# Evaluating improper integral using Residue

My book claims that if a function $f(z)$ such that $z f(z) \rightarrow 0$ uniformly whenever $z \rightarrow 0$ then the line integral along the semi-circular path is zero as $R \rightarrow \infty$. $$\lim\limits_{R \rightarrow \infty} \int_c f(z) = 0$$ The same thing is claimed in here on Wikipedia. I don't understand this particular step

$$M_R:=\max_{\theta\in [0,\pi]} \bigl|g \bigl(R e^{i \theta}\bigr)\bigr| \to 0\quad \mbox{as } R \to \infty\,,\qquad(*)$$ How does the Max value go to zero? And what does $e^{iaz}$ has to do with $zf(z)$ from my book.

-
I assume you mean $z\to\infty$ in the first line? –  mrf Aug 27 '12 at 12:12

First, in Wiki there's a confussion between $\,g\,\,\,and\,\,\,f\,$ in that part, second what you ask is an assumption: if that happens, then...the next line.
Finally, it is the same $$|z|=R\Longleftrightarrow z=Re^{i\theta}\,\,,\,\text{for some }\,\ \theta\in[0,2\pi)$$
can you explain me how the Max value of $g(z)$ is zero? –  Monkey D. Luffy Aug 27 '12 at 8:33
It is not: if the limit of that maximum, when $\,R\to\infty\,$ is zero, then... For example, take on the semicircle $\,\gamma_R:=\{ z=Re^{i\theta}\;\;|\;\;\,0\leq \theta\leq \pi\}\,$ , the function $$f(z):=\frac{e^{iz}}{z^2}\Longrightarrow max_\gamma|f(z)|\cdot \mathcal l(\gamma)=\frac{e^{-R\sin\theta}}{R^2}\cdot \pi R\xrightarrow [R\to\infty]{}0$$since $\,iRe^{i\theta}=iR(\cos\theta+i\sin\theta)=-R\sin\theta+iR\cos\theta\,$ , and $\,|e^{ix}|=1\,\,,\,for\,\,\,x\in\Bbb R\,$ –  DonAntonio Aug 27 '12 at 8:39