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

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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)

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...

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

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

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.

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.

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)

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.

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,

It follows from Jordan’s Lemma with
that Hence,

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).

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 .

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.

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).

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...

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?)

Thus, the limiting version of (5) is
hence, Introduce p > 0: Recalling how the branch of s1/2 was fixed, we have

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

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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