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I want to prove:

$$\frac{1}{\sin z} = \frac{1}{z} + \sum_{n=1}^{\infty} \frac{(-1)^n 2z}{z^2 - \pi^2 n^2}$$

I tried to prove, I got this: $g(z) := \frac{1}{z} + \sum_{n=1}^{\infty} \frac{(-1)^n 2z}{z^2 - \pi^2 n^2} = \sum_{- \infty}^{+ \infty} \frac{(-1)^n}{z+n\pi}$ so $g(z)$ and $h(z) := \frac{1}{\sin z}$ have same pole of same order so $k(z) = h(z)-g(z)$ is holomorphic map. but I can't complete this.

Bernard
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Vahid Shams
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  • Are you allowed to use the infinite product representation of the sine function, namely $$\sin(z)=z\prod_{n=1}^\infty \left(1-\frac{z^2}{n^2\pi^2}\right)$$If so, see THIS – Mark Viola Dec 04 '19 at 18:44
  • Well, $\sin z$ is entire, $2\pi$-periodic, and bounded on $[0,2\pi]$ but not constant. You will need to know also that $k(z)$ has no zeros. – GEdgar Dec 04 '19 at 18:51

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$$k(z)=\frac{1}{\sin z} -\frac{1}{z} - \sum_{n=1}^{\infty} \frac{(-1)^n 2z}{z^2 - \pi^2 n^2}=\frac{1}{\sin z}-\sum_{n=- \infty}^{+ \infty} \frac{(-1)^n}{z+n\pi}$$ is meromorphic, it has no pole so it is entire, it is $2\pi$-periodic, since $k(z)$ is odd then $k(0)=0$ and $$w(z)=\frac{k(z)}{z}=\frac{1}{z\sin z} -\frac{1}{z^2} - \sum_{n=1}^{\infty} \frac{(-1)^n 2}{z^2 - \pi^2 n^2}$$ is entire, by absolute convergence it is bounded on $\Re(z)\in[-\pi,\pi]$, thus $k(z)=O(z)$ on $\Re(z)\in[-\pi,\pi]$, thus $k(z)=O(z)$ on the whole complex plane, thus $w(z)$ is a bounded entire function, thus it is constant $$w(z)=w(2\pi)=\frac{k(2\pi)}{2\pi }=\frac{ k(0)}{2\pi }=0 $$

reuns
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  • I like this but I am curious about one thing - how do you show that $k(0)=0$? The residue is the same so the poles cancel out but how do you show that what remains, cancels out too? Never mind, $k$ is odd! – Conrad Dec 04 '19 at 19:05
  • $k$ is obviously odd - sorry for the silly question – Conrad Dec 04 '19 at 19:08
  • An alternative is to say $1/(z+n)-1/(z+n+1)=\int_n^{n+1} \frac1{(z+t)^2}dt$ from which the boundedness (as $\Im(z)\to \infty$) follows without needing $k(z)/z$ – reuns Dec 04 '19 at 19:11