How can I find the value of this line integral? The counter clockwise, curve $C_r = \{z \vert z=re^{i{\theta}}, 0 \leq \theta \leq \pi \}$
For the $f(z) = \frac{5(1+z)}{z(1+z^5)}$, $\lim\limits_{r\to 0+} \int _{C_r} f(z)dz = 5\pi i$. Find the value of the  $\lim\limits_{R\to \infty} \int _{C_R} \frac{5z^3}{1+z+z^2+z^3+z^4}dz$.
Firstly, I put the $w=\frac{1}{z}$
$\lim\limits_{r\to 0+} \int _{C_r} f(z)dz  = \lim\limits_{R\to \infty}\int_{C_R}f({1\over w}){1\over w^2} dw  = \lim\limits_{R\to \infty}\int_{C_R} \frac{5w^3}{w^4-w^3+w^2-w+1} dw$ (Here the $C_R = \{w \vert w=Re^{i{\theta}}, 0 \leq \theta \leq \pi \} $)
Again, Tried to make the form of the $\frac{5z^3}{1+z+z^2+z^3+z^4}$, I took the $t=-w$
Hence, $\lim\limits_{r\to 0+} \int _{C_r} f(z)dz = \lim\limits_{R\to \infty}\int_{C'_R} \frac{5t^3}{t^4+t^3+t^2+t+1} dt$  (The $C'_R = \{t \vert t=Re^{i{\theta}}, \pi \leq \theta \leq 2\pi \} $)
The problem is I can't find any next steps. The answer requires for the case $0\leq \theta \leq \pi$. But My result is $\pi \leq \theta \leq 2\pi$. plus Is my solution is right? I'm stuck in this question.
 A: I'll follow your notation. Your mistake was you got the wrong $C_R$.
$\lim\limits_{r\to0+}\int_{c_r} f(z) dz = \lim\limits_{R\to \infty}\int_{C_R} f(\frac{1}{w}){\frac{1}{w^2}} dw$ (The $C_R = \{w \vert w = Re^{i\theta}, \pi\leq \theta \leq 2\pi\}$) Not your claim "$0$ to $\pi$"
Let me explain reason for you.
Consider the inversion mapping $\phi(z): z \to w$ by $w=1/z$.
Then the $\phi(C_r)$, image of the $C_r$, maps into lower half circle which its radius $1/r$ and direction $2\pi(=0)$ to $\pi$. Remark the $C_r$ was the upper half circle radius $r$ and direction was $0(=2\pi)$ to $\pi$ in the complex plane. And plus substituting the $w=1/z$, Then we get $dz = -\frac{1}{w^2}dw$. Hence because of the "$-$", the $C_R$ has opposite direction comparing the $\phi(C_r)$. In other words, $C_R$ is a lower half circle $\pi$ to $2\pi(=0)$(same direction with the original curve $C_r$)
$\int_{C_{r}}f(z)dz = \int_{\phi(c_r)}f(\frac{1}{w}) \frac{-1}{w^2}dw \ =\int_{C_{R}}f(\frac{1}{w}) \frac{1}{w^2}dw $ (Substitution $w = 1/z$)
You can easily check these by drawing picture in Complex plane yourself.
Second case for the $t=-w$. This mapping $C_R \to -C_R$. Here the $-C_R$ means $C'_R$ in your post. This mapping rotate $\pi$ to the counter-clockwise direction. Hence the $-C_R = \{t \vert t= Re^{i\theta}, 0 \leq \theta \leq \pi\}$ (The upper half circle) Not yours "$\pi$ to $2\pi$" in the $C'_R$
Consequently the final answer is $5\pi i$
I hope my answer would be helpful. :)
