Evaluation of certain contour integrals: Type I Type I: Integrals of the form 2π F (cos θ, sin θ) dθ If we take z = e iθ, then cos θ = 1 (z + 1 ), sin θ = 1 (z 1 dz ) and dθ = 2 z 2i z iz. Substituting for sin θ, cos θ and dθ the definite integral transforms into the following contour integral 2π F (cos θ, sin θ) dθ = f (z) dz where f (z) = 1 [F ( 1 (z + 1 ), 1 (z 1 ))] iz 2 z 2i z Apply Residue theorem to evaluate f (z) dz. z =1 z =1
Example of Type I Consider 2π 1 1 + 3(cos t) 2 dt = 2π z =1 = 4i = 4i = 4 3 i 1 1 + 3(cos t) 2 dt. 1 dz 1 + 3( 1 (z + 1 2 z ))2 iz z =1 z =1 z =1 z 3z 4 + 1z 2 + 3 dz 3(z + ( 3i) z 3i z ) ( z + ) ) dz i 3 (z i 3 z (z + 3i)(z ( ) ) dz 3i) z + i 3 (z i 3 = 4 3 i 2πi{Res(f, i 3 ) + Res(f, i 3 )}.
Improper Integrals of Rational Functions The improper integral of a continuous function f over [, ) is defined by provided the limit exists. f (x)dx = lim b b f (x)dx If f is defined for all real x, then the integral of f over (, ) is defined by f (x)dx = provided both limits exists. lim a a f (x)dx + lim b b f (x)dx There is another value associated with the improper integral f (x)dx namely the Cauchy Principal value(p.v.) and it is given by R P. V. f (x)dx := lim f (x)dx R R provided the limit exists.
Evaluation of certain contour integrals: Type II If the improper integral f (x)dx converges, then P. V. f (x)dx exists and f (x)dx = P. V. f (x)dx. The P. V. f (x)dx exists the improper integral f (x)dx exists. Take f (x) = x. However if f is an even function (i.e. f (x) = f ( x) for all x R) then P. V. f (x)dx exists = the improper integral f (x)dx exists and their values are equal.
Evaluation of certain contour integrals: Type II Consider the rational function f (z) = P(z) where P(z) and Q(z) are Q(z) polynomials with real coefficients such that Q(z) has no zeros in the real line then P. V. degree of Q(z) > 1+ degree of P(z) f (x)dx can be evaluated using Cauchy residue theorem.
Evaluation of certain contour integrals: Type II Type II Consider the integral 1 (x 2 + 1) 2 dx, To evaluate this integral, we look at the complex-valued function f (z) = 1 (z 2 + 1) 2 which has singularities at i and i. Consider the contour C like semicircle, the one shown below.
Evaluation of certain contour integrals: Type II Note that: C a a Furthermore observe that f (z) dz = a a f (z) dz = f (z) = Then, by using Residue Theorem, C f (z) dz = C C f (z) dz + f (z) dz Arc f (z) dz f (z) dz Arc 1 (z 2 + 1) = 1 2 (z + i) 2 (z i). 2 1 (z+i) 2 (z i) dz = 2πi d ( 2 dz 1 (z + i) 2 ) z=i = π 2
Evaluation of certain contour integrals: Type II Again, So Arc f (z) dz aπ as a. (a 2 1) 2 1 a (x 2 + 1) dx = f (z) dz = lim f (z) dz = π 2 a + a 2. Theorem Suppose f is analytic in C except at a finite number of singular points in C \ R. Let Γ R denote the semi-circle z = R in the upper half plane I(z). If lim z zf (z) = then lim R Γ R f (z)dz =.
Evaluation of certain contour integrals: Type II Exercise. Show that P.V. Show that Show that (x 2 x + 2)dx (x 4 + 1x 2 + 9) = 5π. [Poles are ±i, ±3i.] 12 x 2 dx (x 2 + 9)(x 2 + 4) 2 = π 2. dx x 3 + 1 = 2π 3 3.
Evaluation of certain contour integrals: Type III Type III Integrals of the form where then P. V. P(x) cos mx dx or P. V. Q(x) P(x), Q(x) are real polynomials and m > Q(x) has no zeros in the real line degree of Q(x) > degree of P(x) P(x) sin mx dx, Q(x) P(x) P(x) P. V. cos mx dx or P. V. sin mx dx Q(x) Q(x) can be evaluated using Cauchy residue theorem.
Evaluation of certain contour integrals: Type III Evaluate: Consider the integral cos αx dx or x 2 + 1 e iαx x 2 + 1 dx sin αx x 2 + 1 dx We will evaluate it by expressing it as a limit of contour integrals along the contour C that goes along the real line from a to a and then counterclockwise along a semicircle centered at from a to a. Take a > 1 so that i is enclosed within the curve.
Evaluation of certain contour integrals: Type III ( ) e iαz Res z 2 + 1, i = lim(z i) eiαz z i z 2 + 1 = lim e iαz z i z + i So by residue theorem f (z) dz = (2πi)Res(f, i) = 2πi e α = πe α. 2i C = e α. 2i The contour C may be split into a straight part and a curved arc, so that straight + arc = πe α and thus a a e iαx x 2 + 1 dx = e iαz πe α arc z 2 + 1 dz.
Evaluation of certain contour integrals: Type III Hence, and e iαz arc z 2 + 1 dz arc π a eiαa(cosθ+i sin θ) a 2 1 dθ a π e αa sin θ dθ. a 2 1 e iαz z 2 + 1 dz as a cos αx a P.V. dx = lim x 2 + 1 a a = lim a = πe α. e iαx x 2 + 1 dx [ πe α arc ] e iαz z 2 + 1 dz
Jordan s Lemma Lemma Let f be analytic in C except for finite number of singular points in C \ R. Let Γ R denote the semicircle z = R in the upper half plane. If lim R ( max f (z) z =R ) =, then lim R The result follows from Jordan s inequality: Γ R f (z)e iaz dz =. π e R sin θ dθ < π R (R > ).
Evaluation of certain contour integrals: Type IV Type IV Integrals of the form sin x x dx can be evaluated using Cauchy residue theorem. Before we discuss integrals of Type IV we need the following result. Lemma: Suppose f has a simple pole at z = a on the real axis. If c ρ is the contour defined by c ρ(t) = a + ρe i(π t), t (, π) then lim f (z)dz = iπres(f, a). ρ c ρ Proof: Since f has a simple pole at z = a, the Laurent series expansion of f about z = a is of the form f (z) = Res(f, a) z a + g(z).
Evaluation of certain contour integrals: Type IV Now c ρ f (z)dz = Res(f, a) c ρ z a dz + g(z)dz c ρ = Res(f, a) = iπres(f, a) π iρe i(π t) dt ρei(π t) π π g(a + ρe i(π t) )iρe i(π t) dt g(a + ρe i(π t) )iρe i(π t) dt. Note that f has Laurent series expansion in < z a < R for some R >. The function g is continuous on z a ρ for every ρ < ρ < R. So g(z) < M on z a ρ. So π g(a + ρe i(π t) )iρe i(π t) dt ρmπ as ρ. Hence lim f (z)dz = iπres(f, a). ρ c ρ
Evaluation of certain contour integrals: Type IV Consider the integral sin x x dx. Define f (z) = eiz, (z = is a simple pole on the real axis). z Consider the contour C = [ R, ɛ] τ [ɛ, R] Γ.
Evaluation of certain contour integrals: Type IV By Cauchy s theorem e iz z dz = But So C [ R, ɛ] [ R, ɛ] [ɛ,r] e iz z dz + e iz τ z dz + e iz [ɛ,r] z dz + Γ e iz z dz + e iz [ɛ,r] z dz = [ɛ,r] e ix e ix x e iz dx = τ z dz Γ e ix e ix dx x e iz dz = iπ z e iz dz =. z as ɛ (by the previous Lemma ) and R (by Jordan s inequality) and hence, sin x x dx = π 2.
Evaluation of certain contour integrals: Type V Integration along a branch cut: Consider the improper integral Define f (z) = z a 1 + z x a dx ( < a < 1). 1 + x ( z >, < arg z < 2π). z a The function is a multiple valued function with branch cut 1 + z arg z = (positive real axis). Consider the contour C = [ɛ + iδ, R + iδ] Γ R [R iδ, ɛ iδ] { γ ɛ}.
Evaluation of certain contour integrals: Type V By residue theorem ( Since + [ɛ+iδ,r+iδ] Γ R + f (z) = [R iδ,ɛ iδ] exp( a log z) z + 1 where z = re iθ, it follows that On [ɛ + iδ, R + iδ], θ as δ, ) + f (z)dz = 2πiRes(f, 1) = 2πie iaπ. γ ɛ = exp( a(ln r + iθ)), re iθ + 1 f (z) = exp( a(ln r + i.)) re i. + 1 On [R iδ, ɛ iδ], θ 2π as δ, r a 1 + r as δ. f (z) = exp( a(ln r + i.2π)) re i.2π + 1 r a 1 + r e 2aπi as δ.
Evaluation of certain contour integrals: Type V But and So Γ R z a 1 + z dz R a 2πR 2πR = R 1 R 1 γ ɛ z a 1 + z dz ( R lim R,ɛ ɛ ɛ a ɛ 1 2πɛ = r a ɛ 1 + r dr + R 1 R as R a 2π 1 ɛ ɛ1 a as ɛ. ) r a 1 + r e 2aπi dr = 2πie iaπ That is and hence (1 e 2aπi r a ) dr = 2πie iaπ 1 + r r a 2πie iaπ dr = 1 + r (1 e 2aπi ) = π sin aπ ( < a < 1).
Evaluation of certain contour integrals: Type VI Integration around a branch cut: Consider the improper integral log x 1 + x dx. 2 Define f (z) = log z 1 + z 2 ( z >, π 2 < arg z < 3π 2 ). The function log z is a multiple valued function whose branch cut 1 + z 2 consists of origin and negative imaginary axis. Consider the contour C = [ɛ, R] Γ R [ R, ɛ] { γ ɛ}.
Evaluation of certain contour integrals: Type VI By Cauchy s residue theorem ( Since + [ɛ,r] Γ R + [ R, ɛ] + γ ɛ f (z) = where z = re iθ, it follows that On [ɛ, R], θ =, On [ R, ɛ], θ = π, ) f (z)dz = 2πiRes(f, i) = 2πi π 4 = π2 i 2. log z log z + iθ = z 2 + 1 r 2 e 2iθ + 1, f (z) = f (z) = log x x 2 + 1. log x + iπ. x 2 + 1
Evaluation of certain contour integrals: Type VI But Γ R log z 1 + z dz 2 = Γ R log R + iθ 1 + R 2 e 2iθ ireiθ dθ as R and log z 1 + z dz 2 = γ ɛ log R R R 2 1 π + R R 2 1 γ ɛ log ɛ + iθ 1 + ɛ 2 e 2iθ iɛeiθ dθ π θdθ log ɛ ɛπ ɛ 2 1 + ɛ ɛ 2 1 π θdθ as ɛ.
Evaluation of certain contour integrals: Type VI So That is Hence and lim R,ɛ ( R lim R,ɛ ɛ ( R ɛ log x ɛ x 2 + 1 dx + R ) log x + iπ dx = π2 i x 2 + 1 2 log x R x 2 + 1 dx + log x R ) ɛ x 2 + 1 dx + iπ ɛ x 2 + 1 dx = π2 i 2. log x x 2 + 1 dx = 1 x 2 + 1 dx = π 2.