# Complex Integral Without Using Residue Theorems or Cauchy's Theorem $\int_{\gamma} \frac{1}{(z-a)(z-b)} dz$

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....

• 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. Dec 31, 2020 at 16:14
• @Infinity_hunter Aren't you using Cauchy's theorem? Dec 31, 2020 at 16:38
• Oh Yes, in disguise; but I'm not using residues. Atleast something is better than nothing. Dec 31, 2020 at 16:56

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.