Complex inequality $\int_0^1\frac{1}{|1-e^{2\pi iz}|}dx\leq c\log \frac{1}{y}$

Question

Let $z=x+yi$ with $0<y<0.1$, show that $$\int_0^1\frac{1}{|1-e^{2\pi iz}|}dx\leq c\log \frac{1}{y}$$ for some constant $c>0$.

Answer 1

The wanted integral is $$ \int_{0}^{1}\frac{dx}{\sqrt{(1-e^{-2\pi y}e^{2\pi i x})(1-e^{-2\pi y}e^{-2\pi i x})}}=\int_{0}^{1}\frac{dx}{\sqrt{1-2e^{-2\pi y}\cos x+e^{-4\pi y}}} $$ or $$ \int_{0}^{1}\frac{dx}{\sqrt{(1-e^{-2\pi y})^2+4e^{-2\pi y}\sin^2\frac{x}{2}}}= \frac{e^{\pi y}}{2}\int_{0}^{1/2}\frac{dx}{\sqrt{\sinh^2(\pi y)+\sin^2(x)}}$$ which is bounded by

$$ \frac{e^{\pi y}}{2}\int_{0}^{1/2}\frac{dx}{\sqrt{\sinh^2(\pi y)+4x^2\sin^2\frac{1}{2}}}=\frac{e^{\pi y}}{4}\int_{0}^{1}\frac{dx}{\sqrt{\sinh^2(\pi y)+x^2\sin^2\frac{1}{2}}}. $$

It is not difficult to compute an explicit form for the last integral and to check it is $O(-\log y)$ as wanted.
In a brutal way, the QM-AM (aka Cauchy-Schwarz) inequality implies

$$ \begin{eqnarray*}\int_{0}^{1}\frac{dx}{\sqrt{\sinh^2(\pi y)+x^2\sin^2\frac{1}{2}}}&\leq&\sqrt{2}\int_{0}^{1}\frac{dx}{\sinh(\pi y)+x\sin\frac{1}{2}}\\&=&\frac{\sqrt{2}}{\sin\frac{1}{2}}\left[-\log\sinh(\pi y)+\log\left(\sin\tfrac{1}{2}+\sinh(\pi y)\right)\right]\\&=&O(-\log y)\end{eqnarray*} $$ and for $c=3$ the wanted inequality is fulfilled.
Finding the optimal value for the $c$ constant might require a bit of extra work.