Presentation on theme: "1 2. The Laurent series and the Residue Theorem (continued) Week 8 Let f(z) be continuous on a curve P. Then Theorem 1: The Estimation Lemma, or the M-L."— Presentation transcript:
1 2. The Laurent series and the Residue Theorem (continued) Week 8 Let f(z) be continuous on a curve P. Then Theorem 1: The Estimation Lemma, or the M-L Inequality Proof: Kreyszig, section 14.1 (non-examinable) where and l(P) is the length of P.
2 Example 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...
3 Observe also that, in the limit R → ∞, C 1 tends to the path of the original integral I – hence, C1C1 C2C2 Observe that the integral over the contour C 1 ⋃ C 2 can be evaluated using the Residue Theorem: In what follows, we’ll show that the integral over C 2 vanishes – hence, (1) yields (1)
4 To prove that the integral over C 2 vanishes as R → ∞, we’ll use the M-L Inequality. First, observe that the length of C 2 is Observe also that, if z C 2, we can set z = R e iθ – hence, hence
5 Now, the M-L inequality yields 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. which shows that, as required, M l
6 However, this and some other integrals involving exponential can be evaluated using Jordan’s Lemma. Example 2: Calculate Observe that the M-L inequality cannot be applied to this integral. Solution: (2)
7 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 e iθ ) and Theorem 2: Jordan’s Lemma where a is a positive real constant. Then Proof: Wikipedia (non-examinable)
8 Comment: 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.
9 Example 2 (continued): Close the contour in (2) using a semi-circle in the upper semi-plane, C1C1 C2C2 Clearly, then use the Residue Theorem to calculate
10 Hence, It follows from Jordan’s Lemma with that
11 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 and Comment: Then where a is a real positive constant. This formulation of Jordan’s Lemma will be referred to as “alternative” (we’ll need it later).
12 3. Integrals involving functions with branch points We’ll first calculate some important integrals without branch points. Example 3: Hint: consider and change to polar coordinates (r, φ) : p = r cos , q = r sin . Show that where t > 0 is a real parameter.
13 Example 4: Show that Example 5: Show that where t > 0 and x (of arbitrary sign) are real parameters.
14 The inverse Laplace transformation Theorem 2: Let F(s) be the Laplace transform of f(t). Then where γ is such that the straight line (γ – i∞, γ + i∞) is located to the right of all singular points of F(s). How do we find L –1 [ F(s)] if F(s) isn’t listed in the LT table?... 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). (3)
15 Example 6: Find where the branch of s 1/2 on the complex s plane is fixed by the condition 1 1/2 = +1 and a cut along the negative part of the real axis. Solution: where γ > 0. (4) Close the contour in integral (4) as follows...
16 The integrand is analytic inside the contour – hence, Next, let R → ∞, hence (5) and also (due to Jordan’s Lemma, alt. form.) Let ε → 0, hence (why?)
17 hence, Introduce p > 0 : Recalling how the branch of s 1/2 was fixed, we have Thus, the limiting version of (5) is
18 Then, hence, interchanging the upper ↔ lower limits in both integrals, In the 2 nd integral, let p = –p new, then omit new and ‘merge’ the 1 st and 2 nd integrals together, which yields
19 Finally, using the result of Example 3, obtain Comment: Consider the horizontal segments of the curves C 3 and C 5 (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.