Mittag Leffler partial fraction expansion for $\frac{\sin(z)}{\sin(\pi z)}$

Question

I want to prove the following identity: $$\dfrac{\sin(z)}{\sin(\pi z)} = \dfrac{1}{\pi} + \dfrac{z}{\pi}\sum_{n \neq 0}(-1)^n \dfrac{\sin(n)}{n(z-n)} $$

using Mittag-Leffler. I'm able to show that we have $$\dfrac{\sin(z)}{\sin(\pi z)} = \dfrac{z}{\pi}\sum_{n \neq 0}(-1)^n \dfrac{\sin(n)}{n(z-n)} +g(z)$$ whee $g(z)$ is entire, so the goal would be to show that $g$ is bounded, hence constant. I don't see any periodicity conditions here that would help me.

what kind of observations should I make here?

Answer 1

Look at the limits as $\Im(z) \to \pm \infty$. Since $\pi>1$, the left-hand side tends to zero. If you can show that the sum on the right-hand side is bounded as $\Im(z) \to \pm \infty$, then $g$ is entire and bounded away from the real line, so must be constant.

Answer 2

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Simple poles of $\ds{\sin\pars{z}/\sin\pars{\pi z}}$ are at $\ds{n \in \mathbb{Z}\setminus\braces{0}}$ with correspondent residues $\ds{r_{n} = \lim_{z \to n}\bracks{\pars{z - n}\sin\pars{z}/\sin\pars{\pi z}} = \pars{-1}^{n}\,\sin\pars{n}/\pi}$. Moreover, $\ds{\sum_{{\large n = -\infty} \atop {\large n \not= 0}}^{\infty}{r_{n}/n^{2}}}$ is convergent

such that its Mittag-Leffler Expansion becomes \begin{align} {\sin\pars{z} \over \sin\pars{\pi z}} - {1 \over \pi} & = \sum_{{\large n = -\infty} \atop {\large n \not= 0}}^{\infty} r_{n}\pars{{1 \over z - n} + {1 \over n}} \\[5mm] {\sin\pars{z} \over \sin\pars{\pi z}} & = {1 \over \pi} + {1 \over \pi} \sum_{{\large n = -\infty} \atop {\large n \not= 0}}^{\infty} \pars{-1}^{n}\,{\sin\pars{n} \over z - n}\ +\ \underbrace{{1 \over \pi}\sum_{{\large n = -\infty} \atop {\large n \not= 0}}^{\infty} \pars{-1}^{n}\,{\mrm{sinc}\pars{n}}}_{\ds{\equiv \,\mrm{g}\pars{z} = \color{red}{\large 0}}} \end{align}

Answer 3

$$f(z)= \dfrac{\sin(z)}{\sin(\pi z)} -\frac1\pi \lim_{N\to \infty}\sum_{0<|n|\le N}(-1)^n \dfrac{\sin(n)}{z-n}=g(z)-h(z)$$ is entire.

Differentiating, $g'$ is bounded away from its poles, the series for $h'$ is uniformly convergent thus bounded away from its poles, and the boundedness of $f'$ on $|z-n|=1/2$ plus the Cauchy integral formula implies that $f'$ is bounded on $|z-n|<1/3$, thus constant.

Since $g'(i\infty)=h'(i\infty)=0$ it means that $f' = 0$.

Thus $f$ is constant ie. $$f = f(0) =\frac1\pi $$

The general Mittag-Leffler theorem is mostly the abstraction of this reasonning.