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**Week 8 2. The Laurent series and the Residue Theorem (continued)**

Theorem 1: The Estimation Lemma, or the M-L Inequality Let f(z) be continuous on a curve P. Then where and l(P) is the length of P. Proof: Kreyszig, section 14.1 (non-examinable)

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**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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**Observe that the integral over the contour C1⋃C2 can be evaluated using the Residue Theorem:**

(1) C1 Observe 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

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To prove that the integral over C2 vanishes as R → ∞, we’ll use the M-L Inequality. First, observe that the length of C2 is Observe also that, if z C2, we can set z = R eiθ – hence, hence

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**M 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.

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Example 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.

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**Theorem 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) and Then where a is a positive real constant. Proof: Wikipedia (non-examinable)

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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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**Close 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, C2 C1 then use the Residue Theorem to calculate Clearly,

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**It follows from Jordan’s Lemma with**

that Hence,

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Comment: 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 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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**3. Integrals involving functions with branch points**

We’ll first calculate some important integrals without branch points. Example 3: Show that where t > 0 is a real parameter. Hint: consider and change to polar coordinates (r, φ): p = r cos , q = r sin .

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

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**The 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).

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Example 6: Find where 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...

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**The integrand is analytic inside the contour – hence,**

(5) Next, let R → ∞, hence and also (due to Jordan’s Lemma, alt. form.) Let ε → 0, hence (why?)

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**Thus, the limiting version of (5) is**

hence, Introduce p > 0: Recalling how the branch of s1/2 was fixed, we have

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Then, 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

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**Finally, 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.

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