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I am trying to compute the integral $\int_{\gamma} \frac{1}{(z-a)(z-b)} dz$ without any residue theorems or Cauchy's theorem, where $\gamma$ is a circled center at $0$ of radius $r$ with $|a| < r < |b|$. Using PFD, the integral can be rewritten as

$$\frac{1}{a-b} \left[ \int_{\gamma} \frac{1}{z-a} dz - \int_{\gamma} \frac{1}{z-b} dz \right].$$

If $\gamma$ is parametrized as $\gamma(t) = re^{it}$, then the first integral becomes $$\int_{0}^{2 \pi} \frac{ i r e^{i t}}{re^{it} - a} dt.$$

I'm not sure how to deal with that integral. It looks like the integral of $u'/u$, but that would involve logs and one has to be careful with branch cuts....

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  • $\begingroup$ If $|r| < b$ then $z - b$ will never be $0$ so second integral is simply $0$. For first integral construct a circle centred at $a$ with radius $R$ inside the circle $\gamma$ and use $z-a = Re^{i\theta}$ and principle of deformation of paths. $\endgroup$ Dec 31, 2020 at 16:14
  • $\begingroup$ @Infinity_hunter Aren't you using Cauchy's theorem? $\endgroup$
    – user193319
    Dec 31, 2020 at 16:38
  • $\begingroup$ Oh Yes, in disguise; but I'm not using residues. Atleast something is better than nothing. $\endgroup$ Dec 31, 2020 at 16:56

1 Answer 1

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For $a\in[0,\infty)$ and all $t\in (0,2\pi)$, $re^{it}-a\in\mathbb C\setminus[0,\infty)$, so we can take the branch cut to be at the positive real axis and the integral is $\lim_{t\to 2\pi^-}\log(re^{it}-a)-\lim_{t\to 0^+}\log(re^{it}-a)=2\pi i$. The general case is similar, simply that the branch cut is along a different line.

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