# Complex Analysis: Contour Integral

1. Apr 24, 2015

### nateHI

Here's a link to a professor's notes on a contour integration example.
https://math.nyu.edu/faculty/childres/lec12.pdf

I don't understand where the $e^{i\pi /2} I$ comes from in the first problem. It seems like it should be $e^{i\pi}$ instead since $-C_3$ and $C_1$ are both on the real line.

2. Apr 24, 2015

### Hawkeye18

Judging from the picture in the notes contours $C_1$ and $C_3$ should be interchanged in the calculations, so the limit of the integral over $C_3$ gives us $I$. Since the contour of integration includes half-circles in the upper half plane, a branch cut in the definition of $\sqrt z$ should be somewhere in the lower half-plane, so for $z$ with $\operatorname{Re}z\ge 0$ you have $0\le \operatorname{Arg}z \le\pi$.

In particular, $\sqrt{-x}= i \sqrt x$ for $x>0$.

Using this you can just make the change the variables $x=-z$ in the integral: $$\int_{-R}^{-\varepsilon} \frac{dz}{\sqrt z (z^2+1)} = \int_{R}^{\varepsilon} \frac{d(-x)}{\sqrt{-x} (x^2+1)} = \int^{R}_{\varepsilon} \frac{d x}{i \sqrt{x} (x^2+1)} .$$ As $R\to \infty$ and $\varepsilon \to 0^+$ the integral converges to $\frac1i I$, which agrees with the notes (I computed integral over $C_1$, and in the notes the integral over $-C_1$ was computed).

3. Apr 24, 2015

### nateHI

Cool thanks. Makes sense now.