Update Math416_L25.md
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$$
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$$
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QED
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QED
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## Application ot valuating definite integrals
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Idea:
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It is easy to evaluate intervals around closed contours.
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Choose contour so one side (where you want to integrate).
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Handle the other side by:
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- Symmetry
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- length * supremum of absolute value of integrand
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- Bound function by another function whose integral goes to zero.
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Example:
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Evaluate $\int_0^\infty \frac{\sin x}{x}dx$.
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On the contour $\gamma(t)$ be the semicircle in the upper half plane removed the origin.
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Then let $f(z)=\frac{e^{iz}}{z}=\frac{\cos z+i\sin z}{z}$, by the Cauchy's theorem,
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$$
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\int_\gamma f(z)dz=0
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$$
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So $\frac{\sin z}{z}=0$ on $\gamma$.
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If $x\in \mathbb{R}$, $f(x)=\frac{e^{ix}}{x}=\frac{\cos x+i\sin x}{x}$.
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On the real axis,
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$$
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\begin{aligned}
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\int_{-R}^{-\epsilon}+\int_\epsilon^R f(x)dx&=\int_{-R}^{-\epsilon}\frac{e^{ix}}{x}dx+\int_\epsilon^R \frac{e^{ix}}{x}dx\\
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&=\int_{-R}^{-\epsilon}\frac{\cos x+i\sin x}{x}dx+\int_\epsilon^R \frac{\cos x+i\sin x}{x}dx\\
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&=\int_{-R}^{-\epsilon}\frac{\cos x}{x}dx+i\int_{-R}^{-\epsilon}\frac{\sin x}{x}dx+\int_\epsilon^R \frac{\cos x}{x}dx+i\int_\epsilon^R \frac{\sin x}{x}dx\\
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&=2i\int_0^\infty \frac{\sin x}{x}dx
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\end{aligned}
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$$
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For the clockwise semi-circle around the origin,
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$$
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\int_{S_\epsilon} f(z)dz=\int_{S_\epsilon}\frac{e^{iz}}{z}dz
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$$
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let $\gamma(t)=\epsilon e^{-it}$, $t\in[-\pi,0]$.
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Then $\gamma'(t)=-i\epsilon e^{-it}$,
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CONTINUE NEXT TIME.
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