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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?

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    What standard are you using for "far less than"?2011-04-23
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    $f \ll g$ is an equivalent (now less common, I believe) notation for $f = o(g)$.2011-04-23

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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.

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    Thanks, but I'm having a hard time understanding your way. Why the contribution to the integral is $C{1 \over 1 - r}* 1 - r < C$? Where is the second $1-r$ comes from? And for the second consequence - why is the distance is from $z=r$ to $z = r^{i\theta}$ and how the integral works out to the expression $O({\ln (1 - r) \over r}) $? Sorry for the flood of questions.2011-04-24
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    Second $1-r$ is the size of the domain of integration, first is the magnitude of the integrand. For the second one, i'll edit the answer...2011-04-24
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    Need some more clarification, if possible. When integrating by $|\theta| > 1-r$ it seems like the integration from 1-r to $2\pi$ and then it's not $ln(1-r)$. And I didn't succeed in converting $1-e^{-2\pi y}$ to y using the Taylor series of $e^x$. Thanks in advance.2011-04-24
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    View $\theta$ as going from $-\pi$ to $\pi$ here.. anyhow it's a HW so I should not be giving too many hints.2011-04-24
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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 $$Q\leq -{1\over2}e^{\pi y}\log(2\sinh{\pi y\over2})\ .$$