dQ+Pv dg=0. (4.3) Co+C Here "P.V. " indicates that the integral is computed in the Cauchy principal value sense (see Appendix A). To evaluate the integrals over Co and Co, consider a function f(Z) analytic in the lower half of the Z-plane(Z= Z,+jZi). If the point z lies on the real axis as shown in Figure 4.1, we can calculate the integral F(x)= through the parameterization Z-z=8e/. since dz=jSeJe de we have F(z)=lim f(z+seJ [jseje]de=jf(2) de=jrf(2) Replacing Z by $2 and z by 0 we can compute e(r,)一∈0 to a(r,t We o(r, tdt'= go(r) as the dc conductivity and write mf(r2)-0 Too(r) If we replace Z by S and z by o we get e(r,2) -ds2= jE(r, o) Substituting these into(4.33)we have e(r,) P/(r,9)- If we write E(r, o)=E(r, o)+je(r, a) and equate real and imaginary parts in(4.34) we find that e(r,)-∈0=--PV ec(,sdS2, (4.35) e(r,3)-∈0 go(r) 36) 2001 by CRC Press LLChence C0+Cω ˜ c(r, ) − 0  − ω d + P.V. ∞ −∞ ˜ c(r, ) − 0  − ω d = 0. (4.33) Here “P.V.” indicates that the integral is computed in the Cauchy principal value sense (see Appendix A). To evaluate the integrals over C0 and Cω, consider a function f (Z) analytic in the lower half of the Z-plane (Z = Zr + j Zi). If the point z lies on the real axis as shown in Figure 4.1, we can calculate the integral F(z) = lim δ→0  f (Z) Z − z d Z through the parameterization Z − z = δe jθ . Since d Z = jδe jθ dθ we have F(z) = lim δ→0 0 −π f  z + δe jθ  δe jθ  jδe jθ  dθ = j f (z) 0 −π dθ = jπ f (z). Replacing Z by  and z by 0 we can compute lim →0 C0 ˜ c(r, ) − 0  − ω d = lim →0 C0  1 j  ∞ 0 σ(r, t )e− jt dt + 0  ∞ 0 χe(r, t )e− jt dt  1 −ω  d = −π  ∞ 0 σ(r, t ) dt ω . We recognize ∞ 0 σ(r, t  ) dt = σ0(r) as the dc conductivity and write lim →0 C0 ˜ c(r, ) − 0  − ω d = −πσ0(r) ω . If we replace Z by  and z by ω we get lim δ→0 Cω ˜ c(r, ) − 0  − ω d = jπ˜ c (r,ω) − jπ0. Substituting these into (4.33) we have ˜ c (r,ω) − 0 = − 1 jπ P.V. ∞ −∞ ˜ c(r, ) − 0  − ω d + σ0(r) jω . (4.34) If we write ˜ c(r,ω) = ˜ c (r,ω) + j˜ c(r,ω) and equate real and imaginary parts in (4.34) we find that ˜ c (r,ω) − 0 = − 1 π P.V. ∞ −∞ ˜ c(r, )  − ω d, (4.35) ˜ c(r,ω) = 1 π P.V. ∞ −∞ ˜ c (r, ) − 0  − ω d − σ0(r) ω . (4.36)