# Complex integrals over open semicircular contour

This problem asks about a few integrals over the open curve $C$, which is the bottom half of a circle centered about the origin with radius $\epsilon\rightarrow 0$.

The first integral was $\int_{\tiny{C}} dz\frac{1}{z}$, which I found to be $\pi\,i$ since it should be 1/2 of the same integral if $C$ was a full circle.

The second integral is $\int_{\tiny{C}} dz\frac{1}{z-i}$. I'd like to say this is equal to 0, since the singularity at $i$ is away from $C$, but then again $C$ doesn't enclose anything. Can I do a contour deformation to include $i$? If so, how?

The third integral is $I = \int_{\tiny{C}} dz\frac{1}{z(z-i)}$ Couldn't I just do $I = -\frac{1}{i}\int_{\tiny{C}}dz\frac{1}{z} + \frac{1}{i}\int_{\tiny{C}}dz\frac{1}{z-i}$ and use the answers to the first two integrals?

Thanks a ton.

• Actually, partial fraction expansion is perfectly suitable here. – Mark Viola Nov 23 '16 at 19:53

HINT:

For the integral $\int_C \frac1{z-i}\,dz=\int_{\pi}^{2\pi}\frac1{\epsilon e^{i\phi}-i}\,i\epsilon e^{i\phi}\,d\phi$, what is the length of the contour and what is the maximum of the magnitude of the integrand on $C$ as $\epsilon \to 0$?

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Note that we can write \begin{align}\left|\int_C \frac1{z-i}\,dz\right|&=\left|\int_{\pi}^{2\pi}\frac1{\epsilon e^{i\phi}-i}\,i\epsilon e^{i\phi}\,d\phi\right|\\\\&\le \int_{\pi}^{2\pi}\left|\frac1{\epsilon e^{i\phi}-i}\,i\epsilon e^{i\phi}\right|\,d\phi\\\\&=\int_{\pi}^{2\pi}\frac{\epsilon}{|\epsilon e^{i\phi}-i|}\,d\phi\\\\&\le\int_{\pi}^{2\pi}\frac{\epsilon}{|1-\epsilon|}\,d\phi\\\\&=\frac{\pi\epsilon}{|1-\epsilon|}\end{align}Therefore, $\lim_{\epsilon \to 0}\int_C \frac1{z-i}\,dz=0$.

• The length of the contour is $pi\,\epsilon$ but it goes to 0. And as $\epsilon\rightarrow 0$, wouldn't the maximum magnitude be 0? Is the hint implying that this integral is $=0$? – Spuds Nov 23 '16 at 20:03
• Yes, the second integral is indeed $0$. The maximum magnitude is $\frac{1}{1-\epsilon}$ – Mark Viola Nov 23 '16 at 20:23
• You did help, thank you very much. – Spuds Dec 5 '16 at 0:05
• You're welcome. My pleasure. -Mark – Mark Viola Dec 5 '16 at 0:35
• Of course, sorry I should've done that sooner. – Spuds Dec 5 '16 at 3:01