I'm hoping to come up with an analytical solution for the following definite integral:
$$ \int_{-2/y}^{2/y} \cfrac{u}{\coth\left( \frac{1}{u}\right) - i \cot\left( \frac{z}{2} \right)} du$$
Here, $y$ and $z$ are strictly positive real constants. $i$ is the imaginary number $\sqrt{-1}$.
I tried doing this on Mathematica but it's been 20 minutes and it hasn't given me anything yet. I was wondering if anybody had any tricks/tips for solving this integral?
Thanks!
EDIT: The imaginary component can be eliminated as follows:
$$ \int_{-2/y}^{2/y} \cfrac{u}{\coth\left( \frac{1}{u}\right) - i \cot\left( \frac{z}{2} \right)} du = \int_{-2/y}^{2/y} \cfrac{u\coth\left( \frac{1}{u}\right)}{\coth^2\left( \frac{1}{u}\right) + \cot^2\left( \frac{z}{2} \right)} du + i\int_{-2/y}^{2/y} \cfrac{u\cot\left( \frac{z}{2}\right)}{\coth^2\left( \frac{1}{u}\right) + \cot^2\left( \frac{z}{2} \right)} du$$ The second integral is odd, so it evalutes to zero. Thus, $$ \int_{-2/y}^{2/y} \cfrac{u}{\coth\left( \frac{1}{u}\right) - i \cot\left( \frac{z}{2} \right)} du = \int_{-2/y}^{2/y} \cfrac{u\coth\left( \frac{1}{u}\right)}{\coth^2\left( \frac{1}{u}\right) + \cot^2\left( \frac{z}{2} \right)} du $$ However, I don't think this is any easier to solve...
EDIT2: As noted by the comments, it might be a bit nicer if I set $b = 2/y$ and $c = \cot(z/2)$ so that my integral looks like:
$$ \int_{-b}^{b} \cfrac{u}{\coth\left( \frac{1}{u}\right) - i c} du = \int_{-b}^{b} \cfrac{u\coth\left( \frac{1}{u}\right)}{\coth^2\left( \frac{1}{u}\right) + c^2} du $$
Hint
Try use the substitution $$ I = \int\limits_{-2/y}^{2/y} {u\over\coth{1\over u} - i \cot{z\over2}}\ du = -\int\limits_{-2/y}^{2/y} {u^3\over\coth{1\over u} - i \cot{z\over2}}\ d\left({1\over u}\right)$$ $$ = -\int\limits_{-2/y}^{-0} {u^3\over\coth{1\over u} - i \cot{z\over2}}\ d\left({1\over u}\right) -\int\limits_{+0}^{2/y} {u^3\over\coth{1\over u} - i \cot{z\over2}}\ d\left({1\over u}\right) = \int\limits_{-\infty}^{-y/2} {v^{-3}\over\coth{v} - i \cot{z\over2}}\ dv + \int\limits_{y/2}^{\infty} {v^{-3}\over\coth{v} - i \cot{z\over2}}\ dv.$$