MA 201 Complex Analysis Lecture 17: Evaluation of integrals Lecture 17 Evaluation of integrals Evaluation of certain contour integrals: Type I Type I: Integrals of the form Z 2π F (cos θ; sin θ) dθ 0 dz If we take z = eiθ, then cos θ = 1 (z + 1 ), sin θ = 1 (z − 1 ) and dθ = : 2 z 2i z iz Substituting for sin θ; cos θ and dθ the definite integral transforms into the following contour integral Z 2π Z F (cos θ; sin θ) dθ = f (z) dz 0 jzj=1 1 1 1 1 1 where f (z) = iz [F ( 2 (z + z ); 2i (z − z ))] Apply Residue theorem to evaluate Z f (z) dz: jzj=1 Lecture 17 Evaluation of integrals Example of Type I Consider Z 2π 1 2 dt: 0 1 + 3(cos t) Z 2π 1 Z 1 dz dt = 1 + 3(cos t)2 1 1 2 iz 0 jzj=1 1 + 3( 2 (z + z )) Z z = −4i 4 2 dz jzj=1 3z + 10z + 3 Z z = −4i p p dz jzj=1 3(z + 3i) z − 3i z + pi z − pi 3 3 4 Z z = − i p p dz 3 jzj=1 (z + 3i)(z − 3i) z + pi z − pi 3 3 4 i i = − i × 2πifRes(f ; p ) + Res(f ; − p )g: 3 3 3 Lecture 17 Evaluation of integrals Improper Integrals of Rational Functions The improper integral of a continuous function f over [0; 1) is defined by Z 1 Z b f (x)dx = lim f (x)dx 0 b!1 0 provided the limit exists. If f is defined for all real x, then the integral of f over (−∞; 1) is defined by Z 1 Z 0 Z b f (x)dx = lim f (x)dx + lim f (x)dx −∞ a→−∞ a b!1 0 provided both limits exists. R 1 There is another value associated with the improper integral −∞ f (x)dx namely the Cauchy Principal value(P.V.) and it is given by Z 1 Z R P. V. f (x)dx := lim f (x)dx −∞ R!1 −R provided the limit exists. Lecture 17 Evaluation of integrals Evaluation of certain contour integrals: Type II R 1 R 1 If the improper integral −∞ f (x)dx converges, then P. V. −∞ f (x)dx exists and Z 1 Z 1 f (x)dx = P. V. f (x)dx: −∞ −∞ R 1 R 1 The P. V. −∞ f (x)dx exists 6=) 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 2 R) then R 1 R 1 P. V. −∞ f (x)dx exists =) the improper integral −∞ f (x)dx exists and their values are equal. P(z) Consider the rational function f (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 degree of Q(z) > 1+ degree of P(z) R 1 then P. V. −∞ f (x)dx can be evaluated using Cauchy residue theorem. Lecture 17 Evaluation of integrals Evaluation of certain contour integrals: Type II Type II Consider the integral Z 1 1 2 2 dx; −∞ (x + 1) To evaluate this integral, we look at the complex-valued function 1 f (z) = (z 2 + 1)2 which has singularities at i and −i. Consider the contour C like semicircle, the one shown below. Lecture 17 Evaluation of integrals Evaluation of certain contour integrals: Type II Note that: Z Z a Z f (z) dz = f (z) dz + f (z) dz C −a Arc Z a Z Z f (z) dz = f (z) dz − f (z) dz −a C Arc Furthermore observe that 1 1 f (z) = = : (z 2 + 1)2 (z + i)2(z − i)2 Then, by using Residue Theorem, 1 Z Z 2 d 1 π f (z) dz = (z+i) dz = 2πi = 2 2 C C (z − i) dz (z + i) 2 z=i Lecture 17 Evaluation of integrals Evaluation of certain contour integrals: Type II If we call the arc of the semicircle `Arc', we need to show that the integral over `Arc' tends to zero as a using the estimation lemma Z f (z) dz ≤ ML Arc where M is an upper bound on jf (z)j along the Arc and L the length of `Arc'. Now, Z aπ f (z) dz ≤ 2 2 ! 0 as a ! 1: Arc (a − 1) So Z 1 1 Z 1 Z a π 2 2 dx = f (z) dz = lim f (z) dz = : −∞ (x + 1) −∞ a!+1 −a 2 Lecture 17 Evaluation of integrals Evaluation of certain contour integrals: Type III Type III Integrals of the form Z 1 P(x) Z 1 P(x) P. V. cos mx dx or P. V. sin mx dx; −∞ Q(x) −∞ Q(x) where P(x); Q(x) are real polynomials and m > 0 Q(x) has no zeros in the real line degree of Q(x) > degree of P(x) then Z 1 P(x) Z 1 P(x) P. V. cos mx dx or P. V. sin mx dx −∞ Q(x) −∞ Q(x) can be evaluated using Cauchy residue theorem. Lecture 17 Evaluation of integrals Evaluation of certain contour integrals: Type III π 2θ Jordan's Lemma: If 0 < θ ≤ 2 then π ≤ sin θ ≤ θ: sin θ 0 (θ) Proof: Define φ(θ) = θ : Then φ (θ) = θ2 , where (θ) = θ cos θ − sin θ: 1 0 π Since (0) = 0 and (θ) = −θ sin θ ≤ 0 for 0 < θ ≤ 2 , π decreases as θ increases i.e. (θ) ≤ (0) = 0 for 0 < θ ≤ 2 . (θ) 2 0 π So φ (θ) = θ2 ≤ 0 for 0 < θ ≤ 2 . 3 2 sin θ That means φ is decreasing and hence π ≤ θ ≤ 1 for π 0 < θ ≤ 2 . By Jordan's lemma we have π Z π Z 2 Z π e−a sin θdθ = e−a sin θdθ + e−a sin θdθ π 0 0 2 Z π Z π 2 −a 2θ −a 2(π−θ) ≤ e π dθ + e π dθ π 0 2 both the integrals goes to 0 as a ! 1. Lecture 17 Evaluation of integrals Evaluation of certain contour integrals: Type III Evaluate: Z 1 cos tx Z 1 sin tx 2 dx or 2 dx −∞ x + 1 −∞ x + 1 Consider the integral Z 1 eitx 2 dx −∞ x + 1 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 0 from a to −a. Take a > 1 to be greater than 1, so that i is enclosed within the curve. Lecture 17 Evaluation of integrals Evaluation of certain contour integrals: Type III eitz eitz eitz e−t Res ; i = lim(z − i) = lim = : z 2 + 1 z!i z 2 + 1 z!i z + i 2i So by residue theorem Z −t e −t f (z) dz = (2πi) Resz=i f (z) = 2πi = πe : C 2i The contour C may be split into a "straight" part and a curved arc, so that Z Z + = πe−t straight arc and thus Z a itx Z itz e −t e 2 dx = πe − 2 dz: −a x + 1 arc z + 1 Lecture 17 Evaluation of integrals Evaluation of certain contour integrals: Type III Z itz Z π ita(cosθ+i sin θ) e e 2 dz ≤ a 2 dθ arc z + 1 0 a − 1 Z π a −ta sin θ ≤ 2 e dθ: a − 1 0 Hence, Z itz e dz ! 0 as a ! 1 arc z 2 + 1 and Z 1 cos tx Z a eitx P.V. 2 dx = lim 2 dx −∞ x + 1 a!1 −a x + 1 Z eitz = lim πe−t − dz a!1 arc z 2 + 1 = πe−t : Lecture 17 Evaluation of integrals.