Presentation on theme: "Week 8 2. The Laurent series and the Residue Theorem (continued)"— Presentation transcript:
1Week 8 2. The Laurent series and the Residue Theorem (continued) Theorem 1: The Estimation Lemma, or the M-L InequalityLet f(z) be continuous on a curve P. Thenwhereand l(P) is the length of P.Proof: Kreyszig, section 14.1 (non-examinable)
2Example 1: “Closing the contour” in complex integrals Evaluate the following integral (which is real, but we’ll treat it as complex):Solution:In all complex integrals, one should first locate all singular points of the integrand. In the problem at hand, there exist two simple (order-one) poles at z = ±i.Next, we’ll close the contour of integration, i.e. set the original integral aside and consider the following contour...
3Observe that the integral over the contour C1⋃C2 can be evaluated using the Residue Theorem: (1)C1Observe also that, in the limit R → ∞, C1 tends to the path of the original integral I – hence,In what follows, we’ll show that the integral over C2 vanishes – hence, (1) yields
4To prove that the integral over C2 vanishes as R → ∞, we’ll use the M-L Inequality. First, observe that the length of C2 isObserve also that, if z C2, we can set z = R eiθ – hence,hence
5M l Now, the M-L inequality yields which shows that, as required, Comment:The M-L Inequality works only if the integrand decays as z → ∞ quicker that const/z. Otherwise the product of M and l doesn’t vanish, and we can’t discard the contribution of the ‘infinite’ part of the expanded contour.
6Example 2:Calculate(2)Solution:Observe that the M-L inequality cannot be applied to this integral.However, this and some other integrals involving exponential can be evaluated using Jordan’s Lemma.
7Theorem 2: Jordan’s Lemma Let an arc A of radius R be centred at the origin and located in the upper half of the complex z plane. Let a function f(z) be continuous on A (i.e. for z = R eiθ) andThenwhere a is a positive real constant.Proof: Wikipedia (non-examinable)
8Comment:Jordan’s Lemma is typically used when f(z) decays as z → ∞ like const/z or slower. If it decays quicker than const/z, one can just as well use the M-L Inequality.
9Close the contour in (2) using a semi-circle in the upper semi-plane, Example 2 (continued):Close the contour in (2) using a semi-circle in the upper semi-plane,C2C1then use the Residue Theorem to calculateClearly,
11Comment:Jordan’s Lemma can be re-formulated for integrals with real exponentials (in what follows, the differences between the two formulations are highlighted).Let an arc A of radius R be centred at the origin and located in the left half of the complex z plane. Let a function f(z) be continuous on A andThenwhere a is a real positive constant.This formulation of Jordan’s Lemma will be referred to as “alternative” (we’ll need it later).
123. Integrals involving functions with branch points We’ll first calculate some important integrals without branch points.Example 3:Show thatwhere t > 0 is a real parameter.Hint: considerand change to polar coordinates (r, φ): p = r cos , q = r sin .
13Example 4:Show thatwhere t > 0 and x (of arbitrary sign) are real parameters.Example 5:Show thatwhere t > 0 and x (of arbitrary sign) are real parameters.
14The inverse Laplace transformation How do we find L–1[ F(s)] if F(s) isn’t listed in the LT table?...Theorem 2:Let F(s) be the Laplace transform of f(t). Then(3)where γ is such that the straight line (γ – i∞, γ + i∞) is located to the right of all singular points of F(s).Comment:If F(s) → 0 as s → ∞ and t < 0, integral (3) is identically zero due to Jordan’s Lemma (used in yet another alternative formulation).
15Example 6:Findwhere the branch of s1/2 on the complex s plane is fixed by the condition 11/2 = +1 and a cut along the negative part of the real axis.Solution:(4)where γ > 0.Close the contour in integral (4) as follows...
16The integrand is analytic inside the contour – hence, (5)Next, let R → ∞, henceand also (due to Jordan’s Lemma, alt. form.)Let ε → 0, hence (why?)
17Thus, the limiting version of (5) is hence,Introduce p > 0:Recalling how the branch of s1/2 was fixed, we have
18Then,hence, interchanging the upper↔lower limits in both integrals,In the 2nd integral, let p = –pnew, then omit new and ‘merge’ the 1st and 2nd integrals together, which yields
19Finally, using the result of Example 3, obtain Comment:Consider the horizontal segments of the curves C3 and C5 (let’s call these segments C'3 and C'5), and note that Jordan’s Lemma does not guarantee that their contributions vanish as R → ∞.Note, however, that, for z C'3,5, the integrand decays as R → ∞, whereas the lengths of C'3 and C'5 remain constant (both equal to γ). Hence, the integrals over C'3 and C'5 do decay as R → ∞.Alternatively, we can move γ infinitesimally close to zero: this wouldn’t affect the original integral (why?), but would make the lengths (and, thus, contributions) of C'3 and C'5 equal to zero.