Estimate $ \int_0^x \frac{1}{\sqrt{\cosh(x)-\cosh(y)}} dy$

Question

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

Answer 1

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,

$$\begin{align}\int_0^x\frac1{\sqrt{\cosh(x)-\cosh(y)}}\ dy&\le\frac1{\sqrt2}\sqrt{\operatorname{csch}(x/2)}\int_0^x\frac1{\sqrt{\sinh\left(\frac{x-y}2\right)}}\ dy\\&\le\frac1{\sqrt2}\sqrt{\operatorname{csch}(x/2)}\int_0^x\frac{\sqrt2}{\sqrt{x-y}}\ dy\\&=2\sqrt{x\operatorname{csch}(x/2)}\\&<\infty\end{align}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{\sinh\pars{y} = \sinh\pars{x/2}\sin\pars{t} \iff t = \arcsin\pars{\sinh\pars{y}/\sinh\pars{x/2}}}$.

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

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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.