The Sokhotski-Plemelj Formula

Transcription

1 hysics 25 Winter 208 The Sokhotski-lemelj Formula. The Sokhotski-lemelj formula The Sokhotski-lemelj formula is a relation between the following generalize functions (also calle istributions), ±iǫ = iπ(), () where ǫ > 0 is an infinitesimal real quantity. This ientity formally makes sense only when first multiplie by a function f() that is smooth an non-singular in a neighborhoo of the origin, an then integrate over a range of containing the origin. We shall also assume that f() 0 sufficiently fast as ± in orer that integrals evaluate over the entire real line are convergent. Moreover, all surface terms at ± that arise when integrating by parts are assume to vanish. To establish eq. (), we shall prove that f() ±iǫ = f() where the auchy principal value integral is efine as: f() 0 { f() + iπf(0), (2) f(), (3) assuming f() is regular in a neighborhoo of the real ais an vanishes as. In these notes, I will provie three ifferent erivations of eq. (2). The first erivation is a mathematically non-rigorous proof of eq. (2), which shoul at least provie some insight into the origin of this result. A more rigorous erivation starts with a contour integral in the comple plane, f(z)z. z By efining appropriately, we will obtain two ifferent epressions for this integral. Setting the two resulting epressions equal yiels eq. (2) with the upper sign. omple conjugating this result yiels eq. (2) with the lower sign. Finally, an elegant thir proof makes irect use of the theory of istributions. Finally, a useful check is to consier the Fourier transform of eq. (), as iscusse in Appeni A. Note that eq. () can be generalize as follows, where 0 ±iε = iπ( 0 ), (4) 0 { f() 0 f() f(). (5) 0

2 The corresponing generalization of eq. (2) is straightforwar. Note that eq. (2) an its generalization involve integration along the real ais. These ieas generalize further to the so-calle auchy type integrals as shown in Appeni B, an yiel the lemelj formulae of comple variables theory. 2. A non-rigorous erivation of the Sokhotski-lemelj formula We begin with the ientity, ±iǫ = iǫ 2 +ǫ, 2 where ǫ is a positive infinitesimal quantity. Thus, for any smooth function that is non-singular in a neighborhoo of the origin, f() ±iǫ = f() f() iǫ 2 +ǫ 2 2 +ǫ. (6) 2 The first integral on the right ha sie of eq. (6), f() f() f() f() = ǫ 2 2 +ǫ 2 2 +ǫ 2 2 +ǫ. (7) 2 In the first two integrals on the right han sie of eq. (7), it is safe to take the it ǫ 0. In the thir integral on the right han sie of eq. (7), if is small enough, then we can approimate f() f(0) for values of <. Hence, eq. (7) yiels, However, { f() f() = + 2 +ǫ ǫ 2 = 0, f() +f(0) 2 +ǫ 2. (8) since the integran is an o function of that is being integrate symmetrically about the origin, an { f() f() f() +, 0 efines the principal value integral. Hence, eq. (8) yiels f() 2 +ǫ 2 = f(). (9) Net, we consier the secon integral on the right han sie of eq. (6). Since ǫ is an infinitesimal quantity, the only significant contribution from ǫ f() 2 +ǫ 2 2

3 can come from the integration region where 0, where the integran behaves like ǫ 2. Thus, we can again approimate f() f(0), in which case we obtain ǫ where we have mae use of f() 2 +ǫ 2 ǫf(0) 2 +ǫ = 2 ǫ tan (/ǫ) = πf(0), (0) 2 +ǫ2 = π ǫ. Using the results of eqs. (9) an (0), we see that eq. (6) yiels, which establishes eq. (2). f() ±iǫ = f() iπf(0), () 2. A more rigorous erivation of the Sokhotski-lemelj formula We consier the following path of integration in the comple plane, enote by, shown below. Imz Rez That is, is the contour along the real ais from to, followe by a semicircular path (of raius ), followe by the contour along the real ais from to. The infinitesimal quantity is assume to be positive. Then f() = f() + f(), (2) where the principal value integral is efine in eq. (3). In the it of 0, we can approimate f() f(0) in the last integral on the right han sie of eq. (2). Noting that the contour can be parameterize as = e iθ for 0 θ π, we en up with Hence, 0 f() = f(0) 0 0 π ie iθ θ = iπf(0). eiθ f() = f() iπf(0). (3) 3

4 We can also evaluate the left han sie of eq. (3) by eforming the contour to a contour that consists of a straight line that runs from + iε to + iε, where ε is a positive infinitesimal (of the same orer of magnitue as ). Assuming that f() has no singularities in an infinitesimal neighborhoo aroun the real ais, we are free to eform the contour into without changing the value of the integral. It follows that f() +iε = f() +iε = f(y +iε) y, (4) y +iε where in the last step we have mae a change of the integration variable. Since ε is infinitesimal, we can approimate f(y+iε) f(y). Thus, after relabeling the integration variable y as, eq. (4) yiels f() = Inserting this result back into eq. (3) yiels f() +iε = f(). (5) +iε f() iπf(0). (6) Eq. (6) is also vali if f() is replace by f (). We can then take the comple conjugate of the resulting equation. The en result is 2 in agreement with eq. (). f() ±iε = f() iπf(0), 3. An elegant erivation of the Sokhotski-lemelj formula Starting from the efinition of the auchy principal value given in eq. (3), we integrate by parts to obtain f() f() = f()ln = f()ln f ()ln = f( ǫ)lnǫ f ()ln = f(ǫ)lnǫ f ()ln, f ()ln, More precisely, we can epan f(y+iε) in a Taylor series about ε = 0 to obtain f(y+iε) = f(y)+o(ε). At the en of the calculation, we may take ε 0, in which case the O(ε) terms vanish. 2 Alternatively, we can repeat the above erivation where the contour is replace by a semicircle of raius in the lower half comple plane, which yiels eq. (3) with i replace by i. Finally, after eforming the contour of integration to a new contour that consists of a straight line that runs from iε to iε, one obtains eq. (5) with i replace by i. 4

5 where f () f/ an we have assume that f() 0 sufficiently fast as ± so that the surface terms at ± vanish. Hence, { f() [f( ) f() ] = ln f ()ln f ()ln. 0 (7) Since f() is ifferentiable an well behave, we can efine g() 0 f (t)t = f() f(0) which implies that g() is smooth an non-singular an, f() = f(0)+g(). (8) Inserting eq. (8) back into eq. (7) then yiels { f() [ 2g()ln = f ()ln 0 = f ()ln. f ()ln Note that ln is integrable at = 0, so that the last integral is well-efine. Finally, we integrate by parts an rop the surface terms at ± (uner the usual assumption that f () 0 sufficiently fast as ). The en result is f() = f() ln. That is, we have erive the generalize function ientity, ln =. (9) We can employ eq. (9) to provie a very elegant erivation of eq. (). We begin with the efinition of the principal value of the comple logarithm, Ln z = ln z +iarg z, where arg z is the principal value of the argument (or phase) of the comple number z, with the convention that π < arg z π. In particular, for real an a positive infinitesimal ǫ, Ln(±iǫ) = ln ±iπθ( ), (20) ǫ 0 where Θ() is the Heavisie step function. Differentiating eq. (20) with respect to immeiately yiels the Sokhotski-lemelj formula, 3 where we have use eq. (9) an ±iǫ = iπ(), (2) Θ( ) = Θ() = (). 3 The erivative of the comple logarithm is Ln z/z = /z for z 0. 5

6 Appeni A: Fourier transforms of istributions Eqs. (9) an (2), which we repeat below ln =, (22) ±iε = iπ(), (23) are only meaningful when multiplie by a test function f() an integrate over a region of the real line that inclues the point = 0. In the theory of tempere istributions, test functions must be infinitely ifferentiable an vanish at ± faster than any inverse power of. learly, e ik oes not satisfy this requirement for a test function. Nevertheless, one can efine Fourier transforms of tempere istributions by using the well known property of the Fourier transform, f(k)g(k)k = f(k) g(k), (24) where f(k) f()e ik. If f() is a tempere istribution an g() is a test function, then if follows that g() eists an is well efine. The Fourier transform of f(), enote by g(k), is efine via eq. (24). One can now check the valiity of eqs. (22) an (23) by computing their Fourier transforms. To compute the Fourier transform of eq. (22), we make use of the property of Fourier transforms that Hence, f() eik = ik f(k). ln eik = ik ln e ik. (25) The calculation of the right-han sie of eq. (25) is rather involve, since it only eists in the sense of istributions. One can show that 4 [ ln e ik = π f ] k +2γ(k), (26) where γ is the Euler-Mascheroni constant, an the istribution f(/ k ) is efine as f(k)f k k f(k) f(k) f(0) f(k) k + k + k, (27) k k k for any vali test function f(k). 4 For eample, see Ram. Kanwal, Generalize Functions: Theory an Applications, Thir eition (Birkhäuser, Boston, 2004) pp an pp There are two typographical errors on these pages. In eq. (6.4.33), /u shoul be / u an in the last term in eq. (6.4.57), i(u i0) shoul be +i(u i0). Eq. (26) is a consequence of the correcte eq. (6.4.57). 6

7 Inserting the result of eq. (26) into eq. (25) an using k(k) = 0 an 5 ( k f ) = k = sgn(k), (28) k k the en result is given by, Net, we consier e ik = ln eik = iπsgn(k). (29) cos(k) sin(k) +i. (30) Since cos(k)/ is an o function of (i.e., it changes sign uner ), it immeiately follows from the efinition of the auchy principle value that cos(k) = 0. (3) Net, we observe that 0 sin(k)/ = k; that is, sin(k)/ is regular at = 0. Thus, sin(k) = sin(k) = sgn(k) siny y y = πsgn(k). (32) Note that the symbol has no effect on the integral given by eq. (32), since the integran is regular at = 0. The factor of sgn(k) arises after changing the integration variable, y = k. When k < 0, the integration its must be reverse, which then leas to the etra minus sign. Inserting eqs. (3) an (32) into eq. (30) then yiels, e ik = iπsgn(k). (33) In light of eqs. (29) an (33), we have verifie that the Fourier transform of eq. (22) is satisfie. Likewise, we can verify that the Fourier transform of eq. (23) is satisfie. The following result is require, e ik = 2πiΘ( k), (34) ±iε which was erive in Solution Set in hysics 25. Then, using eq. (33) an employing the two ientities, sgn(k) = Θ(k) Θ( k) an = Θ(k) + Θ( k), it follows that the Fourier transform of eq. (23) is 2πiΘ( k) = iπ [ Θ(k) Θ( k) ] iπ [ Θ(k)+Θ( k) ]. (35) It is a simple matter to check that eq. (35) is satisfie for either choice of sign. Since the Fourier transform of a tempere istribution an its inverse Fourier transform are unique, one can conclue that if the Fourier transforms of eqs. (22) an (23) are satisfie, then eqs. (22) an (23) are vali ientities. Thus, the Fourier transform technique ehibite this Appeni provies a fourth inepenent erivation of the Sokhotski-lemelj formula. 5 When we multiply f(/ k ) by k, the singularity at k = 0 is cancele an the prescription inicate by eq. (27) is no longer require. Noting that k/ k is equal to sign of k for k 0, we en up with eq. (28). 7

8 Appeni B: The lemelj Formulae of omple Variables Theory The Sokhotski-lemelj formula erive in these notes is in fact a special case of a more general result of the theory of comple variables, which is often referre to as the lemelj formulae (an less often as the Sokhotski formulae). In this Appeni, I shall simply state the relevant results, with references for further etails. onsier the auchy type integral, F(z) = 2πi t, (36) where z an t are comple variables, is a smooth curve (which may be an open or a close contour) an is a function efine on that satisfies, f(t 2 ) f(t ) < A t 2 t λ, (37) for any two points t an t 2 locate on the contour, where A an λ are positive numbers. Eq. (37) is calle the Höler conition. 6 For values of z, F(z) is an analytic function. For values of z on the contour, the value of F(z) is not well efine ue to the singularity encountere in the integration along. Nevertheless, F(z) oes have unique value that epens on how z approaches. Inee, there are two ifferent possible bounary values of F(z) epening on whether the contour is approache from the left or right. We therefore introuce F + (z) an F (z) where the former is the it as z approaches from the left an the latter is the it as z approaches from the right. Here, left an right are efine with respect to the positive irection of the contour. 7 The eplicit results for F ± (z) are given by the lemelj formulae, F ± (z) = ± f(z)+ 2 2πi t, for z, (38) an consists of all points of ecluing its enpoints (in the case of a close contour, there are no enpoints to eclue). In eq. (38), we employ the the auchy principal value prescription to treat the singularity in the integran. In this contet, the principal value is a generalization of eq. (3), t = 0 t, (39) where the contour consists of the part of with length 2 centere symmetrically aroun z, an is the contour with the part remove. If is analytic on, then the proof of eq. (38) is a straightforwar generalization of the proof given in Section 2. However, the lemelj formulae are more general an apply 6 If λ >, then it follows that the erivative f (t) must vanish on, in which case is a constant. Thus, one typically assumes that 0 < λ. 7 For eample, for a close counterclockwisecontour, F + (z) is given by the it of F(z) as z approaches from the interior of the region boune by an F (z) is given by the it of F(z) as z approaches from the eterior of the region boune by. 8

9 to any function that satisfies the Höler conition on the contour. In this case, the erivation of eq. (38) is more complicate. We have also siesteppe the case where z in eq. (38) is one of the enpoints of (which is relevant if the contour is open). The reaer is referre to the references below for further etails. One can recast eq. (38) into another form that commonly appears in the literature, F + (z) F (z) = f(z), (40) F + (z)+f (z) = πi t, (4) for values of z locate on all points of the contour not coinciing with its enpoints. In particular, eq. (40) inicates that the function F(z) efine in eq. (36), which is analytic for all comple values of z, has a iscontinuous jump as z crosses the contour. Moreover, the average of the two bounary values of F(z) on is given by eq. (36), where the singularity of the integran is treate by the auchy principal value prescription. Using the lemelj formulae of comple variables theory, one can recover the results of Section as follows. If is a contour that runs along the real ais in the positive irection, then eq. (36) yiels the bounary values, F ± (z), of F(z) as z approaches the real ais from above (i.e., from the left) or below (i.e., from the right), respectively, F ± (z) = F(z ±iǫ) = ǫ 0 ǫ 0 2πi t, (42) iǫ where ǫ > 0 is an infinitesimal real quantity. Hence, eqs. (38) an (42) yiel iǫ t = t±iπf(z). (43) Eq. (43) is equivalent to the ientity involving generalize functions given in eq. (4). As epecte, setting z = 0 in eq. (43) reprouces eq. (2). For further references on the material of Appeni B, see. Mark J. Ablowitz an Athanassios S. Fokas, omple Variables: Introuction an Applications (ambrige University ress, ambrige, UK, 997), hapter A.I. Markushevich, Theory of Functions of a omple Variable, art I (AMS helsea ublishing, rovience, RI, 2005), hapter F.D. Gakhov, Bounary Value roblems (Dover ublications, Inc., New York, NY, 990), hapters N.I. Muskhelishvili, Singular Integral Equations: Bounary problems of functions theory an their applications to mathematical physics, 2n eition (Dover ublications, Inc., New York, NY, 2008), hapter Aleaner O. Gogolin, Lectures on omple Integration (Springer International ublishing, ham, Switzerlan, 204) hapters.4 an