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How to prove that for any complex number $z$ which is not equal to $\pi k + \frac{\pi}{2}$ ($k\in\mathbb Z$) : $$ \tan z = \sum_{n=0}^\infty \frac{8z}{(2n+1)^2\pi^2 - 4z^2} $$ Using complex analysis, I started with the contour intergal $$ \oint_{C_N} \frac{\tan \frac{\pi s}{2}}{s^2-z^2}\,\mathrm ds = \sum_{n=-N}^N \frac{-4i}{(2n+1)^2 - z^2} + \frac{2\pi i \tan \frac{\pi z}{2}}{z}$$ where $C_N$ is the circle centered at 0 of radius $N+1/2$ ($N\in\mathbb N$).
The complex number $z$ is chosen to be non zero & non odd integer.

However, I don't know how proceed to show that the LHS goes to $0$ as $N\to \infty$ :(

Thanks in advance for answers.

Célestin
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  • Are you aware of the identity $$ \cos(z)=\prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2\pi^2}\right)$$ ? If so, just apply $\frac{d}{dz}\log(\cdot)$ to both sides. – Jack D'Aurizio Aug 09 '17 at 21:25
  • Related: https://math.stackexchange.com/questions/581162/how-does-the-herglotz-trick-work – Jack D'Aurizio Aug 09 '17 at 21:30
  • The meromorphic function $\tan \left(\frac{\pi s}{2} \right)$ is periodic in the real direction and tends uniformly to $\pm i$ as $\operatorname{Im}(s) \to \pm \infty$. The magnitude of $\tan \left(\frac{\pi s}{2} \right)$ therefore remains bounded on $C_{N}$ as $N \to \infty$ since the contour stays away from the poles of $\tan \left(\frac{\pi s}{2} \right)$ on the real axis. – Random Variable Aug 11 '17 at 18:16
  • It seems that integrate $s\mapsto \frac{4z,\pi\cot\pi s}{(2s+1)^2\pi^2 - 4z^2}$ over a square contour is a better approach to prove this result :/ – Célestin Aug 11 '17 at 22:12

3 Answers3

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

You can use the Mittag-Leffler Expansion:

$\ds{\tan\pars{z}}$ has single poles at $\ds{p_{n} = \pars{2n + 1}{\pi \over 2}}$, with residues $\ds{r_{n} = -1}$, where $\ds{n \in \mathbb{Z}}$ .

$$ \bbx{\mbox{Note that}\quad p_{-n} = -p_{n - 1}} $$

Then, \begin{align} \tan\pars{z} & = \sum_{n = -\infty}^{\infty}\pars{-1}\pars{{1 \over z - p_{n}} + {1 \over p_{n}}} = \sum_{n = 1}^{\infty}\bracks{% \pars{{1 \over p_{n} - z} - {1 \over p_{n}}} + \pars{{1 \over p_{-n} - z} - {1 \over p_{-n}}}} \\[2mm] & + \pars{{1 \over p_{0} - z} - {1 \over p_{0}}} \\[5mm] & = \lim_{N \to \infty}\sum_{n = 1}^{N}\bracks{% \pars{{1 \over p_{n} - z} - {1 \over p_{n}}} + \pars{{1 \over -p_{n - 1} - z} + {1 \over p_{n - 1}}}} + \pars{{1 \over p_{0} - z} - {1 \over p_{0}}} \\[5mm] & = \lim_{N \to \infty}\bracks{\pars{{1 \over p_{0} - z} - {1 \over p_{0}}} + \sum_{n = 1}^{N}\pars{{1 \over p_{n} - z} - {1 \over p_{n}}} + \sum_{n = 1}^{N}\pars{{1 \over -p_{n - 1} - z} + {1 \over p_{n - 1}}}} \\[5mm] & = \lim_{N \to \infty}\bracks{\sum_{n = 0}^{N - 1} \pars{{1 \over p_{n} - z} - {1 \over p_{n}}} + \pars{{1 \over p_{N} - z} - {1 \over p_{N}}} + \sum_{n = 0}^{N - 1}\pars{-\,{1 \over p_{n} - z} + {1 \over p_{n}}}} \\[5mm] & = \sum_{n = 0}^{\infty}\pars{{1 \over p_{n} - z} - {1 \over p_{n} + z}} = \sum_{n = 0}^{\infty}{2z \over p_{n}^{2} - z^{2}} = \sum_{n = 0}^{\infty}{8z \over \pars{2p_{n}}^{2} - 4z^{2}} \\[5mm] & = \bbx{\sum_{n = 0}^{\infty}{8z \over \pars{2n + 1}^{2}\pi^{2} - 4z^{2}}} \end{align}
Felix Marin
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Short answer: both the LHS and the RHS are meromorphic functions with simple poles at the elements of $\frac{\pi}{2}+\pi\mathbb{Z}$. A simple computation of residues leads to the fact that $$ g(z)\stackrel{\text{def}}{=}\frac{\sin z}{\cos z}-\sum_{n\geq 0}\frac{8z}{(2n+1)^2\pi^2-4z^2} $$ is an entire function. Then we may check that for any $w$ in the region $0<\text{Re}(w)<\frac{\pi}{2}$ we have $$ \int_{0}^{w}g(z)\,dz = 0,$$ hence $g$ constantly equals $0$ on the previous open set. Since $g$ is entire, $g\equiv 0$.

Jack D'Aurizio
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Start by the infinite product expansion of cosine, $$\cos(x)=\prod_{n=0}^\infty\left(1-\frac{4x^2}{\pi^2(2n+1)^2}\right)$$ taking the logarithmic derivative, $$-\frac{\sin(x)}{\cos(x)}=-\sum_{n=0}^\infty\left(\frac{8x}{\pi^2(2n+1)^2-4x^2}\right)$$ so, $$\tan(x)=\sum_{n=0}^\infty\left(\frac{8x}{\pi^2(2n+1)^2-4x^2}\right).$$

Similar identities can also be derived using the same method. For example consider the product for sine, $$\sin(x)=x\prod_{n=1}^\infty\left(1-\frac{x^2}{\pi^2n^2}\right)$$ taking the logarithmic derivative, $$\cot(x)=\frac{1}{x}+2\sum_{n=1}^\infty\left(\frac{x}{x^2-\pi^2 n^2}\right)$$ which is $(4.22.3)$.

bob
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