Evaluation of Definite Integrals Via the Residue Theorem ECE 6382 Evaluation of Definite Integrals Via the Residue Theorem D. R. Wilton ECE Dept. 8/24/10

Review of Cauchy Principal Value Integrals x 1/x Recall for real integrals, but a finite result is obtained if the integral interpreted as because the infinite contributions from the two symmetrical shaded parts shown exactly cancel. Integrals evaluated in this way are said to be (Cauchy) principal value integrals (or “deleted” integrals) and are often written as

Cauchy Principal Value Integrals To evaluate consider the integral Note: Principal value integrals have either symmetric limits extending to infinity or a vanishing, symmetric deleted interval at a singularity. Both types appear in this problem!

Brief Review of Singular Integrals 1 Logarithmic singularities are examples of integrable singularities:

Singular Integrals, cont’d 1/x singularities are examples of singularities integrable only in the principal value (PV) sense. Principal value integrals must not start or end at the singularity, but must pass through to permit cancellation of infinities

Singular Integrals, cont’d Singularities like 1/x2 are non- integrable:

Singular Integrals, cont’d Summary: ln x is integrable at x = 0 1/xa is integrable at x = 0 for 0 < a < 1 1/xa is non-integrable at x = 0 for a = 1, or a > 1 f(x)sgn(x)/|x|a has a PV integral at x = 0 if f(x) is continuous Above results translate to singularities at a point a via the transformation x  x-a

y x Dispersion Relations

Dispersion Relation, Example 1 Im

Dispersion Relation, Example 2 Kramers

Integrals of the form f is finite f is a rational function of Let so the above integral becomes

Integrals of the form (cont.) x iy |z|=1 Example:

Integrals of the form f is analytic in the UHP except for a finite number of poles (can easily be extended to handle poles on the real axis) f is , i.e. , a constant, in the UHP Since Ques: What changes to the problem conditions and result must be made if f is only analytic in the lower half plane (LHP)?

Integrals of the form (cont.) z = 2i z = 3i Example:

Integrals of the form (Fourier Integrals) f is analytic in the UHP except for a finite number of poles (can easily be extended to handle poles on the real axis), (i.e., z in UHP) Choosing the contour shown, the contribution from the semicircular arc vanishes by Jordan’s lemma: since for

Integrals of the form (Fourier Integrals) (cont.) Example: Using the symmetries of and and the Euler formula, , we write and identify . Hence

Exponential Integrals There is no general rule for choosing the contour of integration; if the integral can be done by contour integration and the residue theorem, the contour is usually specific to the problem. Example: Consider the contour integral over the path shown in the figure: The contribution from each contour segment in the limit must be separately evaluated:

Exponential Integrals (cont.)

Exponential Integrals (cont.) Hence

Exponential Integrals (cont.) Finally,

Integration around a Branch Cut A given contour of integration, usually problem specific, must not cross a branch cut. Care must be taken to evaluate all quantities on the chosen branch. Integrand discontinuities are often used to relate integrals on either side of the cut. Usually a separate evaluation of the contribution from the branch point is required. Example: We’ll evaluate the integral using the contour shown

Integration around a Branch Cut (cont.) First, note the integral exists since the integral of the asymptotic forms of the integrand at both limits exists: Define the branch of such that

Integration around a Branch Cut (cont.) Now consider the various contributions to the contour integral

Integration around a Branch Cut (cont.) Hence and finally,