UNIVERSITY OF BRISTOL Department of Physics Level 3 Mathematical Methods 33010 (2009/2010) Definite Integrals by Contour Integration Many kinds of (real) definite integrals can be found using the results we have found for contour integrals in the complex plane. This is because the values of contour integrals can usually be written down with very little diﬃculty. We simply have to locate the poles inside the contour, find the residues at these poles, and then apply the residue theorem. The more subtle part of the job is to choose a suitable contour integral i.e. one whose evaluation involves the definite integral required. We illustrate these steps for a set of five types of definite integral. Type 1 Integrals Integrals of trigonometric functions from 0 to 2 π : I = 2 π 0 (trig function) dθ By “trig function” we mean a function of cos θ and sin θ . The obvious way to turn this into a contour integral is to choose the unit circle as the contour, in other words to write z = exp iθ , and integrate with respect to θ . On the unit circle , both cos θ and sin θ can be written as simple algebraic functions of z : cos θ = 1 2 ( z + 1 /z ) sin θ = 1 2 i ( z − 1 /z ) and making this replacement turns the trigonometric function into an algebraic function of z whose poles can be easily found. Example: I = 2 π 0 dθ 1 + a cos θ where − 1 < a < +1 I = 1 1 + a 2 ( z + 1 z ) dz iz = 2 i dz 2 z + az 2 + a The poles of the function being integrated lie at the roots of the equation az 2 +2 z + a = 0 i.e. at the points z ± = 1 a − 1 ± √ 1 − a 2 1
C Z + Z - Of the poles, only z + lies inside the unit circle, so I = 2 πiR + where R + is the residue at z + To find the residue we note that this is a simple pole and if we write the integrand as f ( z ) = g ( z ) /h ( z ) the residue at z + is: g ( z + ) h ( z + ) = 2 2 i ( az + + 1) = 1 i √ 1 − a 2 Hence the integral required is 2 π/ √ 1 − a 2 Type 2 Integrals Integrals such as I = + ∞ −∞ f ( x ) dx or, equivalently, in the case where f ( x ) is an even function of x I = + ∞ 0 f ( x ) dx can be found quite easily, by inventing a closed contour in the complex plane which includes the required integral. The simplest choice is to close the contour by a very large semi-circle in the upper half-plane. Suppose we use the symbol “ R ” for the radius. The entire contour integral comprises the integral along the real axis from − R to + R together with the integral along the semi-circular arc. In the limit as R → ∞
You've reached the end of your free preview.
Want to read all 6 pages?
- Fall '19
- Calculus, Limit, Line integral, Methods of contour integration, contour integration