Why is the pole of $\frac{x}{\sinh(x)}$ a simple pole and not a removable singularity?

Question

I'm doing my homework for my Complex Analysis class and I'm asked to solve a definite integral of $\frac{xdx}{\sinh(x)}$. This is not a problem for me, however the problem tells us directly that there is a simple pole at $i\pi$ and I want to know why with certainty. I understand that $\sinh(x)$ is in the denominator and a value of $i\pi$ for $x$ will make the denominator zero however, the $x$ in the numerator makes me skeptical of whether this is a real singularity or if it is removable.

Additionally I thought to rewrite the function as $xcsch(x)$ and by expanding $csch(x)$ with a power series of $x^{-1} - \frac {x} {6} +\frac {7x^3} {360} -\frac {31x^5} {15120} + \cdots = x^{-1} + \sum_{n=1}^\infty \frac{ 2 (1-2^{2n-1}) B_{2n} x^{2n-1}}{(2n)!}$. This would make the function $1 - \frac {x^2} {6} +\frac {7x^4} {360} -\frac {31x^6} {15120} + \cdots$. In general, I believe my book but would like to understand for myself and am failing to understand why.

Answer 1

You can see it, for example, by shifting the function to $0$ setting $x=w+i\pi$.

So, using the addition theorem $\sinh(x+y)= \sinh x\cosh y + \cosh x \sinh y$ you get

$$g(w) = \frac{w+i\pi}{\sinh (w+i\pi)}= \frac{w+i\pi}{-\sinh w}=-\frac{w}{\sinh w}-\frac{i\pi}{\sinh w}$$

Obviously $g$ has a simple pole at $w=0$, hence the function we started with has a simple pole at $x=i\pi$.

Answer 2

It must be a pole, because at $x = i\pi$ you get

$$\frac{x}{\sinh x}\ ``=" \frac{i\pi}{0}$$

and with a nonzero numerator, this cannot be a removeable singularity.

Answer 3

Review the basic definition of poles. If $f(x)=(x-x_0)^ng(x)$ where $g$ is a non-vanishing holomorphic function, then $f$ is a holomorphic function near $x_0$ and $1/f$ has a pole at $x_0$.

Back to the question, $(e^x-e^{-x})/2x$ can be expressed by a $1+x^2/3!+\cdots$, so $2x/(e^x-x^{-x})$ has a pole.

Finally, we note that when $x=n\pi i$, $\sinh(x)=0$, hence $n\pi i$ are poles.

Answer 4

1. All the complex zeroes of $f(z)=\sin(z)$ are simple by the Pythagorean theorem, $f(z)^2+f'(z)^2=1$;
2. From the previous point all the zeroes of $g(z)=\sinh(z)=-i\sin(iz)$ are simple;
3. From the previous point all the poles of $\frac{z}{\sinh z}$, located at $E=\pi i\mathbb{Z}\setminus\{0\}$, are simple.

Of course $\pi i$ is not a removable singularity since $\sinh(\pi i)=0$ but $\pi i\neq 0$.

If you already know Weierstrass products, $$ \frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right) $$ implies $$ \frac{z}{\sinh z}=\prod_{n\geq 1}\left(1+\frac{z^2}{n^2\pi^2}\right)^{-1}$$ for any $z\in\mathbb{C}\setminus E$. De l'Hopital's theorem also gives $$\operatorname*{Res}_{z=\pi n i}\frac{z}{\sinh z}=\lim_{z\to \pi n i}\frac{z}{\cosh z} = \frac{\pi n i}{\cos(\pi n)} = (-1)^n \pi n i.$$