Update Math416_L25.md

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Zheyuan Wu
2025-04-17 12:50:16 -05:00
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@@ -93,3 +93,56 @@ $$
$$
QED
## Application ot valuating definite integrals
Idea:
It is easy to evaluate intervals around closed contours.
Choose contour so one side (where you want to integrate).
Handle the other side by:
- Symmetry
- length * supremum of absolute value of integrand
- Bound function by another function whose integral goes to zero.
Example:
Evaluate $\int_0^\infty \frac{\sin x}{x}dx$.
On the contour $\gamma(t)$ be the semicircle in the upper half plane removed the origin.
Then let $f(z)=\frac{e^{iz}}{z}=\frac{\cos z+i\sin z}{z}$, by the Cauchy's theorem,
$$
\int_\gamma f(z)dz=0
$$
So $\frac{\sin z}{z}=0$ on $\gamma$.
If $x\in \mathbb{R}$, $f(x)=\frac{e^{ix}}{x}=\frac{\cos x+i\sin x}{x}$.
On the real axis,
$$
\begin{aligned}
\int_{-R}^{-\epsilon}+\int_\epsilon^R f(x)dx&=\int_{-R}^{-\epsilon}\frac{e^{ix}}{x}dx+\int_\epsilon^R \frac{e^{ix}}{x}dx\\
&=\int_{-R}^{-\epsilon}\frac{\cos x+i\sin x}{x}dx+\int_\epsilon^R \frac{\cos x+i\sin x}{x}dx\\
&=\int_{-R}^{-\epsilon}\frac{\cos x}{x}dx+i\int_{-R}^{-\epsilon}\frac{\sin x}{x}dx+\int_\epsilon^R \frac{\cos x}{x}dx+i\int_\epsilon^R \frac{\sin x}{x}dx\\
&=2i\int_0^\infty \frac{\sin x}{x}dx
\end{aligned}
$$
For the clockwise semi-circle around the origin,
$$
\int_{S_\epsilon} f(z)dz=\int_{S_\epsilon}\frac{e^{iz}}{z}dz
$$
let $\gamma(t)=\epsilon e^{-it}$, $t\in[-\pi,0]$.
Then $\gamma'(t)=-i\epsilon e^{-it}$,
CONTINUE NEXT TIME.