# Evaluating the integral $\int_C \text{Re }z\,dz$ from $-4$ to $4$ via lower half of the circle

I want to evaluate the integral $\int_C \text{Re }z\,dz$ from $-4$ to $4$ via the contour being the lower half of the circle of radius $4$ centered at the origin.

So I can take:

$$z=4e^{i\theta},\quad\pi\leq\theta\leq2\pi$$ $$f(z)=4\cos(\theta),z'(\theta)=4ie^{i\theta}$$ $$\int_\pi^{2\pi}4\cos(\theta)4ie^{i\theta} \,d\theta$$

I then take integration by parts $$I=16\left(\left.-e^{i\theta}\cos\theta\right|_\pi^{2\pi} -\int_{\pi}^{2\pi}ie^{i\theta}\sin(\theta)d\theta\right)$$ $$I=16(-e^{i2\pi}-e^{i\theta}+I)\implies I=0$$

I got a different result via different contours (-4 to -4-4i to 4-4i to 4) ($32i$) and was wondering if I had done the above wrong. I know that the contour choice can effect the result, but I doubt $0$ is correct.

You are right that different contour would possibly give you different result in this case, since the function is not holomorphic.

However, the integration by parts wouldn't work in this case.

You should separate the real part and imaginary part:

$$I=16i\int^{2\pi}_{\pi} (\cos\theta+i\sin\theta)\cos\theta d\theta\\ =16i\int^{2\pi}_{\pi}\cos^2\theta d\theta -16 \int^{2\pi}_{\pi}\cos \theta\sin \theta d\theta$$

I believe you can continue from here.

• is the answer $-8\pi$? – Skies burn May 9 '15 at 12:02
• @SkiesBurn: Should be $8\pi i$. – KittyL May 9 '15 at 18:14
• Oh yes I got that now, and just one more question, sorry if this is a hassle, if I was to take the contour of the top half of the circle, I would take $4e^{i\theta},-\pi \leq \theta \leq 0$ right? And this would give me the same result($8\pi i$)? – Skies burn May 10 '15 at 2:22
• @SkiesBurn: It depends on the direction. If you go counterclockwise, it should be $0\leq \theta \leq \pi$. The result is $8\pi i$. If you go clockwise, it should be $\pi \rightarrow 0$, $-8\pi i$. $-\pi \rightarrow 0$ is still the bottom half. – KittyL May 10 '15 at 9:10
• Oh yes true, but how do I write $\pi\to0$ parametrically? I can't really write it in a $a\leq \theta \leq b$ way can I? – Skies burn May 10 '15 at 11:50