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c
                        Figure 4.3: Integration contour used in Kronig–Kramers relations to find ˜  from ˜  c    for
                        a non-magnetized plasma.


                          The complex permittivity of a plasma (4.76) obviously obeys the required frequency-
                        symmetry conditions (4.27). It also obeys the Kronig–Kramers relations required for
                        a causal material. From (4.76) we see that the imaginary part of the complex plasma
                        permittivity is

                                                                   2
                                                                  0 ω ν
                                                     c             p
                                                    ˜   (ω) =−         .
                                                                      2
                                                                 2
                                                              ω(ω + ν )
                        Substituting this into (4.37) we have
                                                                     2
                                                   2        ∞       0 ω ν
                                                                      p
                                       c
                                      ˜   (ω) −   0 =−  P.V.  −                    d .
                                                                             2
                                                                        2
                                                                    2
                                                   π      0      (  + ν )     − ω 2
                          We can evaluate the principal value integral and thus verify that it produces ˜  c   by
                        using the contour method of § A.1. Because the integrand is even we can extend the
                        domain of integration to (−∞, ∞) and divide the result by two. Thus
                                                                  2
                                                1        ∞       0 ω ν         d
                                                                  p
                                     c
                                    ˜   (ω) −   0 =  P.V.                               .
                                                π      −∞ (  − jν)(  + jν) (  − ω)(  + ω)
                        We integrate around the closed contour shown in Figure 4.3. Since the integrand falls
                                 4
                        off as 1/  the contribution from C ∞ is zero. The contributions from the semicircles C ω
                        and C −ω are given by π j times the residues of the integrand at   = ω and at   =−ω,
                        respectively, which are identical but of opposite sign. Thus, the semicircle contributions
                        cancel and leave only the contribution from the residue at the upper-half-plane pole
                          = jν. Evaluation of the residue gives
                                                          2
                                                 1       0 ω ν      1               0 ω 2 p
                                                          p
                                      c
                                     ˜   (ω) −   0 =  2π j                   =−
                                                                                  2
                                                 π    jν + jν ( jν − ω)( jν + ω)  ν + ω 2
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