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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.

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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...

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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)

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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

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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

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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)

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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)

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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.

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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

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10 Hence, It follows from Jordan’s Lemma with that

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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).

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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.

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13 Example 4: Show that Example 5: Show that where t > 0 and x (of arbitrary sign) are real parameters.

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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)

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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...

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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?)

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17 hence, Introduce p > 0 : Recalling how the branch of s 1/2 was fixed, we have Thus, the limiting version of (5) is

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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

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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.

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