I am trying to modify some work of Henryk Iwaniec involving bounds on the Fourier coefficients of modular forms. There is a bound which he gets that I'm having trouble understanding.
So I want to show that $$ \int_0^1 \vert 1 - e^{2\pi i z} \vert^{-1} dx \ll \log(2 + y^{-1}) $$ where $z = x + iy$ and $y > 0$.
When I write $f(y) \ll g(y)$, I mean that there is some $C$ such that $\vert f(y) \vert \leq C g(y)$ for all sufficiently large $y$. That is, it's a Big Oh notation.
Intuitively, a result of this shape is pretty clear. As $y$ gets large, the term $e^{-2\pi y}$ appearing multiplied by $e^{2\pi i x}$, which has the effect of making the denominator look more and more like a constant.
As a more specific reference, this appears in the proof of Theorem 5.3 in his book Topics in Classical Automorphic Forms.
For first, $$ \left\|1-e^{2\pi i z}\right\| = \left\|1-e^{-2\pi y}\left(\cos(2\pi x)+i\sin(2\pi x)\right) \right\|=\sqrt{(1-e^{-2\pi y})^2+4e^{-2\pi y}\sin^2(\pi x)}$$ hence the integrand function is small if $x$ is not close to the endpoints of $[0,1]$, and can be bounded (by the well-known Taylor series of $\sin(z)$ and $1-e^{-z}$) by something like:
$$\frac{1}{2\pi\sqrt{\min(x^2,(1-x)^2)+y^2}}$$ if $x$ is close to endpoints of $[0,1]$. Notice that: $$ 2\int_{0}^{1/2}\frac{dx}{2\pi\sqrt{x^2+y^2}} = \frac{1}{\pi}\,\log\left(\frac{1+\sqrt{1+4y^2}}{2y}\right)\leq \frac{1}{\pi}\,\log\left(\frac{1+\sqrt{5}}{2}+\frac{1}{y}\right).$$