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TL;DR:

How does one work out the below integral (treating $x$ as a constant)?

$$v=\int\frac{2\cosh(2y)\cos(2x)-2}{(\cosh(2y)-\cos(2x))^2}dy$$

Background:

I need to work out the function $f(z)$, analytic in a suitable part of the Argand diagram, for which:

$$\operatorname{Re}f=\frac{\sin(2x)}{\cosh(2y)-\cos(2x)}$$

Of course, the strategy here is to use the Cauchy-Riemann relations to work out the imaginary part of $f$. If we let $u=Ref$ and $v=Imf$:

$$\frac{\partial{u}}{\partial{x}}=\frac{\partial{v}}{\partial{y}} \\ \frac{\partial{u}}{\partial{y}}=-\frac{\partial{v}}{\partial{x}}$$

Now, if we start with the first relation:

$$\frac{\partial{u}}{\partial{x}}=\frac{2\cosh(2y)\cos(2x)-2}{(\cosh(2y)-\cos(2x))^2}=\frac{\partial{v}}{\partial{y}}$$

$$v=\int\frac{2\cosh(2y)\cos(2x)-2}{(\cosh(2y)-\cos(2x))^2}dy$$

How does one integrate this monster? I've tried a few substitutions but haven't managed to get anywhere. If I've made a conceptual error somewhere then please do point it out.

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Start by multiplying both sides by $\sf \sinh(2y)$: $$\sf v=\int{\frac{\color{blue}{\cosh(2y)\cos(2x)-1}}{\color{red}{(\cosh(2y)-\cos(2x))^2}}}\frac{\color{red}{2\sinh (2y)}}{\color{blue}{\sinh(2y)}}dy$$ Our aim is to integrate by parts (starting by integrating the red thing), but notice that $2\sinh(2y)$ is the derivate of $\cosh(2y)$ thus we can rewrite it as: $$\sf v=\int \color{red}{\left(\frac{1}{\cos(2x)-\cosh(2y)}\right)'}\color{blue}{\frac{\cosh(2y)\cos(2x)-1}{\sinh(2y)}}dy$$ $$\sf \require{cancel}\small={\left(\color{red}{\frac{1}{\cos(2x)-\cosh(2y)}}\right)}\color{blue}{\frac{\cosh(2y)\cos(2x)-1}{\sinh(2y)}}+\color{blue}2\int \color{blue}{\frac{\cancel{\cos(2x)-\cosh(2y)}}{\sinh^2(2y)}}\color{red}{\frac{1}{\cancel{\cos(2x)-\cosh(2y)}}}dy$$ $$\sf =\color{red}{\frac{1}{\cos(2x)-\cosh(2y)}}\color{blue}{\frac{\cosh(2y)\cos(2x)-1}{\sinh(2y)}}-\color{blue}{\frac{\cosh(2y)}{\sinh(2y)}}+K(x)$$ $$\sf=\color{purple}{\frac{\cancel{\cosh(2y)\cos(2x)}-1-\cosh(2y)(\cancel{\cos(2x)}-\cosh(2y))}{(\cos(2x)-\cosh(2y))\sinh(2y)}}+K(x)$$ $$\sf =\color{purple}{\frac{\sinh(2y)}{\cos(2x)-\cosh(2y)}}+K(x)$$

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  • $\begingroup$ Perfect, thank you. $\endgroup$ – Pancake_Senpai May 1 at 22:33
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Hint: $$ i\coth \left( ix+y \right)=\frac{\sin \left( 2x \right)}{\cosh \left( 2y \right)-\cos \left( 2x \right)}+i\frac{\sinh \left( 2y \right)}{\cosh \left( 2y \right)-\cos \left( 2x \right)} $$

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