The estimation for $\int_{\tau}^\infty \frac{e^{t({1/2-\epsilon})}}{\sqrt{\cosh(t)-\cosh(\tau)}}dt$ I wish to get the following estimate for the integral
$$\int_{\tau}^\infty \frac{e^{t({1/2-\epsilon})}}{\sqrt{\cosh(t)-\cosh(\tau)}}dt \le C_\epsilon (1+|\log(\tau)|),$$
as claimed here in the proof of Lemma 1.
The key part is the behavior as $\tau\to 0^+$ and the integration from $\tau$ to $\tau +1$(the integration from $\tau+1$ to $\infty$ is always bounded by a constant). I have tried doing the Taylor series expansion of the square of denominator, which gives:
$$\sinh(\tau)(t-\tau)+\cosh(\tau)(t-\tau)^2/2...$$
but I don't see how $|\log(\tau)|$ comes out from here.
 A: First of all, we have
$$\frac{\sinh(h)}{h} = \sum_{n=0}^\infty \frac{h^{2n}}{(2n+1)!} \ge 1$$
for all $h \in \mathbb{R}$. Now, using the fundamental theorem of calculus, we find that
$$\cosh(t)-\cosh(\tau) = \int_{\tau}^t \sinh(h) \, dh \geq \int_{\tau}^{t} h \, d h = \frac{1}{2} (t^2-\tau^2).$$
Next, we consider the integral over the subregion $[\tau,\tau+1]$: Inserting the latter inequality we obtain
$$
\int_{\tau}^{\tau+1} \frac{e^{t(1/2-\varepsilon)}}{\sqrt{\cosh(t)-\cosh(\tau)}} dt \leq 2^{1/2} e^{(\tau+1)/2} \int_{\tau}^{\tau+1} \frac{1}{\sqrt{t^2-\tau^2}} dt
$$
To evalute the ingegral, we note that $f(t) := \sqrt{t^2 - \tau^2}$ is a bijective mapping from $[\tau,\tau+1]$ onto $[0,\sqrt{2\tau+1}]$ and hence
\begin{align*}
\int_{\tau}^{\tau+1} \frac{1}{\sqrt{t^2-\tau^2}} dt &= \int_0^{\sqrt{2 \tau+1}} \frac{1}{\sqrt{s^2+\tau^2}} ds \\
&= \int_0^{\sqrt{2 \tau+1}/\tau} \frac{1}{\sqrt{s^2+1}} ds = \mathrm{arsinh} \bigg( \frac{\sqrt{2\tau+1}}{\tau} \bigg).
\end{align*}
To end this, we recall that
$$\mathrm{arsinh}(x)= \log \left(x+ \sqrt{x^2+1}\right)$$
and thus
$$ \mathrm{arsinh}\bigg( \frac{\sqrt{2\tau+1}}{\tau} \bigg) = \log \bigg( \sqrt{2 \tau^{-1} + \tau^{-2}} + 1 + \tau^{-1}\bigg) \ll |\log{\tau}|,$$
as can be easily checked. (Here we use Vingroadov's notation $A \ll B$ meaning that $A \le c B$ for some $c>0$.) For $t \geq 1 + \tau$ one has $\sinh(h)/e^h \ge C:= (1-e^{-2})/2$ and therefore
$$\cosh(t)-\cosh(\tau) = \int_{\tau}^t \sinh(h) \, dh \geq C( e^{t} - e^{\tau}),$$
Hence, we get
$$\int_{\tau+1}^\infty \frac{e^{t(1/2-\varepsilon)}}{\sqrt{\cosh(t)-\cosh(\tau)}} dt  \le \frac{1}{\sqrt{C}} \int_{\tau+1}^\infty \frac{e^{-t\varepsilon}}{\sqrt{1-e^{\tau-t}}} dt  \ll \frac{e^{-\varepsilon(\tau+1)}}{\varepsilon}.$$
Taking both together, we obtain that
$$\tag{1}\int_{\tau}^{\infty} \frac{e^{t(1/2-\varepsilon)}}{\sqrt{\cosh(t)-\cosh(\tau)}} dt  \leq C_\varepsilon (1+ |\log\tau|)$$
for all, say, $\tau \leq 1$.
If $\tau \ge 1$, then the second inequality holds here as well. For the first region we have $t \ge 1$ and thus $\cosh(h) - \cosh(\tau) \gg e^t - e^\tau$. So, we can apply the second argument in the form
$$
\int_{\tau}^{\tau+1} \frac{e^{t(1/2-\varepsilon)}}{\sqrt{\cosh(t)-\cosh(\tau)}} dt \ll \int_\tau^{\tau+1} \frac{e^{t(1/2-\varepsilon)}}{\sqrt{e^t - e^\tau}} dt \leq \int_\tau^{\tau+1} \frac{1}{\sqrt{1- e^{\tau-t}}} dt
$$
and this can further rewritten by
$$\int_0^1 \frac{1}{\sqrt{1-e^{-t}}} dt = \int_{e^{-1}}^1 \frac{1}{x \sqrt{1-x}} dx \ll \int_{e^{-1}}^1 \frac{1}{\sqrt{1-x}} dx \ll 1.$$
Thus, the inequality (1) holds for all $\tau >0$ as claimed.
