz dz = e iπ/4 e ar dr = e iπ/4, (9) where we have employed eq. (1) in the final step above.

3 It is amusing to note that if we interpret i 1/ = e iπ/4 (which implies that we are defining the complex square root function as its principal value on the first iemann sheet), then, one can rewrite eq. (9) as iπ e iax dx =. (1) Note that eq. (1) coincides with eq. (1) if a is replaced by ia. That is, it appears that eq. (1) is valid even when ea =. 1 Thus, we have established the formula, e iax dx = e iπ/4, for a >. (11) The corresponding result for a < can be obtained simply by complex conjugating the above result, Eqs. (11) and (1) can be combined into one formula, e iax dx = e iπ/4, for a >. (1) e iax dx = e iπsgn(a)/4, where a is a real constant, (13) 4 a andthesignfunction isdefined assgn(a) = a/ a forreal a. Taking thereal andimaginary parts of eq. (13) yields the Fresnel integrals, π sin(ax )dx = sgn(a) 8 a, cos(ax )dx = 8 a. It is instructive to employ eq. (1) in evaluating the Fourier transform of e iax, F(k) = e iax e ikx dx, assuming that a >. By completing the square, one can write ( i(ax +kx) = ia x+ k ) + ik a. Hence, e iax e ikx dx = e ik /() exp { ia ( x+ k ) } dx. a hanging integration variables by defining x = x+k/(a), it follows that e iax e ikx dx = e ik /() e iax dx = e ik /() e iax dx, 1 This conclusion is correct as long as one remembers to interpret i 1/ = e iπ/4 in eq. (1). One can also obtain eq. (1) by repeating the calculation of eq. (3), where the closed contour is now located in the fourth quadrant of the complex plane with an arc that makes an angle of 45. 3

4 where the final step makes use of the fact that the integrand is an even function of x. Using eq. (1), we arrive at the desired result, F(k) = e iax e ikx dx = a e iπ/4 e ik /(), for real a >. (14) Additional results can be obtained by differentiating eq. (14). For example, It then follows that i F k = xe iax e ikx dx = 1 xe iax e ikx dx. (15) a 3 e iπ/4 e ik /(), for real a >. (16) Eq. (16) was employed in class in evaluating the free particle propagator in three dimensions. It should be noted that strictly speaking, eq. (15) is false, which then casts doubt on the validity of eq. (16). One is permitted to compute F/ k by differentiating under the integral sign only if certain conditions are satisfied. 3 Unfortunately, eq. (15) does not satisfy the necessary conditions. Nevertheless, if we regard e iax e ikx as a generalized function (also called a distribution), then one can justify eq. (15) in the sense of distributions. An alternative (andultimatelyequivalent) approachistoinsertaconvergence factor, e εx, intotheintegrand of F(k), where ε is a real positive infinitesimal quantity. In this case, one simply modifies eqs. (14) (16) by replacing a a iε. Eq. (15) is then valid as long as ε. At the end of the calculation, one takes ε to recover eq. (16). APPENIX A: Proof of Jordan s inequality In this appendix, we shall prove Jordan s inequality, x π sinx x, for x 1 π. (17) The simplest proof is a graphical one. Plot the following three functions in the x-y plane: (i) y = x; (ii) y = sinx; and (iii) y = x/π for values of x 1 π. It is easy to see that the the graph of y = sinx lies below the graph of y = x over the entire range of x 1π. Likewise, it is easy to see that the graph of y = sinx lies above the graph of y = x/π over the same range. All three functions meet at x = and the graphs of y = sinx and y = x/π intersect againat x = 1 π. (I leave the explicit graphing of these three functions to the reader). Thus, eq. (17) is verified. 4 3 For a discussion of the precise conditions, see, e.g., p. 7 of T.W. Körner, A ompanion to Analysis (American Mathematical ociety, Providence, I, 4). 4 More details on the graphical proof of eq. (17) can be found on pp of ebabrata Basu, Introduction to lassical and Modern Analysis and their Application to Group epresentation Theory (World cientific, ingapore, 11). 4

5 For the more analytically minded student, here is another proof. First, we define a function, f(x) = x sinx, for x 1 π. Then, f (x) (df/dx) = 1 cosx. In particular, one achieves f (x) = at x = in the x range of interest. ince f() = and the slope of f(x) is positive for < x 1 π, it follows that f(x) > for < x 1 π. It then follows that sinx < x for < x 1 π, with equality achieved only at x =. Next, we consider a different function, omputing the first and second derivatives, g(x) = πsinx x. g (x) = πcosx, g (x) = πsinx <, for < x 1 π. In particular, g (x) is decreasing as x increases from to 1π. ince g () = π > and g ( 1π) =, there must exist some value of x (call it x ) in the range < x < 1 π such that g (x ) =. It followsthat g(x)isanmonotonicallyincreasing functionintherange < x < x and g(x) is a monotonically decreasing function for x < x < 1 π. ince g() = g(1 π) =, we can conclude that g(x) > for < x < 1 π. That is, g(x) = πsinx x >, for < x < 1 π, with equality achieved only at the two endpoints x = and x = 1 π. It therefore follows that sinx x/π for < x 1 π. The proof of eq. (17) is now complete. 5