Inequality involving log and e I need to show that $\displaystyle{\int_{0}^{1} \frac{dx}{|1-e^{2\pi i\tau}|}} \ll -\log y$ where $\tau = x + iy$ and  $0 < y < \frac{1}{10}$.   
I began showing it by using the lower estimate of triangle inequality, i.e., 
$$
\frac{1}{|1-e^{2\pi i\tau}|} \leq \frac{1}{|1-e^{-2\pi y}|} \> .
$$ 
Then, as the integration is with respect to $x$, it seems that I need to show that the latter is far less than $\log(\frac{1}{y})$.  
Am I doing it right? How do I proceed from here?
 A: What you are doing is integrating ${1 \over |1 - re^{i\theta}|}$ over the unit circle, where $r = e^{-2\pi y}$. Geometrically $|1 - re^{i\theta}|$ is the distance between $z = re^{i\theta}$ and $z = 1$. When $|\theta| < 1 - r $, this distance is $O(1 - r)$, so the contribution to the integral for $|\theta| < 1 - r$ will be $C{1 \over 1 - r}* 1 - r < C$.
When $|\theta| > 1 - r$, the distance is of the same order of magnitude as the distance to
 $z = r$ to $z = r^{i\theta}$. Since $r$ is near 1, your integral is the same order of magnitude as 
$$ \int_{|\theta| > 1 - r}{1 \over |1 - e^{i\theta}|}$$
$$= \int_{|\theta| > 1 - r}\frac{1}{|e^{-i{\theta \over 2}} - e^{i{\theta \over 2}}|}$$
$$= \int_{|\theta| > 1 - r}\frac{1}{2|\sin{\theta \over 2}|}$$
Since $\sin(\theta) \sim \theta$, you are integrating ${1 \over \theta}$ basically and 
your term becomes $O(\ln (1 - r))$. So the overall integral is $O(\ln (1 - r))$. Plugging back in $r = e^{-2\pi y}$ this is $O(\ln (1 - e^{-2\pi y}))$ or $O (\ln(y))$ if you plug in the Taylor expansion. 
If you want to go from this to $<< |\ln(y)|$ instead of $O(\ln(y))$, you can use the fact the first term is a lot smaller than the second. So instead of breaking up at $|\theta| = 1 - r$, break it up at $|\theta| = N(1 - r)$ for some large $N$. You will get a correspondingly smaller constant in front of $|\ln(y)|$ in the second term, while the first term is still much smaller as $y$ gets small. 
A: Note that
$$|1-\exp(2\pi i(x+i y)|^2=4e^{-2\pi y}(\sinh^2(\pi y)+\sin^2(\pi x))\ .$$
Now for $|t|\leq{\pi\over2}$ one has $|\sin t|\geq{2|t|\over\pi}$. It follows that for your integral Q you get the estimate
$$Q\leq {1\over4}e^{\pi y} \int_{-1/2}^{1/2}{dx\over\sqrt{a^2+x^2}}\ , \qquad a:={\sinh(\pi y)\over 4}\ .$$
Here the right side can be done by Mathematica, and after simplification you will obtain the result
$$Q\leq -{1\over2}e^{\pi y}\log(2\sinh{\pi y\over2})\ .$$
