Mathematical Methods for Physics and Engineering - Matematica.NET

APPLICATIONS OF COMPLEX VARIABLESIdeally, we would like the contribution to the integral from the circular arc Γ totend to zero as its radius R →∞. Using a modified version of Jordan’s lemma,it may be shown that this is indeed the case if there exist constants M>0andα>0 such that on Γ|¯f(s)| ≤ M R α .Moreover, this condition always holds when ¯f(s) hastheform¯f(s) = P (s)Q(s) ,where P (s) andQ(s) are polynomials and the degree of Q(s) is greater than thatof P (s).When the contribution from the part-circle Γ tends to zero as R →∞,wehave from the residue theorem that the inverse Laplace transform (25.25) is givensimply byf(t) = ∑ (residues of ¯f(s)e sx at all poles ) . (25.27)◮Find the function f(x) whose Laplace transform iss¯f(s) =s 2 − k , 2where k is a constant.It is clear that ¯f(s) is of the form required for the integral over the circular arc Γ to tendto zero as R →∞, and so we may use the result (25.27) directly. Now¯f(s)e sx se sx=(s − k)(s + k) ,and thus has simple poles at s = k and s = −k. Using (24.57) the residues at each pole canbe easily calculated asR(k) = kekxand R(−k) = ke−kx2k2k .Thus the inverse Laplace transform is given by(f(x) = 1 2 e kx + e −kx) =coshkx.This result may be checked by computing the forward transform of cosh kx. ◭Sometimes a little more care is required when deciding in which half-plane toclose the contour C.◮Find the function f(x) whose Laplace transform is¯f(s) = 1 s (e−as − e −bs ),where a and b are fixed and positive, with b>a.From (25.25) we have the integralf(x) = 12πi∫ λ+i∞λ−i∞e (x−a)s − e (x−b)ssds. (25.28)886