📂Complex Anaylsis

Summation Formulas for All Integers Using Residue Theorem

Formulas

The ratio of a polynomial function, that is, a rational function $f$, assumes $f(n) \ne 0$ at $n \in \mathbb{Z}$ while being $\lim_{z \to \infty} z f(z) = 0$. When $f$ has a finite singularity $z_{1}, \cdots , z_{m}$, $$ \sum_{n=-\infty}^{\infty} f(n) = - \sum_{n = 1}^{m} \text{Res}_{z_{n}} (\pi f(z) \cot \pi z) $$

Explanation

It’s meaningful not just to sum up all natural numbers but to represent the sum of all integers within a finite sum. Certainly, if the given $f$ is an even function, taking half of it could also be applied to finding the sum over natural numbers. Cotangent is involved here and there up to $\pi$ in a complex form due to multiple multiplications, which might be hard to memorize but it’s good to know such a tool exists.

Derivation ¹

Part1. Boundedness on a rectangle of $\cos \pi z$

Considering a path $\mathscr{C}_{k}$ like in the picture above for a natural number $k \in \mathbb{N}$. Let’s assume there exists a natural number $k_{0}$ such that when $k>k_{0}$, rational function $f$ is continuous on $\mathscr{C}_{k}$ and $\displaystyle \lim_{z \to \infty} z f(z) = 0$.

For each $\mathscr{C}_{k}$, given that it’s continuous and $\displaystyle |z| \ge k + {{1 } \over {2}} > k$ for all $\varepsilon> 0$, $$ {{1} \over {|z|} } < \delta \implies |z f(z) | < \varepsilon $$ there exists a $\delta > 0$ that satisfies. Selecting $\displaystyle k > {{1} \over {\delta}}$, there exists a $k$ that satisfies $\displaystyle |f(z)| < {{\varepsilon} \over {k + 1/2}}$ about $\varepsilon > 0$ on $\mathscr{C}_{k}$.

• Addition formulas for trigonometric functions: $$ \sin\left( \alpha +\beta \right) =\sin\alpha \cos\beta +\cos\alpha \sin\beta \\ \sin\left( \alpha -\beta \right) =\sin\alpha \cos\beta -\cos\alpha \sin\beta \\ \cos\left( \alpha +\beta \right) =\cos\alpha \cos\beta -\sin\alpha \sin\beta \\ \cos\left( \alpha -\beta \right) =\cos\alpha \cos\beta +\sin\alpha \sin\beta \\ \tan\left( \alpha +\beta \right) =\frac { \tan\alpha +\tan\beta }{ 1-\tan\alpha \tan\beta } \\ \tan\left( \alpha -\beta \right) =\frac { \tan\alpha -\tan\beta }{ 1+\tan\alpha \tan\beta } $$
• Relationship between trigonometric functions and hyperbolic functions: $$ \begin{align*} \sinh (iz) =& i \sin z \\ \sin (iz) =& i \sinh z \\ \cosh (iz) =& \cos z \\ \cos (iz) =& \cosh z \end{align*} $$
• Relationship between trigonometric functions and exponential functions: $$ \sin z = { {e^{iz} - e^{-iz}} \over 2 i } \\ \cos z = { {e^{iz} + e^{-iz}} \over 2 } $$

If we conveniently set $\alpha := k + 1/2$, then $\cos \alpha \pi = 0$ and $\sin \alpha \pi = (-1)^{k}$. On $\mathscr{C}_{k}$, $|\cot \pi z|$ is $z = \pm \alpha + iy$ on the vertical line of $\mathscr{C}_{k}$ and $\left| y \right| \le \alpha$, therefore $$ \begin{align*} |\cot \pi z| =& \left| {{ \cos \pi z } \over { \sin \pi z }} \right| \\ =& \left| {{ \cos \alpha \pi \cos i \pi y - \sin \pm \alpha \pi \sin i \pi y } \over { \sin \pm \alpha \pi \cos i \pi y + i \cos \alpha \pi \sin i \pi y }} \right| \\ =& \left| {{ \cos \alpha \pi \cosh \pi y \mp i \sin \alpha \pi \sinh \pi y } \over { \pm \sin \alpha \pi \cosh \pi y + i \cos \alpha \pi \sinh \pi y }} \right| \\ =& \left| {{ 0 + 1 \cdot \sinh \pi y} \over { 1 \cdot \cosh \pi y + 0 }} \right| \\ =& \left| \tanh \pi y \right| \\ <& 1 \end{align*} $$ and on the horizontal line of $\mathscr{C}_{k}$, it’s $z = x \pm i \alpha$ and similarly $\left| x \right| \le \alpha$. Since the magnitude of the imaginary root of a real number is always $1$, $$ \begin{align*} \left| e^{i \pi x} e^{\mp \alpha \pi} + e^{ - i \pi x} e^{\pm \alpha \pi} \right| \le& \left| e^{i \pi x} e^{\mp \alpha \pi} \right| + \left| e^{ - i \pi x} e^{\pm \alpha \pi} \right| = e^{\alpha \pi} + e^{ - \alpha \pi} \\ \left| e^{i \pi x} e^{\mp \alpha \pi} - e^{ - i \pi x} e^{\pm \alpha \pi} \right| \ge& \left| \left| e^{i \pi x} e^{\mp \alpha \pi} \right| - \left| e^{ - i \pi x} e^{\pm \alpha \pi} \right| \right| = e^{\alpha \pi} - e^{ - \alpha \pi} \end{align*} $$ from which we obtain the following: $$ \begin{align*} |\cot \pi z| = \left| {{ \cos \pi z } \over { \sin \pi z }} \right| \\ =& \left| {{ e^{i \pi z} + e^{ - i \pi z} } \over { e^{i \pi z} - e^{ -i \pi z} }} \right| \\ =& \left| {{ e^{i \pi x} e^{\mp \alpha \pi} + e^{ - i \pi x} e^{\pm \alpha \pi} } \over { e^{i \pi x} e^{\mp \alpha \pi} - e^{ -i \pi x} e^{\pm \alpha \pi} }} \right| \\ \le & {{e^{\alpha \pi} + e^{- \alpha \pi}}\over {e^{\alpha \pi} - e^{- \alpha \pi}}} \\ =& \cot \alpha \pi \\ \le& \max \left\{ \pm \cot {{1} \over {2}} \pi, \pm \cot {{3} \over {2}} \pi \right\} & \because \sin \alpha \pi = (-1)^{k} \\ \le& \cot {{3} \over {2}} \pi \\ <& 2 \end{align*} $$ Eventually, $|\cot \pi z|$ is always bounded on the rectangle $\mathscr{C}_{k}$.

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Part 2. $\lim_{k \to \infty} \int_{\mathscr{C}_{k}} f(z) \cot \pi z dz = 0$

The length of $\mathscr{C}_{k}$ is $$ 8 \left( k + {{1 } \over {2}} \right) $$ thus, by the ML lemma, $$ {{1} \over {k}} < \delta \implies \left| \int_{\mathscr{C}_{k}} f(z) \cot \pi z dz \right| \le {{8 (k + 1/2) 2 \varepsilon } \over { k + 1/2}} = 16 \varepsilon $$ Therefore, $$ \lim_{k \to \infty} \int_{\mathscr{C}_{k}} f(z) \cot \pi z dz = 0 $$

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Part 3. $\sum_{n=-\infty}^{\infty} f(n) = - \sum_{n = 1}^{m} \text{Res}_{z_{n}} (\pi f(z) \cot \pi z)$

Now, defining $F(z) := \pi f(z) \cot \pi z$ as $f(n) \ne 0$, $n \in \mathbb{Z}$s are all simple poles $F$. Calculating the residues, $$ \text{Res}_{n} F(z) = {{ \pi f(z) \cos \pi z} \over { (\sin \pi z)' }} = \left. {{ \pi f(z) \cos \pi z} \over { \pi \cos \pi z }} \right|_{z = n} = f(n) $$ According to the assumption, $F$ still possess singularities $z_{1} , z_{2} , \cdots , z_{m}$, hence by the residue theorem, $$ \begin{align*} \lim_{k \to \infty} \int_{\mathscr{C}_{k}} F(z) dz =& \lim_{k \to \infty} 2 \pi i \left( \sum_{n = -k} ^{k} f(n) + \sum_{n = 1} ^{m} \text{Res}_{z_{n}} F(z) \right) \\ =& 2 \pi i \left( \sum_{n=-\infty}^{\infty} f(n) + \sum_{n = 1}^{m} \text{Res}_{z_{n}} (\pi f(z) \cot \pi z) \right) \end{align*} $$ We’ve already shown above that $\displaystyle \lim_{k \to \infty} \int_{\mathscr{C}_{k}} f(z) \cot \pi z dz = 0$, therefore $$ \sum_{n=-\infty}^{\infty} f(n) = - \sum_{n = 1}^{m} \text{Res}_{z_{n}} (\pi f(z) \cot \pi z) $$

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Formula for Alternating Series

Formula for the sum of an alternating series over all integers: For the rational function $f$, $$ \lim_{z \to \infty} z f(z) = 0 $$ assuming $f(n) \ne 0$ at $n \in \mathbb{Z}$ while being $\lim_{z \to \infty} z f(z) = 0$. When $f$ has a finite singularity $z_{1}, \cdots , z_{m}$, $$ \sum_{n=-\infty}^{\infty} (-1)^{n}f(n) = - \sum_{n = 1}^{m} \text{Res}_{z_{n}} (\pi f(z) \csc \pi z) $$

Try deriving it similarly for alternating series by writing it down manually.