Mathematical Methods for Physics and Engineering - Matematica.NET

24.13 DEFINITE INTEGRALS USING CONTOUR INTEGRATIONWe note that result (24.60) is a special case of (24.63) in which θ 2 is equal toθ 1 +2π.24.13 Definite integrals using contour integrationThe remainder of this chapter is devoted to methods of applying contour integrationand the residue theorem to various types of definite integral. However, threeapplications of contour integration, in which obtaining a value for the integral isnot the prime purpose of the exercise, have been postponed until chapter 25. Theyare the location of the zeros of a complex polynomial, the evaluation of the sumsof certain infinite series and the determination of inverse Laplace transforms.For the integral evalations considered here, not much preamble is given since,for this material, the simplest explanation is felt to be via a series of workedexamples that can be used as models.Suppose that an integral of the form24.13.1 Integrals of sinusoidal functions∫ 2π0F(cos θ, sin θ) dθ (24.64)is to be evaluated. It can be made into a contour integral around the unit circleC by writing z =expiθ, and hencecos θ = 1 2 (z + z−1 ), sin θ = − 1 2 i(z − z−1 ), dθ = −iz −1 dz. (24.65)This contour integral can then be evaluated using the residue theorem, providedthe transformed integrand has only a finite number of poles inside the unit circleandnoneonit.◮Evaluate∫ 2πcos 2θI =dθ, b > a > 0. (24.66)0 a 2 + b 2 − 2ab cos θBy de Moivre’s theorem (section 3.4),cos nθ = 1 2 (zn + z −n ). (24.67)Using n = 2 in (24.67) and straightforward substitution for the other functions of θ in(24.66) givesI =i ∮z 4 +12ab C z 2 (z − a/b)(z − b/a) dz.Thus there are two poles inside C, a double pole at z = 0 and a simple pole at z = a/b(recall that b>a).We could find the residue of the integrand at z = 0 by expanding the integrand as aLaurent series in z and identifying the coefficient of z −1 . Alternatively, we may use the861