Isolated Singularities and Residue Theorem

First of all, I want to apologize for the names I'm going to use on this wiki, because many of them probably have different names when written in books. We'll suppose that \(\Omega \subseteq \mathbb{C}\) is an open set throughout, unless otherwise stated, and \(f \in \mathcal{H}\big(\Omega\big)\) means \(f\) is a holomorphic function in \(\Omega\). We are going to deal with integrals, series, Bernouilli numbers, Riemann zeta function, and many interesting problems, as well as many theories. It's very important to tell everything is very joined and connected inside of complex analysis, so we'll use much knowledge of complex analysis and all the branches of mathematics.

Relevant wiki: Contour Integration

If \(\gamma : [0,1] \to \mathbb{C}\) is a closed piecewise smooth path and \(a \notin \gamma\big([0,1]\big)\), the winding number of \(\gamma\) with respect to \(a\) \((\)number of turns around \(a\) by \(\gamma\) anticlockwise) is

\[\displaystyle \text{ Ind(}\gamma , a) = \frac{1}{2i \pi} \cdot \int_{0}^{1} \frac{\gamma '(s)}{\gamma(s) - a} \, ds = \frac{1}{2i \pi} \oint_{\gamma} \frac{1}{z - a} \, dz.\]

If \(\gamma (x) = e^{2i \pi x}\) for \(x \in [0, 1],\) the winding number around \(0\) is

\[\displaystyle \text{ Ind(}\gamma , 0) = \frac{1}{2i \pi} \cdot \int_{0}^{1} \frac{2i \pi \cdot e^{2i \pi x}}{e^{2i \pi x}} \, dx = 1.\]

If \(\gamma_1 (x) = e^{i \pi x}\) for \(x \in [0, 1],\)

\[\displaystyle \frac{1}{2i \pi} \cdot \int_{0}^{1} \frac{i \pi \cdot e^{i \pi x}}{e^{i \pi x}} \, dx = \frac{1}{2}.\]

Nevertheless, we can't say

\[\displaystyle \text{ Ind(}\gamma_1 , 0) = \frac{1}{2}\]

because \(\gamma_1\) is not closed.

The winding number of a path \(\gamma\) with respect to \(a\) is constant in the connected component containing (or not) \(a\). This is, the winding number is 0 if \(\gamma\) doesn't turn around \(a,\) and it's constant if \(\gamma\) turns around \(a\). (Think about the winding number as the number of turns around...)

• Proposition: (generalization of the previous definition)
  If \(C_r (t) = a + re^{it}\) for \(t \in [0, 2\pi],\) then \(\text{ Ind(C}_r, z) = 1\) if \(|z - a| < r\) and \(\text{ Ind(C}_r, z) = 0\) if \(|z - a| > r.\)

if \(|z - a| < r,\) then \(\text{ Ind(C}_r, z) =\displaystyle \frac{1}{2i \pi} \cdot \int_{0}^{2 \pi} \frac{ir e^{it}}{re^{it}} \,dr = \frac{1}{2i \pi} \oint_{C_r} \frac{1}{w - a} \,dw =1.\)

if \(|z - a| > r,\) then \(\text{ Ind(C}_r, z) = 0\) due to the previous argument. \(_\square\)

If \(\gamma_i\) with \(i =1, ... , m\) are closed paths, the direct sum of these paths is called a cycle \(\Gamma = \gamma_1 \bigoplus \gamma_2 \bigoplus \ldots \bigoplus \gamma_m\) and \(\displaystyle \text{Image }(\Gamma) = \cup_{j = 1}^{m} \text{Image } (\gamma_j). \)

Two cycles \(\Gamma, \Delta\) in the open set \(\Omega \subseteq \mathbb{C}\) are \(\Omega\)-homologue if

\[\displaystyle \text{ Ind (}\Gamma , z) = \sum_{j = 1}^m \text{ Ind (}\gamma_j , z) = \text{ Ind (}\Delta , z) \space \forall z \notin \Omega.\]

A cycle \(\Gamma \subset \Omega\) is \(\Omega\)-homologue to \(0\) if \(\text{Ind (}\Gamma , z) = 0 \space \forall z \notin \Omega\).

Relevant wiki: Cauchy Integral Formula

Two cycles \(\Gamma, \Delta\) in the open set \(\Omega \subseteq \mathbb{C}\) are \(\Omega\)-homologue if \[\displaystyle \text{ Ind (}\Gamma , z) = \sum_{j = 1}^m \text{ Ind (}\gamma_j , z) = \text{ Ind (}\Delta , z) \space \forall z \notin \Omega.\] A cycle \(\Gamma \subset \Omega\) is \(\Omega\)-homologue to \(0\) if \(\text{ Ind (}\Gamma , z) = 0 \space \forall z \notin \Omega\).

• Proposition: (homologous version of Cauchy's integral formula)
  If \(\Gamma\) is a piecewise smooth cycle in the open set \(\Omega \subset \mathbb{C}\), \(\Omega\)-homologue to \(0\), then \(\forall f \in \mathcal{H}(\Omega)\) and every \(z \in \Omega \text{-Image(}\Gamma)\), \[\displaystyle \text{ Ind (}\Gamma , z) \cdot f(z) = \frac{1}{2i \pi} \oint_{\Gamma} \frac{f(w)}{w - z} \, dw.\]

• Proposition: (homologous version of Cauchy's theorem)
  If \(\Gamma, \Delta\) are 2 piecewise smooth cycles \(\Omega\)-homologue in the open set \(\Omega \subset \mathbb{C}\), and \(f \in \mathcal{H}(\Omega)\), then \[\displaystyle \oint_{\Gamma} f(z) \,dz = \oint_{\Delta} f(z) \,dz .\] In particular, if \(\Gamma\) is \(\Omega\)-homologue to \(0\), then \(\oint_{\Gamma} f(z) \,dz = 0.\)

a) \(\mathbb{C}_{\infty} = \mathbb{C} \cup \{\infty\}\) is compacting of Alexandroff in \(\mathbb{C}.\)

b) Let \(\Omega \subseteq \mathbb{R}^2\) be an open set in \(\mathbb{R}^2\) and \(u : \Omega \to \mathbb{R}\) be a function such that \(u \in \mathcal{C}^2 (\Omega)\), then the Laplacian operator for \(u\) is \(\Delta u = \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2}\). Then \(u\) is a harmonic funcion if \(\Delta u = 0\).

If \(\Omega \subseteq \mathbb{C}\) is an open connected set, then the following are equivalent:

a) \(\mathbb{C}_{\infty} - \Omega\) is a connected set.
b) Each piecewise smooth cycle is \(\Omega\)-homologue to \(0\).
c) If \(\Gamma\) is a piecewise smooth cycle in the open set \(\Omega \subset \mathbb{C}\), then \(\forall f \in \mathcal{H}(\Omega)\) and every \(z \in \Omega \text{-Image(}\Gamma)\), \[\displaystyle \text{ Ind (}\Gamma , z) \cdot f(z) = \frac{1}{2i \pi} \oint_{\Gamma} \frac{f(w)}{w - z} \, dw.\] d) For each piecewise smooth cycle \(\Gamma\) in \(\Omega\) and each \(f \in \mathcal{H}(\Omega)\), \(\oint_{\Gamma} f(z) \, dz = 0.\)
e) \(\Omega\) is holomorphically connected, i.e. \(\forall f \in \mathcal{H}(\Omega) \space \exists F \in \mathcal{H}(\Omega\)) such that \( F' = f.\)
f) \(\forall f \in \mathcal{H}(\Omega)\) with \(0 \notin f(\Omega), \space \exists g \in \mathcal{H}(\Omega)\) such that \(e^{g} = f.\)
g) \(\forall u\) harmonic and analytical function in \(\Omega,\) there exists a harmonic and analytical function \(v\) in \(\Omega\) such that \(u + iv \in \mathcal{H}(\Omega)\).

Definition 1.

Let \(f \in \mathcal{H}(\Omega)\). If \(a \notin \Omega\) and \(D^{*}(a, r) = \big(D(a, r) - \{a\}\big) \subseteq \Omega,\) it is said that \(a\) is an isolated singularity of \(f\).

a) If there exists \(\displaystyle \lim_{z \to a} f(z) = b \in \mathbb{C},\) it is said that \(a\) is an avoidable isolated singularity of \(f\).
b) If there exists \(\displaystyle \lim_{z \to a} f(z) = \infty,\) it is said that \(a\) is a pole of \(f\).
c) In \(\mathbb{C}_{\infty} = \mathbb{C} \cup \{\infty\}\), if there doesn't exist \(\displaystyle \lim_{z \to a} f(z),\) it is said that \(a\) is an essential singularity of \(f\).

Proposition 1.

Let \(f \in \mathcal{H}(\Omega)\) and \(a \notin \Omega\) an isolated singularity of \(f\). Then the following are equivalent:

a) \(a\) is an avoidable isolated singularity of \(f\).
b) \(f\) is bounded in some \(D^{*}(a, r) \subseteq \Omega\).
c) \(f\) supports a holomorphic extension to open \(\Omega_a= \Omega \cup \{a\}\).

---

Proof.

Since \(c) \Rightarrow a) \Rightarrow b)\) is trivial, let's prove \(b) \Rightarrow c).\)

Let \(g(z) = f(z)(z - a)^2\) and \(g(a) = 0\). This implies that \(g \in \mathcal{H}\big(\Omega \cup \{a\}\big)\) and \(g'(a) = 0\), and then \[\displaystyle g(z) = \sum_{n = 2}^\infty a_n (z - a)^n = (z - a)^2 \cdot h(z), \space \forall z \in D(a, r),\] where \(h(z) = a_2 + a_3 (z - a) + \cdots +a_{n + 2} (z - a)^n + \cdots \) converges in \(D(a, r)\).

If we define \(f(a) = h(a) = a_2,\) we obtain a holomorphic extension of \(f\) in \(\Omega \cup \{a\}\) \(\big(\)due to \(f(z) = h(z), \space \forall z \in D(a, r)\big)\). \(_\square\)

Whenever a holomorphic function \(f\) has an avoidable singularity at a point \(a\), this function will be (generally) assigned at this point the value \[\displaystyle f(a) = \lim_{z \to a} f(z).\]

\(f(z) = \frac{\sin z}{z}\) has an avoidable singularity at \(z = 0,\) which is eliminated by defyning \(f(0) = 1.\)

\(f(z) = \dfrac{1}{e^z - 1} - \dfrac{1}{z}\) has an avoidable singularity at \(z = 0,\) which is eliminated by defyning \(f(0) = -\frac12.\)

Relevant wiki: Liouville's Theorem

What is the relationship between two entire functions \(f, g:\mathbb{C} \to \mathbb{C}\) such that \(f, g \in \mathcal{H}(\mathbb{C}),\) which verifies \(\big|f(z)\big| \leq \big|g(z)\big|, \space \forall z \in \mathbb{C}?\)

---

If \(g = 0,\) then \(f = 0\). If \(g \neq 0\) due to a consequence of identity theorem and this (proposition 16.1), each \(a \in \mathcal{Z}(g)\) \((\)each zero of \(g)\) is an isolated singularity for \(\frac{f}{g}\); hence \(\frac{f}{g}\) is a bounded function and applying proposition 1, every isolated singularity of \(\frac{f}{g}\) is avoidable. Eliminating them, \(F = \frac{f}{g}\) is an entire function fulfilling \(\big|F(z)\big| \leq 1, \space \forall z \in \mathbb{C} \Rightarrow\) (apllying Liouville's theorem) \(\exists c \in \mathbb{C}\) such that \[f = cg.\ _\square\]

Proposition 2.

If \(f \in \mathcal{H}(\Omega)\) and \(a \notin \Omega\) is an isolated singularity of \(f\), then the following are equivalent:

a) \(a\) is a pole of \(f\).
b) \(\exists D^{*}(a,\delta) \subseteq \Omega\) and \(F \in \mathcal{H}\big(D(a, \delta)\big)\) such that \(a\) is a zero isolated of \(F\) and \(f(z) = \frac{1}{F(z)}, \space \forall z \in D^{*}(a,\delta)\).
c) \(\exists m \in \mathbb{N}\) such that \((z - a)^m f(z)\) has a finite and non-zero limit as \(z \to a\).

---

Proof.

1. \(a) \Rightarrow b)\)
   Due to \(\displaystyle \lim_{z \to a} f(z) = \infty, \space \exists \space D^{*}(a,\delta) \subseteq \Omega\) such that \(|f(z)| > 1,\space \forall z \in D^{*}(a,\delta)\). The function \(F(z) = \frac{1}{f(z)}\) is defined and holomorphic in \(D^{*}(a,\delta)\) and \(|F| < 1\). Because of proposition 1, \(F\) has an avoidable singularity at \(a,\) which is eliminated by defining \(F(a) = 0 \Rightarrow\) \(F \in \mathcal{H}\big(D(a, \delta)\big)\) such that \(a\) is a zero isolated of \(F\), fulfilling b).\[\]

2. \(b) \Rightarrow c)\)
   Let \(m = \mathcal{v}(F, a) \in \mathbb{N}\) be the multiplicity of \(a\) being a zero of \(F\). This implies that \[\frac{F(z)}{(z - a)^m} = \frac{1}{(z - a)^m f(z)}\] has a finite non-zero limit as \(z \to a\). The same thing happens to the function \((z - a)^m f(z)\). \[\]

3. \(c) \Rightarrow a)\)
   The function \(h(z) = (z - a)^m f(z)\) has an avoidable singularity at \(a,\) which is eliminated by defining \(\displaystyle h(a) = \lim_{z \to a} (z - a)^m f(z) \neq 0\). Then \[\displaystyle \lim_{z \to a} f(z) = \lim_{z \to a} \frac{h(z)}{(z - a)^m} = \infty \implies a \text{ is a pole of } f.\]

Hence all proved. \(_\square\)

Definition 2.

Under the conditions of proposition 2, it is said that the pole \(a\) of the function \(f \) has multiplicity \(m = \mathcal{v}(f, a) \in \mathbb{Z}^{+}\), if \(m = 1\) \(a\) is a simple pole.

Proposition 3.

if \(f \in \mathcal{H}(\Omega)\) and \(a \notin \Omega\) is an isolated singularity of \(f\), then the following are equivalent:

a) \(a\) is a pole of \(f\) of multiplicity \(m\).
b) There exists the only polynomial \(P(z)\) of degree \(m\) with \(P(0) = 0\) such that \(a\) is an avoidable singularity of \(f(z) - P\left(\frac{1}{z - a}\right).\)

---

Proof.

1. \(a) \Rightarrow b)\)\[\] Let \(D^{*}(a, r) \subseteq \Omega\), then using c) of proposition 2, \(g(z) = (z - a)^m f(z)\) has at \(a\) an avoidable singularity, which is eliminated by defining \(\displaystyle g(a) = \lim_{z \to a} (z - a)^m f(z) = b_0 \neq 0\). This implies that \(g\) is a holomorfic function in the open set \(\Omega_{a} = \Omega \cup \{a\} \supset D(a, r)\) \(\Big(g \in \mathcal{H}\big(\Omega_{a} = \Omega \cup \{a\}\big) \supset D(a, r)\Big),\) and supports the development of a series of powers \[\displaystyle g(z) = \sum_{n = 0}^\infty b_n (z- a)^n,\] which implies that if \( 0 < | z - a| < r\), \[f(z) = \frac{g(z)}{(z - a)^m} = \frac{b_0}{(z - a)^m} + \frac{b_1}{(z - a)^{m - 1}} + \cdots + \frac{b_{m - 1}}{z -a} + h(z)\] with \(h(z) = b_m + b_{m+1} (z - a) + \cdots + b_{m + k} (z - a)^k + ...\).\[\] The polynomial \(P(w) = b_0 w^m +b_1 w^{m + 1} + \cdots + b_{m - 1} w\) fulfills \(P(0) = 0\) and \[\lim_{z \to a} \left(f(z) - P\Big(\frac{1}{z - a}\Big)\right) = \lim_{z \to a} h(z) = h(a) = b_m.\] Let's see that \(P\) is unique. If \(Q(z)\) is another polynomial satisfying b), then \(a\) is an avoidable singularity of \[R(z) = P\left(\frac{1}{z - a}\right) - Q\left(\frac{1}{z - a}\right) = \left(f(z) - Q\Big(\frac{1}{z - a}\Big)\right) - \left( f(z) - P\Big(\frac{1}{z - a}\Big)\right).\] Eliminating the singularity, we can consider that \(R\) is an entire function. Because of \(P(0) = Q(0) = 0\), \(\displaystyle \lim_{z \to \infty} R(z) = 0\) and this imply that \(R\) is bounded. Liouville's theorem tells us that \(R\) is constant, and obviously \(R = 0\).\[\]

2. \(b) \Rightarrow a)\)\[\] If b) is true, then c) of proposition 2 is true. \(_\square\)

Let \(\Omega \subset \mathbb{C}\) be a connected open set, both \(f\) and \(g \in \mathcal{H}(\Omega),\) and \( g \neq 0,\) then each \(a \in \mathcal{Z}(g)\) is an isolated singularity of \(\frac{f}{g}\). If \(a\) is a zero of \(g\) of multiplicity \(m > 0\) and \(f(a) \neq 0,\) then the singularity is a pole of \(f\) with multiplicity \(m\).

Consequence: If \(a\) is a zero of \(f\) of multiplicty \(p\), then \(a\) is a pole of \(\frac{f}{g}\) \(\iff\) \(m - p > 0.\)

\(f(z) = \dfrac{\sin z}{z^3}\) has at \(z = 0\) a double pole (pole of multiplicity 2) since \[ \frac{\sin z}{z^3} = \frac{z - \left(\frac{1}{3!}\right)z^3 + \left(\frac{1}{5!}\right)z^5 - \cdots}{z^3} = \frac{1}{z^2} - \frac{1}{3!} + \frac{z^2}{5!} - \cdots.\]

Theorem. (Casorati-Weierstrass)

If \(f \in \mathcal{H}(\Omega)\) and \(a \notin \Omega\) is an isolated singularity. Then the following are equivalent:

a) \(a\) is an essential singularty of \(f\).
b) \(\forall D^{*}(a, \epsilon) \subset \Omega\), \(f\big(D^{*}(a, \epsilon)\big)\) is dense in \(\mathbb{C}\).

1. \(b) \Rightarrow a)\)\[\] This is trivial (reductio ad absurdum).\[\]

2. \(a) \Rightarrow b)\)\[\] If not b), then \(\exists \epsilon > 0\) such that \(\overline{f\big(D^{*}(a, \epsilon)\big)} \neq \mathbb{C} \Rightarrow \exists D(b, R) \neq \emptyset\) \((R > 0)\) such that \(D(b, R) \cap f\big(D^{*}(a, \epsilon)\big) = \emptyset\), i. e, \(|f(z) - b| \ge R,\space \forall z \in D^{*}(a, \epsilon)\). Then, the function \(h(z) = \frac{1}{f(z) - b}\) is holomorphic and bounded in \(D^{*}(a, \epsilon)\), and therefore has an avoidable singularity at \(a\) due to proposition 1, i.e. \(\displaystyle \exists \space \lim_{z \to a} h(z) = \mu \in \mathbb{C} \Rightarrow\) \[f(z) = b + \frac{1}{h(z)}, \space \forall z \in D^{*}(a, \epsilon) \Rightarrow\] in \(\mathbb{C}_{\infty}\) \[\displaystyle \begin{cases} \lim_{z \to a} f(z) = b + \frac{1}{\mu}, &\text { if } \mu \neq 0 \\ \\ \lim_{z \to a} f(z) = b + \frac{1}{\mu} = \infty, &\text { if } \mu = 0.\ _\square \end{cases}\]

If \(f \in \mathcal{H}\big(D^{*}(0, 1)\big)\) such that \(\text{Real f(z)} < \alpha, \space \forall z \in D^{*}(0, r), (\alpha > 0, \alpha \in \mathbb{R}, 0 < r < 1),\) what kind of singularity presents \(f\) at \(0?\)

---

\(0\) can`t be an essential sigularity of \(f\) due to Casorati-Weierstarass theorem.

If \(0\) was a pole of \(f\), then due to proposition 2, there would exist \(D^{*}(0, \delta)\) and \(F \in \mathcal{H}\big(D(0, \delta)\big)\) such that \(0\) would be a zero of \(F(z) = \frac{1}{f(z)}\) and \(D(0, r) \supset F(D(0, \delta)) \Rightarrow\) \[f\big(D^{*}(0, \delta)\big) \supset \left\{z: |z| > \frac{1}{r}\right\},\] which is a contradiction. Therefore \(0\) is an avoidable singularity of \(f\). \(_\square\)

Prove that 0 is an a essential singularity of \(f(z) = e^{\frac{1}{z}}.\)

---

1. We have \[\lim_{z \to 0^{-}} e^{\frac{1}{z}} = 0\quad \text{ and }\quad \lim_{z \to 0^{+}} e^{\frac{1}{z}} = \infty.\]
2. Using Casorati-Weierstrass theorem
3. Due to the corollary of the following proposition:

Proposition 4.

If \(f \in \mathcal{H}(\Omega),\) \(a \notin \Omega\) is an isolated singularity of \(f,\) \(\displaystyle \sum_{n = -\infty}^{\infty} a_n (z - a)^n\) is Laurent series of \(f\) in \(D^{*}(a, R) \subseteq \Omega,\) and \(M = \{n \in \mathbb{N}; a_n \neq 0\}\), then

a) \(a\) is an avoidable singularity of \(f\) \(\iff\) \(\text{ inf(M)} \ge 0\)
b) \(a\) is a pole of multiplicity \(m\) of \(f\) \(\iff\) \(\text{ inf(M)} = -m < 0\)
c) \(a\) is an essential singularity of \(f\) \(\iff\) \(\text{ inf(M)} = - \infty\).

---

Proof.

a) Trivial consequence after eliminating the singularity, due to proposition 1.
b) A consequence of proposition 3 b).
c) By exclusion. \(_\square\)

Corollary. If \(f \in \mathcal{H}(\Omega)\), then \(a\) is an essential singularity of \(f\left(\frac{1}{z - a}\right)\) \(\iff\) \(f\) is not a polynomial.

Proof. Consequence of proposition 4 c).

Show that the function \(h(z) = \left(e^\frac{1}{z}\right)^z\) defined in \(\mathbb{C} - \{0\}\) has an essential singularity at \(z = 0.\)

---

We have

\[\displaystyle \lim_{z \to 0} \left(e^{\frac{1}{z}}\right)^{z} = \lim_{z \to \infty} (e^z)^{\frac{1}{z}}.\]

This limit doesn't exist in \(\mathbb{C}\), because if \(R \in \mathbb{R}\) with \(R >0\),

\[\displaystyle \lim_{R \to 0^{+}} \left(e^{\frac{1}{R}}\right)^{R} = \lim_{\frac{1}{R} \to + \infty} \left(e^R\right)^{\frac{1}{R}} = e,\]

and taking \(iR,\) such that \(R \in \mathbb{R}\) and \(R >0\),

\[\displaystyle \lim_{R \to 0^{+}} \left(e^{\frac{1}{iR}}\right)^{iR} = \lim_{R \to 0^{+}} e^{iR(i k)} = 1\]

since\(k \in (\pi, \pi],\) \(k\) is the main argument of \(e^{\frac{1}{iR}}\), \(\text{Arg }e^{\frac{1}{iR}},\) and \(\ln e^{\frac{1}{iR}} = \ln 1 + i \cdot \text{Arg }e^{\frac{1}{iR}} = ik.\) \(_\square\)

Definiton 3.

If \(f \in \mathcal{H}(\Omega)\) and \(D^{*}(\infty, r) = \big\{ z: \frac{1}{r} < |z| < + \infty\big\} \subset \Omega,\) it is said that \(\infty\) is an isolated singularity of \(f\).

a) \(\infty\) is an avoidable singularity of \(f\) \(\iff\) \(\displaystyle \exists \space \lim_{z \to \infty} f(z) \in \mathbb{C}\) \(\iff\) \(f\left(\frac{1}{z}\right)\) defined in \(D^{*}(0, r)\) has an avoidable singularity at \(0\).
b) \(\infty\) is a pole of \(f\) \(\iff\) \(\displaystyle \space \lim_{z \to \infty} f(z) = \infty\) \(\iff\) \(f\left(\frac{1}{z}\right)\) defined in \(D^{*}(0, r)\) has a pole at \(0\).
c) \(\infty\) is an essential singularity of \(f\) \(\iff\) there doesn't exist \(\displaystyle \space \lim_{z \to \infty} f(z) \) in \(\mathbb{C}_{\infty}\) \(\iff\) \(f\left(\frac{1}{z}\right)\) defined in \(D^{*}(0, r)\) has an essential singularity at \(0\).

If \(f \in \mathcal{H}(\mathbb{C})\), then due to Liouville's theorem and Weierstrass theorem 3.7.,

\[\infty \text{ is an avoidable singularity of } f \iff f = \text{constant}.\]

If \(f \in \mathcal{H}(\mathbb{C})\), then prove the following:

\[\infty \text{ is a pole of multiplicity } m \ge 1 \text{ of } f \iff f \text{ is a polynomial of degree } m \ge 1.\]

---

Proof. You can see the solutions in the problem below; or

\(\Leftarrow\))
Suppose that \(f\) is a polynomial of degree \(m \ge 1\) with coefficients in \(\mathbb{C}\). Let's also suppose that \(f(z) = a_0 + a_1 z + \cdots + a_m z^m.\) Then using polar coordinates of a complex number, \[\displaystyle \lim_{z \to \infty} f(z) = \lim_{z \to \infty} \frac{z^m ( a_0 + a_1 z + \cdots + a_m z^m )}{z^m} = \infty \in \mathbb{C}_{\infty},\] which implies that \(\infty\) is a pole of multiplicity \(m \ge 1\) of \(f.\)

\(\Leftrightarrow\))
\(\infty\) is a pole of \(f\) of multiplicity \(m \)
\(\iff\) \(f\left(\frac{1}{z}\right)\) defined in \(D^{*}(0, r)\) has a pole at \(0\) of multiplicity \(m\)
\(\iff\) (using propoisition 3)
\(\hspace{1.2cm}\) there exists the only polynomial \(P(z)\) of degree \(m\) with \(P(0) = 0\) such that \(0\) is an avoidable singularity of \(f\left(\frac{1}{z}\right) - P\left(\frac{1}{z}\right)\)
\(\iff\) (using propoisition 1) \(f\left(\frac{1}{z}\right) - P\left(\frac{1}{z}\right)\) is bounded in a disk \(D^{*}(0, r)\)
\(\iff\) \(f(z) - P(z)\) is bounded in some disk \(D^{*}(\infty, R)\)
\(\iff\) (using Liouville's theorem and a Weierstrass theorem) \(f(z) = P(z) + C,\) where \(C\) is a constant. \(_\square\)

By exclusion,

\[\infty \text{ is an essential singularity of } f \in \mathcal{H}(\mathbb{C}) \iff f \text{ is not a polynomial}.\]

Definition 2.1.

Let \(f \in \mathcal{H}\big(D^{*}(a, r)\big)\), then residue of \(f\) at \(a\) \(\big(\text{Res}(f, a)\big)\) is the complex number \[\text{Res}(f, a) = \frac{1}{2\pi i} \cdot \oint_{C_{\rho}} f(z) \, dz,\] where \(0 < \rho < r\) and \(C_{\rho} (\theta) = a + \rho e^{i \theta}.\)

Note: The value of this integral is independent of \(0 < \rho < r\); i.e. in \(\Omega = D^{*}(a, r)\) \[\oint_{C_{\rho}} f(z) \, dz = \oint_{C_{\rho^{ '}}} f(z) \, dz\] if \( 0 < \rho^{ '} < r\) due to \(C_{\rho}\) and \(C_{\rho^{ '}}\) being \(\Omega\)-homologue, due to the homologus version of Cauchy's theorem.

Proposition 2.1.

If \(f \in \mathcal{H}(\Omega)\) and \(a \in \mathbb{C}\) is an isolated singularity of \(f\), then there exists one and only one \(\alpha \in \mathbb{C}\) such that \(f(z) - \frac{\alpha}{z -a}\) has a primitive (anti-derivative) for each \( D^{*}(a, r) \subset \Omega\).

---

Proof.

I'm going to use the previous note and this theorem:

• If \(f : \Omega \to \mathbb{C}\) is a continuous function, then the following are equivalent:

  • a) \(\exists F \in \mathcal{H}(\Omega)\) such that \(F' = f.\)

  • b) \(\oint_{\gamma} f(z) \,dz = 0\) for each piecewise smooth path \(\gamma\) in \(\Omega\).

Let \(\alpha = \frac{1}{2\pi i} \oint_{C_{\rho}} f(z) \, dz\) with \(0 < \rho < r\), then It will suffice to prove that for each piecewise smooth path \(\gamma\) in \(D^{*}(a, r) \subset \Omega\), \[I(\gamma) = 0, \ \text{ where } \ I(\gamma) = \frac{1}{2\pi i} \oint_{\gamma} \left(f(z) - \frac{\alpha}{z -a}\right) \, dz = 0.\] If \(m = \text{ Ind(}\gamma , a)\) and \(0 < \rho < r\), then the cycles \(\gamma, \space mC_{\rho}\) are \(\Omega\)-homologue, and this implies that \[I(\gamma) = I(mC_{\rho}) = \frac{1}{2\pi i} \oint_{mC_{\rho}} f(z) \, dz - \frac{1}{2\pi i} \oint_{mC_{\rho}} \frac{\alpha}{z -a} \, dz = m \alpha - m \alpha = 0.\ _\square\]

Definition 2.2.

If \(f \in \mathcal{H}(\Omega)\) and \(a \in \mathbb{C}\) is an isolated singularity of \(f\), then \(\text{Res}(f, a) = \alpha,\) where \(\alpha \in \mathbb{C}\) is the unique complex number such that \(f(z) - \frac{\alpha}{z -a}\) has a primitive (anti-derivative) for each \( D^{*}(a, r) \subset \Omega\). If \(\infty\) is an isolated singularity of \(f\), then \(\text{Res}(f, \infty) = \text{Res}(g, 0),\) where \(g(z) = - \frac{f\left(\frac1z\right)}{z^2}\).

Calculation of Residues:

Let \(f \in \mathcal{H}\big(D^{*}(a, r)\big)\), and the Laurent series of \(f\) \[\displaystyle f(z) = \sum_{n = \infty}^{\infty} a_n (z -a)^n, \quad 0 < |z - a| < r,\] then \(f(z) - \frac{a_{-1}}{z -a}\) has a primitive (anti-derivative) in \( D^{*}(a, r) \Rightarrow \text{Res}(f, a) = a_{-1}\).

a) If \(f\) has an avoidable singluarity at \(a \in \mathbb{C},\) then \(\displaystyle \text{Res}(f, a) = 0 = \lim_{z \to a} (z - a) f(z)\). The reciprocal is not true \(\big(\)e.g. \(f(z) = \frac{1}{z^3} \) at \(z = 0\big).\)
b) If \(f\) has a simple pole \(m = 1\) at \(a \in \mathbb{C}\), then \(\displaystyle \text{Res}(f, a) = a_{-1} = \lim_{z \to a} (z - a) f(z)\).
c) If \(f\) has a pole of multiplicity \(m > 1\) at \(a \in \mathbb{C}\), then \[\displaystyle \text{Res}(f, a) = a_{-1} = \frac{1}{(m - 1)!} \cdot \dfrac{d^{m - 1}}{dz^{m -1}} \Big((z - a)^m f(z) \Big)\Big|_{z = a},\] because \( (z - a)^m f(z) = a_{-m} + a_{-m + 1} (z -a) + \cdots + a_{-1} (z - a)^{m - 1} + \ldots\).

Let \(f \in \mathcal{H}\big(D^{*}(a, r)\big)\) with \(a\) a simple pole and \(0 < \rho < r\). Let's call \(C_{\rho} (\theta) = a + \rho e^{i\theta}, \space \theta \in [\alpha, \beta], \space 2\pi \ge \beta > \alpha \ge 0\). Prove that, for a fixed \(0 < \rho_{1} < r\), there exists \[\displaystyle \lim_{\rho \to 0,\space \rho < \rho_{1}} \oint_{C_{\rho}} f(z) \, dz = i (\beta - \alpha) \text{Res}(f , a).\]

---

If \(0 < |z - a| < r \longrightarrow f(z) = \frac{\text{Res(f, a)}}{z - a} + g(z)\), such that \(g \in \mathcal{H}\big(D(a, r)\big)\).

\[\begin{align} \overline{D(a, \rho_{1})} \subset D(a, r) &\Rightarrow \exists M > 0 \text{ / } |g(z)| \leq M, \space \forall z \in \overline{D(a, \rho_{1})}.\\\\ \displaystyle \left| \oint_{C_{\rho}} g(z) \, dz \right| &\leq M \cdot \text{length (}C_{\rho}) \longrightarrow 0, \text{ as } \rho \to 0. \end{align}\]

Therefore,

\[\begin{align} \displaystyle \lim_{\rho \to 0,\space \rho < \rho_{1}} \oint_{C_{\rho}} f(z) \, dz &=\displaystyle \lim_{\rho \to 0,\space \rho < \rho_{1}}\left(\oint_{C_{\rho}} \frac{\text{Res}(f, a)}{z - a} \, dz + \oint_{C_{\rho}} g(z) \, dz\right) \\ &= \oint_{C_{\rho}} \frac{\text{Res}(f, a)}{z - a} \, dz \\ &= \text{Res}(f , a) \int_{\alpha}^{\beta} \frac{i \rho e^{i\theta}}{\rho e^{i\theta}} \,d\theta \\ &= i (\beta - \alpha) \text{Res}(f , a).\ _\square \end{align}\]

Theorem. (Residue Theorem)

Let \(f \in \mathcal{H}(\Omega)\), \(S \subseteq \delta \Omega\) \((\)boundary of \(\Omega)\) a set of isolated singularities of \(f\), and \(\Omega_{S} = \Omega \cup S\). If \(\Gamma\) is a piecewise smooth cycle in \(\Omega\) and \(\Omega_{S}\)-homologue to \(0,\) then \[\displaystyle \frac{1}{2i \pi} \oint_{\Gamma} f(z) \, dz = \sum_{a \in S} \text{ Ind(}\Gamma , a) \cdot \text{Res} (f, a),\] where this sum has a mounting (support) finite.

The set \(V = \big\{z \in \mathbb{C} - \text{Image(}\Gamma); \space \text{Ind(}\Gamma, z) = 0\big\}\) is the union of some connected components of \(\mathbb{C} - \text{Image(}\Gamma)\) including the unbounded connected component. Therefore, \(K = \mathbb{C} - V\) is a compact set because it's closed and bounded, and this implies that \[K \cap S = \{a \in S; \space \text{Ind(}\Gamma, a)\neq 0\}\] is finite (by reductio ad absurdum, all bounded sequence in \(\mathbb{R}^n\) has a convergent subsequence... ).

If \(\{a_1, a_2, \ldots , a_p\} = \{a \in S; \space \text{Ind(}\Gamma, a)\neq 0\}\) choosing \(r_j > 0\) for \(1 \leq j < p\) small enough, we get all disks \(D^{*}(a_j, r_j)\) which are contained in \(\Omega\) and are disjoints two to two. For \(j = 1, 2, ..., p\), take \(C_j (t) = a + \rho_j ae^{it}\) such that \(t \in [0, 2\pi]\), whith \(0 < \rho_j < r_j\), and take the cycle \[\Delta = m_1 C_1 \bigoplus m_ 2C_2 \bigoplus \cdots \bigoplus m_3 C_p,\] where \(m_j = \text{Ind(}\Gamma, a_j)\). Obviously, \(\text{ Ind(}\Gamma , a) = \text{ Ind(}\Delta , a), \space \forall a \in S\), and \(\text{Ind(}\Gamma , z) = 0 =\text{Ind(}\Delta , z), \space \forall z \in \mathbb{C} - \Omega_{S} \subset \mathbb{C} - \Omega \Rightarrow\) \(\Gamma\) and \(\Delta\) are \(\Omega_{S}\)-homologue, and applying the homologus version of Cauchy's theorem, \[\displaystyle \frac{1}{2\pi i} \oint_{\Gamma} f(z) \, dz = \displaystyle \frac{1}{2\pi i} \oint_{\Delta} f(z) \, dz = \frac{1}{2\pi i} \sum_{j = 1}^p m_j \oint_{C_j} f(z) \, dz\] and \(\frac{1}{2\pi i} \oint_{C_j} f(z) \, dz = \text{Res} (f, a_j)\). \(_\square\)

Evaluate \[\large I = \int_{|z| = 1} \frac{e^{z}}{z} \, dz, \] where \(z \in \mathbb{C}\).

---

We have \[f(z) = \frac{e^z}{z} \in \mathcal{H}\big(\mathbb{C} - \{0\}\big),\] and \(C(x) = e^{ix}\), such that \(x \in [0, 2\pi]\). Applying residue theorem, \[ I = \int_{|z| = 1} \frac{e^{z}}{z} \, dz = \oint_{C} f(z) \, dz = 2i \pi \big(\text{Res}(f, 0) \cdot \text{Ind}(C, 0) \big) = 2i \pi.\ _\square\]

Show that

\[\displaystyle I = \int_0^{2\pi} \cot (x + a) \, dx = -2\pi i \cdot \big(\text{sign (Im a)}\big), \text{ such that } a \notin \mathbb{R}.\]

---

Let \(a = a_1 + i\cdot a_2\) with \(a_1, a_2 \in \mathbb{R}\) and \(a_2 \neq 0\). Note that \(\sin (x + a) \neq 0\) due to \(a \notin \mathbb{R}\) and \(\big| e^{-i (a_1 + i \cdot a_2)} \big| = e^{a_2}\). Then

\[\begin{align} \cot (x + a) &= \frac{\cos (x + a)}{\sin (x + a)} \\\\ &= \frac{i\big(e^{ix} e^{ia} + e^{-ix} e^{-ia}\big)}{e^{ix} e^{ia} - e^{-ix} e^{-ia}} \\\\ &= \frac{i\big(e^{2ix} + e^{-2ia}\big)}{e^{2ix} - e^{-2ia}}. \end{align} \]

Applying residue theorem and the path \(C(x) = e^{ix}, \space x\in [0, 2\pi], z = e^{ix} \Rightarrow \frac{dz}{iz} = dx\) and \(g(z) = \dfrac{z^2 + e^{-2ia}}{z (z - e^{-ia}) (z + e^{-ia}) },\) which gives

\[\begin{align} I = \int_0^{2\pi} \cot (x + a) \, dx &=\displaystyle \int_{C} \frac{i(z^2 + e^{-2ia})}{iz(z^2 - e^{-2ia})} \, dz \\ &= 2i \pi \left(\text{Res}(g, 0) + \text{Res}(g, e^{-ia}) \cdot \text{Ind} (C, e^{-ia}) + \text{Res}(g, -e^{-ia}) \cdot \text{Ind} (C, -e^{-ia})\right). \end{align} \]

Therefore,

\[\displaystyle \left.\begin{cases} \text{Res}(g , 0) = \displaystyle \lim_{z \to 0} z g(z) = \frac{e^{-2ia}}{- e^{-2ia}} = - 1 \\ \\ \text{Res}(g, e^{-ia}) = \displaystyle \lim_{z \to e^{-ia}} (z - e^{-ia}) g(z) = 1 \\ \\ \text{Res} (g, -e^{-ia}) = 1 \end{cases}\right) \Rightarrow I = -2\pi i \cdot \text{sign} \big(\text{Im} (a)\big),\]

where \(\text{sign} \big(\text{Im} (a)\big) = 1\) if \(\text{Im(a)} = a_2 > 0,\) and \(\text{sign} \big(\text{Im} (a)\big) = -1\) if \(\text{Im(a)} = a_2 < 0.\ _\square\)

Let \(P\) be a polynomial of degree \(n \ge 1\) such that \(P(0) = 0\). Calculate the residue of \(f\) at \(a\) of the function:

\[f(z) = \frac{1}{z - b} P\left(\frac{1}{z - a}\right),\]

where \(b \neq a\).

---

Let \(P(z) = a_1 z + a_2 z^2 + \cdots + a_n z^n\) with \(P(0) = 0,\) then it's sufficient to calculate the residue at \(z = 0\) of \(f(a + z) = \frac{1}{z - c} P(\frac{1}{z}),\) where \( c = b - a\) . If \(|z| < |c|,\) then

\[f(a + z) = \left(\frac{1}{1 - \frac zc}\right) P\left(\frac{1}{z}\right) \left(\frac{-1}{c}\right) = \frac{-1}{c} \cdot \left(\frac{a_1}{z} + \frac{a_2}{z^2} + \cdots + \frac{a_n}{z^n}\right) \cdot \left(1 + \frac{z}{c} + \frac{z^2}{c^2} + \cdots \right).\]

Therefore, the residue of \(f(a + z)\) at \(z = 0\) is the coefficient of \(\frac{1}{z}\), which is

\[ \frac{-1}{c} \cdot \left(a_1 + \frac{a_2}{c} + \cdots + \frac{a_n}{c^{n - 1}}\right) = -P\left(\frac{1}{c}\right) = -P\left(\frac{1}{b - a}\right).\ _\square\]

Show that

\[\displaystyle \int_{0}^{\pi} \frac{1}{(a + b \cos \theta)^2} \, d\theta = \int_{0}^{\pi} \frac{1}{(a - b \cos \theta)^2} \, d\theta = \frac{\pi a}{\sqrt{(a^2 - b^2)^3}}\]

sometimes...

---

We have

\[\begin{align} \int_{a}^{b} f(\theta) \, d\theta &= \int_{a}^{b} f(a +b - \theta) \, d\theta\\\\ \int_{0}^{\pi} \frac{1}{(a + b \cos \theta)^2} \, d\theta &= \frac{1}{2} \cdot \int_{0}^{2\pi} \frac{1}{(a + b \cos \theta)^2} \, d\theta \\ &= \frac{1}{2} \cdot \int_{0}^{2\pi} \frac{1}{\left(a + b\left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)\right)^2} \, d\theta. \end{align}\]

Applying residue theorem, \(e^{i\theta} = z \rightarrow \frac{dz}{iz} = d\theta\) and \(C(\theta) = e^{i\theta}\) such that \(\theta \in [0, 2\pi],\) this is equal to

\[\begin{align} \frac{1}{2i} \cdot \int_{C} \dfrac{4z^2}{\big(2az + b(z^2 +1)\big)^2 \cdot z} \, dz &= \frac{1}{2i} \cdot \int_{C} \frac{4z}{\big(2az + b(z^2 +1)\big)^2 } \, dz \\ &= \frac{1}{2i} \left(2\pi i \cdot \sum_{|s| < 1} \text{Res}\left( g(z) = \frac{4z}{\big(2az + b(z^2 +1)\big)^2 }, s\right)\right)\\ &= \pi \cdot \sum_{|s| < 1} \text{Res} \left(g(z) = \frac{4z}{\big(2az + b(z^2 +1)\big)^2 }, s\right) \qquad (\text{being s a singularity of }g) \\ &=\frac{\pi a}{\sqrt{(a^2 - b^2)^3}} \end{align}\]

sometimes... because

\[g(z) = \frac{4z}{\big(2az + b(z^2 +1)\big)^2} = \frac{4z}{b^2 \cdot \left(z + \frac{a + \sqrt{a^2 - b^2}}{b}\right)^2 \cdot \left(z + \frac{a - \sqrt{a^2 - b^2}}{b}\right)^2}\]

depends on their poles and the modulus of these poles. \(_\square\)

A function \(f: \Omega \to \mathbb{C}_{\infty}\) defined in an open set \( \Omega \subset \mathbb{C}_{\infty}\) is said to be a meromorphic function if it is a continuous function and satisfies

a) \(\mathcal{P}(f) = \{z \in \Omega ; \space f(z) = \infty\}\) is a set of isolated points;
b) \(f\) defined in \(\Omega_{0} = \Omega - \mathcal{P}(f)\) is a holomorphic function.

The set containing the meromorphic functions in \(\Omega\) will be denoted \(\mathcal{M}(\Omega)\).

Let \(R(z) = \frac{P(z)}{Q(z)}\) be a rational function where \(P, Q\) are polynomials without common zeros of degree \(n, m\) respectively.
Also let us define \(R(a) = \infty\) when \(Q(a) = 0\) and \(\displaystyle R(\infty) = \lim_{z \to \infty} R(z) \in \mathbb{C}_{\infty}.\)

Then we'll obtain a meromorphic function \(R \in \mathcal{M}(\mathbb{C}_{\infty})\). If \(n > m\), \(R\) has a pole at \(z = \infty\) of multiplicity \(n - m.\)

Theorem. (Argument Principle)

Let \(f \in \mathcal{M}(\Omega)\), \(f \neq 0\), \(\Omega \subseteq \mathbb{C}\) be a connected open set and \(S = \mathcal{P}(f) \cup \mathcal{Z}(f)\). If \(\Gamma\) is a piecewise smooth cycle in \(\Omega - S\), which is \(\Omega_{S}\text{-homologue}\) to \(0,\) then \[\displaystyle \frac{1}{2\pi i} \oint_{\Gamma} \frac{f'(z)}{f(z)} \, dz = \sum_{a \in \mathcal{Z}(f)} \text{ Ind( }\Gamma, a) \cdot \mathcal{v} \text{(f, a)} - \sum_{a \in \mathcal{P}(f)} \text{ Ind( }\Gamma, a) \cdot \mathcal{v} \text{(f, a)},\] where there is only a finite number of non-zero summands.

\(\frac{f'(z)}{f(z)} \in \mathcal{H}(\Omega - S)\) with isolated singularities in \(S\) \((\)due to identity principle again, and \(\Omega \subseteq \mathbb{C}\) being a connected open set\().\) If \(a \in S, \space \exists D^{*}(a, r) \subseteq \Omega\) and a function \(F \in \mathcal{H}\big(D(a, r)\big)\) such that \(0 \notin F\big(D(a, r)\big)\) veryfing \(f(z) = (z - a)^m F(z), \space \forall z \in D^{*}(a, r),\) where \(m = \mathcal{v} \text{(f, a)} \) if \(a \in \mathcal{Z}(f)\) and \(m = - \mathcal{v} \text{(f, a)} \) if \(a \in \mathcal{P}(f)\). Then, \[\frac{f'(z)}{f(z)} - \frac{m}{z - a} = \frac{F'(z)}{F(z)}, \space \forall z \in D^{*}(a, r).\] Since \(\frac{F'(z)}{F(z)}\) has a primitive (antiderivative) in \(D(a,r) \) \((\)because it is a homolomorphic function in \(D(a,r),\) or due to one of the most important theorem in complex analysis\(),\) we get \(\text{Res}\Big(\frac{f'(z)}{f(z)}, a\Big) = m,\) and applying residue theorem, we get the result (outcome). \(_\square\)

Show that

\[\displaystyle I = \int_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx = \frac{\pi}{\sin \pi a},\space \text{ if }\ 0 < a < 1.\]

---

Let's first see that the above integral converges: \((\)remember that \(0 < a < 1)\)

\[ \begin{cases} { \ \ \int_{0}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx = \int_{0}^{\infty} \frac{e^{(a - 1)x}}{e^{-x} + 1} \,dx < \int_{0}^{\infty} e^{(a - 1)x} \,dx < + \infty \\ \int_{- \infty}^{0} \frac{e^{ax}}{e^x + 1} \,dx < \int_{- \infty}^{0} e^{ax} \,dx < + \infty} \end{cases} \Rightarrow \displaystyle I = \int_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx < + \infty.\]

Let \(f(x) = \frac{e^{ax}}{e^x + 1} \), then \(e^x + 1 = 0\)
\( \iff x \in \{i( \pi + 2k \pi) \text{ / } k \in \mathbb{Z}\}\), i.e. the poles of \(f\) are in \(\{i( \pi + 2k \pi) \text{ / } k \in \mathbb{Z}\}\), i.e. \(f \in \mathcal{H}\big(\mathbb{C} - \{i( \pi + 2k \pi) \text{ / } k \in \mathbb{Z}\}\big).\)

Let \(\gamma_R = \gamma_1 \bigoplus \gamma_2 \bigoplus \gamma_3 \bigoplus \gamma_4\) be the following cycle (see picture), direct sum of piecewise smooth paths, with \(R > 0\) very big...

• \(\gamma_1 (t) = t, \text{ such that }\ t \in [-R, R]\)
• \(\gamma_2 (t) = R + i t, \text{ such that }\ t \in [0, 2 \pi]\)
• \(\gamma_3 (t) = t + 2i \pi, \text{ such that }\ t \in [-R , R ]\)
• \(\gamma_4 (t) = -R + i t, \text{ such that }\ t \in [0, 2 \pi]\).

Then \(\text{Ind(}\gamma_R, i \pi) = 1\) and, applying residue theorem, we get \[\displaystyle I = \int_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx = \lim_{R \to \infty} \int_{-R}^{R} \frac{e^{ax}}{e^x + 1} \,dx = 2 \pi i \cdot \text{Res}(f, i \pi) \cdot \text{Ind(}\gamma_R, i \pi) = - 2 \pi i \cdot e^{i \pi a}\] since \(\text{Res}(f, i \pi) = - e^{i \pi a}\) because \[\displaystyle \text{Res}(f, i \pi) = \lim_{z \to i \pi} \frac{e^{az} \cdot (z - i\pi)}{e^z +1} = \lim_{z \to i \pi} \frac{e^{az} \cdot (z - i\pi)}{-(z - i\pi)} = - e^{i \pi a}\] due to power series of \(g(z) = e^z\) around \(z = i \pi\). Now, \[\begin{align} \oint_{\gamma_R} f(t) \, dt &= \oint_{\gamma_1} f(t) \,dt + \oint_{\gamma_2} f(t) \, dt - \oint_{\gamma_3} f(t) \, dt - \oint_{\gamma_4} f(t) \, dt \\ &= \displaystyle \int_{-R}^{R} f(t) \,dt + \int_{0}^{2\pi} f(R + it) \cdot i \,dt - \int_{-R}^{R} f(t + 2\pi i) \,dt - \int_{0}^{2\pi} f(-R + it) \cdot i \,dt. \qquad \text{(A)} \end{align}\] On the other hand, \[\begin{align} \left| \oint_{\gamma_2} f(t) \, dt \right| &= \left| \int_{0}^{2\pi} f(R + it) \cdot i \,dt \right| \\ &= \left| \int_{0}^{2\pi} \frac{e^{a(R + it)} \cdot i}{1 + e^{R + it}} \,dt \right| \\ &\leq \int_{0}^{2\pi} \frac{e^{aR}}{e^R - 1} \, dt \\ &\leq 2\pi \frac{e^{aR}}{e^R - 1} \to 0 \end{align}\] as \(R \to +\infty\), because \(\Big| 1 + e^{R + it}\Big| \ge \Big| \big|e^{R + it}\big| - |-1| \Big| = e^R -1 \space (R > 0)\). In the same way, \[ \left| \oint_{\gamma_4} f(t) \, dt \right| \to 0 \text{ as } R \to \infty.\] Therefore, \[ \oint_{\gamma_R} f(t) \, dt = \oint_{\gamma_1} f(t) \,dt - \oint_{\gamma_3} f(t) \, dt = \int_{-R}^{R} \frac{e^{at}}{e^t + 1} \,dt - \int_{-R}^{R} \frac{e^{a(t + 2i \pi)}}{e^{t + 2i \pi} + 1} \,dt\] and \[\begin{align} - \int_{-R}^{R} \frac{e^{a(t + 2i \pi)}}{e^{t + 2i \pi} + 1} \,dt = - e^{a2\pi i} \cdot \int_{-R}^{R} \frac{e^{at}}{e^t + 1} \,dt &\Rightarrow I\big(1 - e^{a2\pi i}\big) = - 2\pi i \cdot e^{\pi i a} \\ &\Rightarrow I = \frac{- 2\pi i \cdot e^{\pi i a}}{1 - e^{a2\pi i}} = \frac{\pi}{\sin \pi a}.\ _\square \end{align}\]

Proposition.

Let \(P, Q\) be polynomials with \(\text{degree} (Q) - \text{degree} (P) \ge 2\). If the rational function \(f = \frac{P}{Q}\) has no poles on the real axis, the next improper integral is convergent and \[\displaystyle \int_{- \infty}^{\infty} f(x) \, dx = 2\pi i \cdot \sum_{\text{Im}(a) > 0} \text{Res}(f, a)\] with \(a\) being a pole of \(f.\)

---

Proof.

By assumption, \(z^2 f(z)\) has a finite limit as \(z \to \infty\). Therefore, \(\exists \rho > 0\) and \(M > 0\) such that \(\big|z^2 f(z)\big| \leq M\) if \(|z| \ge \rho\).

Since \(|f(z)| \leq \frac{M}{z^2}\) if \(|z| \ge \rho\), taking into account the convergence of the following integrals: \[\displaystyle \int_{- \infty}^{-\rho} \frac{1}{x^2} \, dx \quad \text {and} \quad \int_{\rho}^{\infty} \frac{1}{x^2} \, dx,\] and taking into account the fact that \(f = \frac{P}{Q}\) has no poles on the real axis, the above improper integral \(\displaystyle \int_{- \infty}^{\infty} f(x) \, dx \) is convergent, due to the criterion of comparison of integrals.

\(\exists R > \rho\) such that all poles of \(f\) are contained in \(D(0, R),\) and if \(\Gamma_{R}\) is the border of \(\{z: \space |z| \leq R \text{ such that Im(z) } \ge 0\}\) traveled counter-clockwise, applying residue theorem gives \[\displaystyle 2\pi i \cdot \sum_{\text{Im}(a) > 0} \text{ Res}(f, a) = \oint_{\Gamma_{R}} f(z) \, dz = I(R) + J(R),\] where \[I(R) = \int_{-R}^{R} f(x) \, dx,\quad J(R) = \int_0^{\pi} f\big(Re^{it}\big) iRe^{it} \,dt.\] Since \(\big|f(Re^{it})\big| \leq \frac{M}{R^2} \Rightarrow |J(R)| \leq \frac{\pi M}{R} \Rightarrow \displaystyle \lim_{R \to \infty} J(R) = 0\), \[\displaystyle \int_{- \infty}^{\infty} f(x) \, dx = \lim_{R \to \infty} I(R) = \lim_{R \to \infty} \oint_{\Gamma_{R}} f(z) \, dz = 2\pi i \cdot \sum_{\text{Im}(a) > 0} \text{Res}(f, a).\ _\square\]

Exercise: Let \(P, Q\) be polynomials with \(\text{degree} (Q) - \text{degree} (P) \ge 2\). If the rational function \(f = \frac{P}{Q}\) has no poles on the real axis, the next improper integral is convergent and \[\displaystyle \int_{- \infty}^{\infty} f(x) \, dx = -2\pi i \cdot \sum_{\text{Im}(a) < 0} \text{Res}(f, a)\] with \(a\) being a pole of \(f.\)

Show that \[\int_{0}^{\infty} \dfrac{dx}{(x^2 + a^2)^2} = \frac{\pi}{4a^3} \ \text{ for } \ a > 0.\]

---

Since \(f(x) = \frac{1}{(x^2 + a^2)^2},\) we have \[\begin{align} \text{Res}(f , ia) &= \left.\left(\frac{d}{dz} \frac{(z - ia)^2}{(z - ia)^2 \cdot (z + ia)^2} \right) \right|_{z = ia} \\ &= \left.\frac{-2}{(z + ia)^3} \right|_{z = ia} \\ &= \frac{-2}{-8ia^3} \\ &= \frac{1}{4ia^3}, \end{align}\] which implies \[\begin{align} \int_{0}^{\infty} \dfrac{dx}{\big(x^2 + a^2\big)^2} &= \frac{1}{2} \int_{-\infty}^{\infty} \dfrac{dx}{\big(x^2 + a^2\big)^2} \\ &= \frac{1}{2} \big(2\pi i \, \text{Res}(f , ia) \big) \\ &= \frac{\pi}{4a^3}.\ _\square \end{align}\]

Proposition.

Let \(P,Q\) be polynomials with \(\text{degree}(Q) - \text{degree}(P) \ge 1\). If \(f = \frac{P}{Q}\) doesn't have any poles at the real axis, the following improper integral converges and its value is \[\displaystyle \int_{- \infty}^{\infty} f(x) \cdot e^{ix} \, dx = 2i\pi \cdot \sum_{\text{Im}(a) > 0} \text{Res}\left(f(x)\cdot e^{ix},a\right)\] with \(a\) being a pole of \(f\).

---

Proof.

By assumption \(z f(z)\) has a finite limite as \(z \to \infty\). The same thing happens to \(z^2 f'(z)\), because \(f' = \frac{P'Q - Q'P}{Q^2}\) and \(\text{degree}(P'Q - Q'P) - \text{degree}(Q^2) \leq -2\). Therefore, \(\exists \rho > 0\) and \(\exists M > 0\) such that \[\text{if}\ |z| \ge \rho \Rightarrow |zf(z)| \leq M \ \text{ and } \ \big|z^2 f'(z)\big| \leq M.\] Since \(|f'(x)| \leq \frac{M}{x^2}\) when \(|x| \ge \rho,\) the integral \(\int_{\rho}^{\infty} f'(x) e^{ix} \, dx\) is absolutely convergent. Taking into account \(\displaystyle \lim_{y \to \infty} f(y) = 0\), \[\displaystyle \int_{\rho}^{y } f(x) e^{ix} \, dx = i \rho f(\rho) e^{i\rho} - iy f(y) e^{iy} + i \int_{\rho}^{y } f'(x) e^{ix} \, dx.\] As \(y \to \infty\), the integral \(\int_{\rho}^{\infty} f(x) e^{ix} \, dx\) is convergent. Similarly, it is proved that the improper integral \(\int_{-\infty}^{-\rho} f(x) e^{ix} \, dx\) is convergent, and because of continuity of \(f(x) e^{ix}\) in \([-\rho, \rho]\) \((f\) doesn't have any poles at the real axis\()\) \[\displaystyle \int_{- \infty}^{\infty} f(x) \cdot e^{ix} \, dx \] is convergent.

Take \(R > 0\) such that all poles of \(f\) are in \(D(0, R)\) and if \(\Gamma_{R}\) is the border of \(\{z: \space |z| \leq R \text{ such that Im(z) } \ge 0\}\) traveled counter-clockwise, applying residue theorem gives \[\displaystyle 2\pi i \cdot \sum_{\text{Im}(a) > 0} \text{Res}\left(f(z)e^{iz}, a\right) = \oint_{\Gamma_{R}} f(z) e^{iz} \, dz = I(R) + J(R),\] where \[I(R) = \int_{-R}^{R} f(x) e^{ix} \, dx,\quad J(R) = \int_0^{\pi} f(Re^{it}) e^{iRe^{it}} iRe^{it} \,dt.\] When \(0 \leq t \leq \pi\), \(\big|f(Re^{it}) e^{iRe^{it}} iRe^{it}\big| \leq M e^{-R \sin t},\) which implies \[\displaystyle J(R) \leq M \int_0^{\pi} e^{-R \sin t} \, dt = \alpha(R).\] If \(\alpha(R) \to 0\) as \(R \to \infty\), then \[\displaystyle 2\pi i \cdot \sum_{\text{Im}(a) > 0} \text{Res}\big(f(z)e^{iz}, a\big) = \lim_{R \to \infty} I(R) = \int_{- \infty}^{\infty} f(x) \cdot e^{ix} \, dx \] and \[\alpha(R) = M \int_0^{\pi} e^{-R \sin t} \, dt \to 0\] as \(R \to \infty\), because given \(\epsilon > 0\) there exists \( 0 < \delta < \text{minimum}\big\{\frac{\epsilon}4, \frac{\pi}2\big\}\), and there exists \(C > 0\) such that if \(R > C\) then \[\displaystyle \int_0^{\pi} e^{-R \sin t} \, dt \leq 2\delta + \int_{\delta}^{\pi - \delta} e^{-R \sin \delta} \, dt < 2\delta + \pi e^{-R \sin \delta}< \frac\epsilon2 + \frac\epsilon2 = \epsilon. \left(\pi e^{-R \sin \delta} < \frac\epsilon2\right) \ _\square\]

Let \( a, b \in \mathbb{R} \space\) such that \(a , b > 0.\) Show that

\[ I = \int_{0}^{\infty} \frac{\cos bt}{t^2 +a^2} \, dt = \frac{\pi}{2ae^{ab}}, \qquad J = \int_{0}^{\infty} \frac{t \sin bt}{t^2 +a^2} \, dt = \frac{\pi}{2e^{ab}}.\]

---

\(I\) and \(J\) are both integrals of even functions, so we can first calculate \(I\) as follows:

\[\displaystyle I = \int_{0}^{\infty} \frac{\cos bt}{t^2 +a^2} \, dt = \frac{b}{2}\left( \int_{-\infty}^{\infty} \frac{\cos x}{x^2 +a^2 b^2} \, dx \right) = \frac{b}{2} \left (\text{ Real }\int_{-\infty}^{\infty} \frac{e^{ix}}{x^2 +a^2 b^2} \, dx \right).\]

Then we are going to apply proposition 1. The only pole of \( \frac{e^{ix}}{x^2 +a^2 b^2}\) in the upper-half plane is \(z = ab i\) and its residue value is \(\frac{e^{-ab}}{2ab i},\) so

\[\displaystyle \int_{-\infty}^{\infty} \frac{e^{ix}}{x^2 +a^2 b^2} \, dx = 2\pi i \frac{e^{-ab}}{2ab i} = \frac{\pi}{ab e^{ab}} \Rightarrow I = \frac{\pi}{2a e^{ab}}.\]

Now we can calculate \(J\) as follows:

\[\displaystyle J = \int_{0}^{\infty} \frac{t \sin bt}{t^2 +a^2} \, dt = \frac{1}{2}\left( \int_{-\infty}^{\infty} \frac{x \sin x}{x^2 +a^2 b^2} \, dx \right) = \frac{1}{2} \left (\text{Image}\int_{-\infty}^{\infty} \frac{xe^{ix}}{x^2 +a^2 b^2} \, dx \right).\]

Then we are going to apply proposition 1. The only pole of \( \frac{xe^{ix}}{x^2 +a^2 b^2}\) in the upper-half plane is \(z = ab i\) and its residue value is \(\frac{e^{-ab}}{2},\) so

\[\displaystyle \int_{-\infty}^{\infty} \frac{xe^{ix}}{x^2 +a^2 b^2} \, dx = 2\pi i \frac{e^{-ab}}{2} = \frac{\pi i}{ e^{ab}} \Rightarrow J= \frac{\pi}{2 e^{ab}}.\ _\square\]

Proposition.

Let \(P,Q\) be polynomials with \(\text{degree}(Q) - \text{degree}(P) \ge 1\). If \(f = \frac{P}{Q}\) doesn't have any poles at the real axis, except at the origin, where \(f\) is able to have a simple pole, then \(\forall \epsilon > 0\) the integrals \[\int_{- \infty}^{-\epsilon} f(x)\cdot e^{ix} \, dx, \quad \int_{\epsilon}^{\infty} f(x)\cdot e^{ix} \, dx\] converge as \(\epsilon \to 0\), and \[\displaystyle \lim_{\epsilon \to 0} \int_{|x| \ge \epsilon} f(x)\cdot e^{ix} \, dx = 2i\pi \cdot \sum_{\text{Im}(a) > 0} \text{Res}\left(f(x)\cdot e^{ix},a\right) + i\pi \cdot \text{Res}\left(f(x)\cdot e^{ix},0\right).\]

---

Proof.

"Coming soon..."

Show that \[\displaystyle \int_{0}^{\infty} \frac{\sin t}{t} \, dt = \frac{\pi}{2}.\]

---

Applying the above proposition, \(\text{Res}\left(\frac{e^{ix}}{x}, 0\right) = 1,\) which implies\[\displaystyle \int_{-\infty}^{\infty} \frac{e^{ix}}{x} \, dx = \pi i \Rightarrow \int_{0}^{\infty} \frac{\sin x}{x} \, dx = \frac{\pi}{2}.\ _\square\]

We will call summation function to a meromorphic function \(\alpha \in \mathcal{M}(\mathbb{C})\) with poles \(\mathcal{P}(\alpha) = \mathbb{Z}\), all simple poles, that remains bounded in \(\cup_{n \in \mathbb{N}} \{z: |z| = R_n\}\), where \(R_n\) is a sequence satisfying \(n < R_n < n + 1, \space \forall n \in \mathbb{N}\).

Here are some of the most important examples:

a) \(\alpha (z) = \pi \cot \pi z\)
b) \(\beta(z) = \dfrac{\pi}{\sin \pi z}\)

These two functions meet the above conditions with \(R_n = n + \frac12,\) and

\[\displaystyle \text{Res}(\alpha, k) = \lim_{z \to k} \pi(z - k) \cot \pi z =1,\space \forall k \in \mathbb{Z}\quad \text{ and }\quad \text{Res}(\beta, k) = \lim_{z \to k} \frac{\pi(z - k)}{\sin \pi z} = (-1)^k, \space \forall k \in \mathbb{Z}.\]

Proposition.

Let \(\alpha \in \mathcal{M}(\mathbb{C})\) be a summation function and \(\alpha_k = \text{Res}(\alpha, k),\space \forall k \in \mathbb{Z}\). If \(P, Q\) are polynomials \((\)with coefficients in \(\mathbb{C})\) and \(f = \frac{P}{Q}\) doesn`t have any poles in \(\mathbb{Z}\), each one of the following conditions:

a) \(\text{degree}(Q) - \text{degree}(P) \ge 2\)
b) \(\text{degree(Q) - degree(P) } \ge 1\) and the function \(\alpha\) is an odd function

implies that \[\displaystyle \lim_{n \to \infty} \sum_{|k| \leq n} \alpha_k f(k) = - \sum_{a \in \mathcal{P}(f)} \text{Res}(\alpha f, a).\] If a) is fulfilled and the sequence \(\alpha_k\) is bounded, then the series \(\displaystyle \sum_{k = - \infty}^{\infty} \alpha_k f(k)\) is absolutely convergent.

Note: It may happen that there exists \(\displaystyle \lim_{n \to \infty} \sum_{|k| \leq n} \alpha_k f(k)\), even if the series \(\displaystyle \sum_{k = - \infty}^{\infty} \alpha_k f(k)\) is not convergent. In this case, the main value (p. v.) of the sum is, by definition, \[\text{p.v.} =\sum_{k = - \infty}^{\infty} \alpha_k f(k) = \lim_{n \to \infty} \sum_{|k| \leq n} \alpha_k f(k) = \alpha_0 f(0) + \sum_{ n = 1}^\infty \big(\alpha_k f(k) + \alpha_{-k} f(-k)\big).\]

---

Sketch of Proof in Spanish:

\(\)
Using the above proposition, and taking \(\alpha(z) = \pi \cot \pi z\) or \(\beta(z) = \frac{\pi}{\sin \pi z}\), we are going to calculate the following sums, with \(a \notin \mathbb{Z}\):

[A] \(\displaystyle \sum_{n = - \infty}^{\infty} \frac{1}{(a + n)^2}; \ \sum_{n = - \infty}^{\infty} \frac{(-1)^n}{(a + n)^2 }\quad \quad\) [B] \(\displaystyle \sum_{n = 1}^{\infty} \frac{1}{a^2 + n^2}; \ \sum_{n = 1}^{\infty} \frac{(-1)^n}{a^2 + n^2 } \quad \quad \) [C] \( \displaystyle \text{ p. v. } \sum_{n = - \infty}^{\infty} \frac{1}{a - n}; \ \text{ p. v. } \sum_{n = - \infty}^{\infty} \frac{(-1)^n}{a - n}.\)

\(\)

Get the following 2 sums: \[\sum_{n = - \infty}^{\infty} \frac{1}{(a + n)^2}; \quad \sum_{n = - \infty}^{\infty} \frac{(-1)^n}{(a + n)^2 }.\]

---

First Sum:
Take \(f(z) = \frac{1}{(z + a)^2}\) and \(\alpha (z) = \pi \cot \pi z\): \[\displaystyle \sum_{n = - \infty}^{\infty} \frac{1}{(a + n)^2} = - \text{Res}(\alpha f, - a) = - \lim_{z \to -a} \frac{\pi (z + a)\cdot \cot z \pi }{(z + a)^2} = \frac{\pi^2}{\sin^2 \pi a}.\]

Second Sum:
Take \(f(z) = \frac{1}{(z + a)^2}\) and \(\beta(z) = \frac{\pi}{\sin \pi z}\): \[\displaystyle \sum_{n = - \infty}^{\infty} \frac{(-1)^n}{(a + n)^2} = - \text{Res}(\beta f, - a) = - \lim_{z \to -a} \frac{\pi (z + a)}{(z + a)^2 \cdot \sin \pi z} = \frac{\pi^2 \cdot \cos \pi a}{\sin^2 \pi a}.\]

In both cases the series are absolutely convergent. \(_\square\)

Get the following 2 sums: \[\sum_{n = 1}^{\infty} \frac{1}{a^2 + n^2}; \quad \sum_{n = 1}^{\infty} \frac{(-1)^n}{a^2 + n^2 }.\]

---

First Sum:
Let's call \[\displaystyle S = \sum_{n = 1}^{\infty} \frac{1}{a^2 + n^2} \Rightarrow \frac{1}{a^2} + 2S = \sum_{n = - \infty}^{\infty} \frac{1}{a^2 + n^2}.\] Then applying the above proposition and taking \(f(z) = \frac{1}{z^2 + a^2}\) and \(\alpha (z) = \pi \cot \pi z\)) gives \[\begin{align} \sum_{n = - \infty}^{\infty} \frac{1}{a^2 + n^2} &= -\text{Res}(\alpha f, a\cdot i) - \text{Res}(\alpha f, -a\cdot i) \\ &= \text{Res}(\alpha f, a\cdot i) \\ &= \lim_{z \to a\cdot i} \frac{\pi (z - ai)\cdot \cot a \pi i}{(z - ai)(z + ai)} \\ &= \frac{\pi \cdot \cot a \pi i}{2a\cdot i} = \frac{\pi \cdot (1 + e^{2\pi a})}{2a\cdot (1 -e^{2\pi a})} \\ &= \text{Res}(\alpha f, -a\cdot i)\\\\ \Rightarrow S &= \sum_{n = 1}^{\infty} \frac{1}{a^2 + n^2} = \frac{\pi \cdot (1 + e^{2\pi a})}{2a\cdot (e^{2\pi a} - 1)} - \frac{1}{2a^2}. \end{align}\]

Second Sum:
Take \(f(z) = \frac{1}{z^2 + a^2}\) and \(\beta(z) = \frac{\pi}{\sin \pi z}\). Then doing the same procedure as above, we obtain \[\displaystyle \sum_{n = 1}^{\infty} \frac{(-1)^n}{a^2 + n^2 } = \frac{ \pi e^{\pi a}}{a(e^{2\pi a} - 1)} - \frac{1}{2a^2}.\ _\square\]

Get the following 2 sums: \[\text{ p. v. } \sum_{n = - \infty}^{\infty} \frac{1}{a - n}; \quad \text{ p. v. } \sum_{n = - \infty}^{\infty} \frac{(-1)^n}{a - n}.\]

---

Take \(f(z) = \frac{1}{a - z}, \alpha (z) = \pi \cot \pi z,\) and \(\beta(z) = \frac{\pi}{\sin \pi z}\).

First Sum:
\[ \text{ p. v. } \sum_{n = - \infty}^{\infty} \frac{1}{a - n} = - \text{Res}\big(\alpha(z) \cdot f(z), a\big) = \pi \cot \pi a.\]

Second Sum:
\[ \text{ p. v. } \sum_{n = - \infty}^{\infty} \frac{(-1)^n}{a - n} = - \text{Res}\big(\beta(z) \cdot f(z), a\big) = \frac{\pi}{\sin \pi a}.\ _\square\]

\(\)
\(\underline{\text{Relationship between Bernoulli numbers and Riemann's zeta function:}}\)

Relevant link: Bernoulli Numbers
Relevant wiki: Riemann-zeta Function

Here is "Exercise 2.2.1."

Find the relationship between \(\displaystyle \oint_{C_n} \frac{\pi \cot \pi z}{z^{2m}} \, dz,\) where \(C_n (t) = \left(n + \frac12\right) \cdot e^{it}\) with \(t \in [0, 2\pi],\) and \(\displaystyle \zeta(2m) = \sum_{k = 1}^{\infty} \frac{1}{k^{2m}}\) with \(m \in \mathbb{Z}^{+}.\)

---

The function \(f(z) = \pi z \cot \pi z \in \mathcal{M}(\mathbb{C})\), with simple poles in \(\mathbb{Z} - \{0\}\) and an avoidable singularity at \(z = 0,\) which vanishes defining \(\displaystyle f(0) = \lim_{z \to 0} f(z) = 1\). Then the meromorphic function \[f_m (z) = \frac{\pi \cot \pi z}{z^{2m}}\] has simple poles in \(\mathbb{Z} - \{0\}\), and at \(z = 0\) has a pole of multiplicity \(2m + 1:\) \[\displaystyle \forall k \in \mathbb{Z} - \{0\}, \space \text{Res}(f, k) = \lim_{z \to k} (z - k) \pi z \cot \pi z = k.\] The integral \[I_n = \oint_{C_n} f_m (z) \,dz \longrightarrow 0, \text{ as } n \to 0\] because \(\alpha (z) = \pi \cot \pi z\) is bounded in \(\cup_{n \in \mathbb{N}} \big\{z ; |z| = n + \frac12\big\}\) (see exercise 2.2.3 below), and then \(\exists M > 0\) if \[\begin{align} |z| = n + \frac12 &\Rightarrow | f_m (z) | = \Big| \frac{\pi \cot \pi z}{z^{2m}} \Big| \leq \frac{M}{\left(n + \frac12\right)^{2m}} \\ &\Rightarrow | I_n | \leq 2\pi \left(n + \frac12\right) \cdot \frac{M}{\left(n + \frac12\right)^{2m}} \\ &\Rightarrow \lim_{n \to \infty} I_n = 0. \end{align}\] Applying residue theorem gives \[\displaystyle \oint_{C_n} f_m (z) \,dz = \sum_{0 < |k| \leq n} \text{Res}(f_m, k) + \text{Res}(f_m, 0) = 2 \sum_{ k = 1}^n \frac{1}{k^{2m}} + \text{Res}(f_m, 0).\] As \(n \to \infty\) \[\displaystyle \sum_{k = 1}^{\infty} \frac{1}{k^{2m}} = - \frac{1}{2} \text{Res}(f_m, 0),\] if \(\sum_{n = 0}^{\infty} a_{2n} z^{2n}\) is the power series around \(0\) of \(f(z) = \pi z \cot \pi z\), we get \(\text{Res}(f_m, 0) = a_{2m}\) and because of exercise 2.2.2 below \[a_{2m} = \frac{(-1)^m (2\pi)^{2m} B_{2m}}{(2m)!},\] where \(B_{2m}\) are Bernoulli's numbers. \(_\square\)

The sequence of Bernoulli numbers \(B_n\) is defined by recurrence as follows: \[B_0 = 1, \quad {n \choose 0} B_0 + {n \choose 1} B_1 + \cdots + {n \choose n-1} B_{n - 1} = 0.\]

Here is "Exercise 2.2.2."

A. If \(\sum_{n = 0}^{\infty} a_{n} z^{n}\) is the power series around \(0\) of \(g(z) = \frac{z}{e^z - 1},\) prove that \(a_n = \frac{B_n}{n!},\) where \(B_n\) are Bernoulli numbers.
B. \(B_n\) is not bounded and \(B_{2n + 1} = 0, \space \forall n \in \mathbb{Z}^{+}\).
C. Get the power series around \(0\) of \(f(z) = \pi z \cot \pi z\), in terms of \(B_n\).

---

Applying the definition of Bernoulli numbers, \(B_0 = 1,\) and take \(n = 2\) then \(B_1 = -\frac12 \to B_2 = \frac16 \to B_3 = 0 \to B_4 = -\frac1{30},\ldots\)

\[\zeta(2) = \frac{\pi^2}{6}, \quad \zeta(4) = \frac{\pi^4}{90}, \quad \ldots.\]

A. \(g(z) = \frac{z}{e^z - 1}\) has an avoidable singularity at \(z = 0,\) which vanishes defining \(g(0) = 1\). Using the method of indeterminate coefficients, \[1 = g(z) \frac{e^z - 1}{z} = \big(a_0 + a_1 z + a_2 z^2 + \cdots\big) \cdot \left(1 + \frac{z}{2!} + \frac{z^2}{3!} + \cdots\right),\] which implies \[1 = a_0, \ 0 = a_1 + \frac{a_0}{2!}, \ 0 = a_2 + \frac{a_1}{2!} + \frac{a_0}{3!}, \ ..., \ 0 = a_{n - 1} + \frac{a_{n - 2}}{2!} + \frac{a_{n - 3}}{3!} + \cdots + \frac{a_0}{n!}.\] Replacing \(a_n = \frac{B_n}{n!}\space \forall n \ge 0\) and multiplying by \(n!,\) we get the formula \[B_0 \frac{n!}{n!} + B_1 \frac{n!}{1!(n - 1)!} + \cdots + B_{n - 2} \frac{n!}{2!(n - 2)!} + B_{n - 1} \frac{n!}{1!(n - 1)!} = 0.\]

B. If \(B_n\) was bounded, then \(g(z) = \frac{z}{e^z - 1}\) would have a radius of converge \(\infty\) around \(0\). Contradiction. \(\Big(\)The radius of convergence around \(0\) of \(g\) is \(2\pi\) \(\big(g \in \mathcal{H}\big(D(0, 2\pi)\big)\) because of C below\(\big).\)\(\Big)\)

For proving that \(B_{2n + 1} = 0 \space \forall n \in \mathbb{Z}^{+}\), observe that \(a_1 = - \frac12 \Rightarrow\) \[h(z) = g(z) - a_1 z = \frac{z}{e^z - 1} + \frac{z}{2} = \frac{2z + z(e^z - 1)}{2(e^z -1)} = \frac{z(e^z + 1)}{2(e^z -1)} = h(-z),\] which implies that \(h\) is an even function, and this then implies that \(a_{2n + 1} = 0 \space \forall n \in \mathbb{Z}^{+} \Rightarrow B_{2n + 1} = 0 \space \forall n \in \mathbb{Z}^{+}\).

C. We have \[ f(z) = \pi z \cot \pi z = i \pi z \frac{e^{i \pi z} + e^{-i \pi z}}{e^{i \pi z} - e^{-i \pi z}} = i \pi z \frac{e^{2i \pi z} + 1}{e^{2i \pi z} - 1} = h(2i \pi z)\] as above, i.e. \[\displaystyle h(z) = \sum_{n = 0}^{\infty} \frac{B_{2n}}{(2n)!} z^{2n} \Rightarrow f(z) = \pi z \cot \pi z = \sum_{n = 0}^{\infty} \frac{(-1)^n (2\pi)^{2n} B_{2n}}{(2n)!} z^{2n}.\ _\square\]

Here is "Exercise 2.2.3."

Let \(A_{\epsilon} \subset \mathbb{C}\) be the complementary set of \(\displaystyle \cup_{n \in \mathbb{Z}} D(n\pi, \epsilon), \text{ and } 0 < \epsilon < 1\). Prove that \(\exists C_{\epsilon} > 0\) such that \(|\sin z| \ge C_{\epsilon}\) and \(|\tan z| \ge C_{\epsilon}\) for all \(z \in A_{\epsilon}.\)

---

Since the function \(|\sin z|\) is periodic of period \(\pi\), it will suffice to prove that there exists \(C_{\epsilon} > 0 \) such that \(|\sin z| \ge C_{\epsilon}\) for each \(z \in H(\epsilon) = \left\{ z:-\frac{\pi}2 \leq \text{ Real (z) } \leq \frac{\pi}2, |z| > \epsilon\right\}:\) \[|\sin z| = \left|\frac{e^{iz} - e^{-iz}}{2i}\right| \ge \frac{1}{2} \Big| |e^{iz}| - |e^{-iz}| \Big| = \frac{1}{2} \Big| e^{-y} - e^{y} \Big|, \space z = x + iy.\] The function \(h(y) = \frac{1}{2} \Big| e^{-y} - e^{y} \Big|\) is an even function and an increasing function in \([1, \infty] \Rightarrow h(y) \ge h(1)\) if \(|y| \ge 1\). Then \[|\sin z| \ge h(1)\ \text{ if }\ z = x + iy \in H(\epsilon)\ \text{ and }\ |y| \ge 1.\] On the other hand, \[ K(\epsilon) = \{ x + iy \in H(\epsilon); \space |y| \leq 1\}\] is a compact set where \(|\sin z| \neq 0\).Therefore, \(\exists a \in K(\epsilon)\) such that \(|\sin z| \ge |\sin a| > 0, \space \forall z \in K(\epsilon)\). Taking \(C_{\epsilon} = \text{ minimum } \{h(1), |\sin a|\}\), we get our goal for \(|\sin z|\). Make the same reasoning with \(|\tan z|\)... \(_\square\)