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Source: https://www.maths.ed.ac.uk/~jmf/Teaching/MT3.html by Jose Figueroa-O'Farrill

Background:

p.v. is Cauchy's principal value

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which can be "complexified" to the form,

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To make use Cauchy's residue theorem, we close the contour with the following curve,

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My confusion:

How the integral over the $C^+_\rho$ part of the contour vanishes...

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Most of my confusion here seems to lie in understanding the referenced equation 2.28, which reads,

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where equation 2.24 reads,

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for completeness, equation 2.36 reads,

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Any pointers are appreciated :(

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    $\begingroup$ Don't you understand why 2.28 holds? Or don't you understand how it is applied here? $\endgroup$ Commented Dec 1, 2019 at 22:48
  • $\begingroup$ Ah of course, thank you for the question. I do not understand how 2.28 holds. I find the step from the 2nd line to the final line to be unclear. Using 2.24 is straight forward, absolute value of each factor of the integrand, followed by dt. I find the pulling of the f(z(t)) out of the integral and this "max" function confusing. Moreover, I am not sure how the integral in the 3rd line is clearly the arc length. And finally, I am confused about why dz appears in the middle term of the boxed inequality line. $\endgroup$
    – Lopey Tall
    Commented Dec 2, 2019 at 10:27
  • $\begingroup$ There is no need to wrote "p.v." since the integral converges absolutely. $\endgroup$
    – GEdgar
    Commented Dec 2, 2019 at 11:36

1 Answer 1

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The inequality (2.28) states that the absolute value of the integral $\int_\gamma f(z)\,\mathrm dz$ of a function $f$ along a path $\gamma\colon[a,b]\longrightarrow\mathbb C$ is smaller than or equal to the product of two numbers:

  • the maximum of $\lvert f\rvert$ restricted to $\gamma\bigl([a,b]\bigr)$;
  • the length of $\gamma$.

Let us now apply this to your problem. The length of $C_\rho^+$ is $\pi\rho$. Furthermore, if $z$ belongs to the semicircle, then $\lvert z^2+4\rvert\geqslant\rho^2-4$ and therefore (if $\rho>1$), $\max\left\lvert\frac1{z^2+4}\right\rvert\leqslant\frac1{\rho^2-4}$. So,$$\left\lvert\int_{C_\rho^+}\frac{\mathrm dz}{z^2+4}\right\rvert\leqslant\frac{\pi\rho}{\rho^2-4}.$$Since $\lim_{\rho\to\infty}\frac{\pi\rho}{\rho^2-4}=0$,$$\lim_{\rho\to\infty}\left\lvert\int_{C_\rho^+}\frac{\mathrm dz}{z^2+4}\right\rvert=0.$$

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