Estimate $ \int_0^x \frac{1}{\sqrt{\cosh(x)-\cosh(y)}} dy$ Consider the following integral for $x>0$:
$$ \int_0^x \frac{1}{\sqrt{\cosh(x)-\cosh(y)}} dy.$$
I would like to show that this integral is actually convergent. Is it possible to find an integrable upper bound for the function $y\mapsto \frac{1}{\sqrt{\cosh(x)-\cosh(y)}}$? I could not accomplish anything in this direction.
Best wishes
 A: Simple sum/difference of two hyperbolic trig functions:
$$\cosh(x)-\cosh(y)=2\sinh\left(\frac{x+y}2\right)\sinh\left(\frac{x-y}2\right)$$
Thus, the real question is whether or not
$$\int_0^x\frac1{\sqrt{\sinh\left(\frac{x-y}2\right)}}\ dy$$
converges. We may use $\sinh(x)\ge x$ for $x\ge0$ to compare this to
$$\int_0^x\frac1{\sqrt{x-y}}\ dy$$
which converges.  That is,
$$\begin{align}\int_0^x\frac1{\sqrt{\cosh(x)-\cosh(y)}}\ dy&\le\frac1{\sqrt2}\sqrt{\operatorname{csch}(x/2)}\int_0^x\frac1{\sqrt{\sinh\left(\frac{x-y}2\right)}}\ dy\\&\le\frac1{\sqrt2}\sqrt{\operatorname{csch}(x/2)}\int_0^x\frac{\sqrt2}{\sqrt{x-y}}\ dy\\&=2\sqrt{x\operatorname{csch}(x/2)}\\&<\infty\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\int_{0}^{x\ >\ 0}
{\dd y \over \root{\cosh\pars{x} - \cosh\pars{y}}} =
\int_{0}^{x}
{\dd y \over \root{\bracks{2\sinh^{2}\pars{x/2} + 1} - \bracks{2\sinh^{2}\pars{y/2} + 1}}}
\\[5mm] = &\
\root{2}\int_{0}^{x/2}
{\dd y \over \root{\sinh^{2}\pars{x/2} - \sinh^{2}\pars{y}}}
\end{align}

Lets $\ds{\sinh\pars{y} = \sinh\pars{x/2}\sin\pars{t} \iff
t = \arcsin\pars{\sinh\pars{y}/\sinh\pars{x/2}}}$.

$$
\mbox{Note that}\quad
\left\{\begin{array}{l}
\ds{\partiald{y}{t} =
{\sinh\pars{x/2}\cos\pars{t} \over \cosh\pars{y}} =
{\sinh\pars{x/2}\cos\pars{t} \over \root{1 + \sinh^{2}\pars{y}}} =
{\sinh\pars{x/2}\cos\pars{t} \over \root{1 + \sinh^{2}\pars{x/2}\sin^{2}\pars{t}}}}
\\[3mm]
\ds{\root{\sinh^{2}\pars{x \over 2} - \sinh^{2}\pars{y}} = \sinh\pars{x \over 2}\cos\pars{t}}
\\[3mm]
\ds{y = {x \over 2} \implies t = {\pi \over 2}}
\end{array}\right.
$$

Then,
\begin{align}
&\int_{0}^{x}
{\dd y \over \root{\cosh\pars{x} - \cosh\pars{y}}} =
\root{2}\int_{0}^{\pi/2}
{\dd t \over \root{1 + \sinh^{2}\pars{x/2}\sin^{2}\pars{t}}}
\\[5mm] = &\
\root{2}\int_{0}^{\pi/2}
{\dd t \over \root{1 - \bracks{\ic\sinh\pars{x/2}}^{\,2}\sin^{2}\pars{t}}} =
\bbx{\ds{\root{2}\,
\mrm{K}\pars{\ic\sinh\pars{x \over 2}}}}
\end{align}


where $\ds{\mrm{K}}$ is the
  Complete Elliptic Integral of the First Kind.

