Prove that $$\lim_{\rho\to0}\int_{C_\rho} \frac{z^{1/3}\log z}{z^2+1}dz=0$$ $(|z|>0, -\pi/2<\arg z<3\pi/2)$

where $C_\rho$ is the upper half of the circle with radius $\rho<1$ centered at the origin oriented clockwise.

My attempt:

Along $C_\rho$,


$\displaystyle\left|\frac{z^{1/3}\log z}{z^2+1}\right|=\frac{\rho^{1/3}|\log z|}{|z^2+1|}\le \frac{\rho^{1/3}|\log z|}{1-\rho^2}$

I need to find an upper bound $M$ for the above expression in terms of $\rho$, so that I can use

$\displaystyle\left|\int_{C_\rho} \frac{z^{1/3}\log z}{z^2+1}dz\right|\le M\pi\rho$

and then prove that $$\lim_{\rho\to0}M\pi\rho=0$$

  • $\begingroup$ What branch of the logarithm are you using? $\endgroup$ – Michael Joyce Apr 15 '16 at 1:44
  • $\begingroup$ @MichaelJoyce Thanks, I added the branch. $\endgroup$ – Thomas Apr 15 '16 at 1:48

Let $f(z)=\frac{z^{1/3}\log(z)}{z^2+1}$ and choose the branch of the logarithm for which

$$-\pi/2 \le \arg(z) < 3\pi/2$$

Note that on $C_{\rho}$, $z=\rho e^{i\phi}$ where $\phi$ begins at $\pi$ and ends at $0$. Then, we can write

$$\begin{align} \left|\int_{C_{\rho}}\frac{z^{1/3}\log(z)}{z^2+1}\,dz\right|&=\left|\int_\pi^0\frac{\rho^{1/3}e^{i\phi/3}(\log(\rho)+i\phi)}{\rho^2e^{i2\phi}+1}\,i\rho e^{i\phi}\,d\phi\right|\\\\ &\le \rho^{4/3} \int_0^\pi \frac{\log^2(\rho)+\phi^2}{\sqrt{\rho^4+2\rho^2\cos(2\phi)+1}}\,d\phi\\\\ &\le \frac{\pi \rho^{4/3}(\log^2(\rho)+\pi^2)}{1-\rho^2} \tag 1 \end{align}$$

Now, it is evident that $\lim_{\rho \to 0}\frac{\pi \rho^{4/3}(\log^2(\rho)+\pi^2)}{1-\rho^2} =0$. If we need to find a number $M$ such that

$\frac{\pi \rho^{4/3}(\log^2(\rho)+\pi^2)}{1-\rho^2} \le M \pi \rho$

we must first restrict $\rho$. We arbitrarily restrict $\rho \le 1/2$. Then, we have

$$\frac{\pi \rho^{4/3}(\log^2(\rho)+\pi^2)}{1-\rho^2} \le \left(\frac{(1/2)^{1/3}(\log^2(2)+\pi^2)}{3/4}\right)\,\pi \rho$$

Therefore, for $\rho\le 1/2$, we can write

$$\frac{\pi \rho^{4/3}(\log^2(\rho)+\pi^2)}{1-\rho^2} \le M\,\pi \rho$$

where $M$ is given by

$$M=\frac{2^{5/3}}{3} (\log^2(2)+\pi^2)$$


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