MATH3401
Complex Analysis
Lecture 35
Application: contour integrals
Suppose \(f\) is even and "nice" on \(\R\). Our aim is to evaluate \(\ds\int_{-\infty}^{\infty}f\).
Applications contour integrals
Suppose \(f\) is analytic in and on \(C = \Gamma_1+ \Gamma_2\), except maybe for isolated singularities in \(\text{Int}(C)\).
\(\ds\int_C f = \int_{\Gamma_1}f + \int_{\Gamma_2}f\)
\(\;\;\qquad(\text{I})\) \(\qquad (\text{II})\)
LHS: hopefully can evaluate via residue theorem. Then let \(R \ra \infty\).
Applications contour integrals
\((\text{II}) \ra\) \(\text{PV}\ds\int_{-\infty}^{\infty}f = \int_{-\infty}^{\infty}f\) since \(f\) is even.
Only remains to deal with \(\ds\lim_{R\ra \infty}\int_{\Gamma_1}f\).
Hopefully = 0, e. g. via \(M\)-\(\ell\) estimate.
Applications contour integrals
Example 1: Evaluate \(I = \ds \int_{0}^{\infty}\frac{x^2}{x^6+1}dx\).
Note \(f\) is even and continuous.
Also \(f \backsim 1/x^4\) as \(x\ra
\pm\infty\).
So \(I\) converges (\(p\)-test, \(p\gt
1\)).
Note \(f(z)=\dfrac{z^2}{z^6+1}\) is analytic on \(\C\), except for the 6 zeros of the denominator \(z^6+1\).
Applications contour integrals
Example (cont): \(f\) is analytic in and on \(C= \Gamma_1 + \Gamma_2\), except for the 3 zeros of \(z^6+1\) in the UHP.
Applications contour integrals
Example 1 (cont): Note that \(z_1 = e^{\pi i/6}\), \(z_2 = i\), \(z_3 = e^{5\pi i/6}\).
Residue Theorem \(\Rightarrow \) \[\int_C f= 2 \pi i \sum_{j=1}^{3}\underset{z=z_j}{\res} f(z)\qquad (1)\]
Also note \(f\) has the form \(p/q\): for each \(z_j\), there holds: \[p(z_i)\neq 0, q(z_j)=0, q'(z_i)\neq 0.\]
Applications contour integrals
Example 1 (cont): Thus we have simple poles at each \(z_j\) and so Theorem 3 (Lec 34) implies
\(\text{RHS} = 2 \pi i \ds \sum_{j=1}^{3} \left.\frac{z^2}{(z^6+1)^{\prime}}\right|_{z=z_j}\) \(= 2 \pi i \ds \sum_{j=1}^{3} \frac{z_j^2}{6z_j^5}\) \(= 2 \pi i \ds \sum_{j=1}^{3} \frac{1}{6z_j^3}\)
\(= 2 \pi i \left( \ds \frac{1}{6i} -\frac{1}{6i}+ \frac{1}{6i} \right)\) \(= \dfrac{\pi}{3}.\qquad \qquad \qquad\)
Applications contour integrals
Example 1 (cont): Note
\(\ds\int_C f = \int_{\Gamma_1}f + \int_{\Gamma_2}f \quad (2)\)
\(\quad(\alpha)\) \(\qquad (\beta)\)
As \(R\ra \infty\), \((\beta) \ra 2I\).
Claim: \((\beta) \ra 0\) as \(R\ra \infty\).
Applications contour integrals
Example (cont):
Claim: \((\beta) \ra 0\) as \(R\ra \infty\).
Note \(\big|(\alpha)\big| \leq M_R \ell_R\; \,(*)\,\) where \( \ell_R = \text{length } \Gamma_1 = \pi R \)
and \(M_R = \ds\max_{z\in \Gamma_1 } \left| \,f(z) \right|\) \(\leq \ds \max_{|z|=R } \left| \frac{z^2}{z^6+1} \right|\) \(\leq \ds \frac{R^2}{R^6-1} \).
(via reverse triangle inequality: \(|a+b|\geq \big||a|-|b|\big|\)).
Applications contour integrals
Example 1 (cont): So \((*)\) implies
\(\big|(\alpha)\big| \leq \ds \frac{\pi R \cdot R^2}{R^6-1}\) \(=\ds\frac{\pi}{R^3 -\frac{1}{R^3}}\)
this espression \(\ra 0\) as \(R\ra \infty\). So \(\ds\lim_{R\ra \infty}\) in \((2)\) \(\Rightarrow\)
\(\ds \frac{\pi}{3} = 2I + 0 \;\) \( \Rightarrow \; I = \ds\frac{\pi}{6}\).
Applications contour integrals
Example 2: Evaluate \(I = \ds \int_{0}^{\infty}\frac{\sin x}{x}dx\).
Note \(f\) is even, so if \(I\) exists
\(I = \ds \frac{1}{2} \int_{-\infty}^{\infty}\frac{\sin x}{x}dx \) \(= \ds \frac{1}{2} \text{PV}\int_{-\infty}^{\infty}\frac{\sin x}{x}dx \).
Applications contour integrals
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Example 2 (cont): 
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Take
\(f(z)=\dfrac{e^{iz}}{z}\). |
Applications contour integrals
Example 2 (cont): Now \(\ds \int_{\gamma_1 } f + \int_{C_{\rho} } f+ \int_{\gamma_2} f+ \int_{ C_R} f = 0 \quad(*)\)
\(\;\qquad I_1\qquad I_2 \qquad I_3 \qquad I_4\)
\(\qquad\qquad I_1 = \ds \int_{-R}^{-\rho} \frac{e^{ix}}{x}dx \) \(= \ds- \int_{\rho}^{R} \frac{e^{i w}(-1)}{-w}dw \).
\(\qquad\qquad I_3 = \ds \int_{\rho}^{R} \frac{e^{ix}}{x}dx \).
Applications contour integrals
Example 2 (cont): \( \Rightarrow I_1 + I_3 = \ds \int_{\rho}^{R} \frac{e^{ix}- e^{-ix}}{x}dx \)
\( \;\quad\qquad= \ds 2 i \int_{\rho}^{R} \frac{\sin x}{x}dx \)
Need limit as \( \rho \downarrow 0 \) and \(R\ra \infty\).
\(\qquad\)