import Mathlib

open Complex Filter Function Metric Finset MeasureTheory intervalIntegral
open scoped BigOperators Topology

Show that $\int_{0}^{\infty} \frac{\sin x}{x} d x=\frac{\pi}{2}$.

theorem Shakarchi_exercise_2_2 :
    Tendsto (fun y => ∫ x in (0 : ℝ)..y, Real.sin x / x) atTop (𝓝 (Real.pi / 2)) := by
  sorry

We have $\int_0^{\infty} \frac{\sin x}{x} d x=\frac{1}{2 * i} \int_0^{\infty} \frac{e^{i * x}-e^{-i * x}}{x} d x=\frac{1}{2 * i}\left(\int_0^{\infty} \frac{e^{i * x}-1}{x} d x-\int_0^{\infty} \frac{e^{-i * x}-1}{x} d x=\right.$ $\frac{1}{2 * i} \int_{-\infty}^{\infty} \frac{e^{i * x}-1}{x} d x$. Now integrate along the big and small semicircles $C_0$ and $C_1$ shown below. For $C_0$ : we have that $\int_{C_0} \frac{1}{x} d x=\pi * i$ and $\left|\int_{C_0} \frac{e^{i * x}}{x} d x\right| \leq$ $2 *\left|\int_{C_{00}} \frac{e^{i * x}}{x} d x\right|+\left|\int_{C_{01}} \frac{e^{i * x}}{x} d x\right|$ where $C_{00}$ and $C_{01}$ are shown below $\left(C_{01}\right.$ contains the part of $C_0$ that has points with imaginary parts more than $a$ and $C_{00}$ is one of the other 2 components). We have $\left|\int_{C_{00}} \frac{e^{i * x}}{x} d x\right| \leq$ $\sup _{x \in C_{00}}\left(e^{i * x}\right) / R * \int_{C_{00}}|d x| \leq e^{-a} * \pi$ and $\left|\int_{C_{01}} \frac{e^{i * x}}{x} d x\right| \leq\left|\int_{C_{01}} \frac{1}{x} d x\right| \leq$ $\frac{1}{R} * C * a$ for some constant $C$ (the constant $C$ exists because the length of the curve approaches $a$ as $a / R \rightarrow 0)$. Thus, the integral of $e^{i * x} / x$ over $C_0$ is bounded by $A * e^{-a}+B * a / R$ for some constants $A$ and $B$. Pick $R$ large and $a=\sqrt{R}$ and note that the above tends to 0 . About the integral over $C_1$ : We have $e^{i * x}-1=1+O(x)$ for $x \rightarrow 0$ (this is again from $\sin (x) / x \rightarrow 1$ ),
4
so $\left|\int_{C_1} \frac{e^{i * x}-1}{x} d x\right| \leq O(1) *\left|\int_{C_1} d x\right| \rightarrow 0$ as $x \rightarrow 0$. Thus, we only care about the integral over $C_{00}$ which is $-\pi * i$. Using Cauchy's theorem we get that our integral equals $\frac{1}{2 * i}(-(\pi * i))=\pi / 2$.


import Mathlib

open Complex Filter Function Metric Finset
open scoped BigOperators Topology

Suppose $f$ is an analytic function defined everywhere in $\mathbb{C}$ and such that for each $z_0 \in \mathbb{C}$ at least one coefficient in the expansion $f(z) = \sum_{n=0}^\infty c_n(z - z_0)^n$ is equal to 0. Prove that $f$ is a polynomial.

theorem Shakarchi_exercise_2_13
    {f : ℂ → ℂ}
    (hf_diff : Differentiable ℂ f)
    (hf_vanish : ∀ z₀ : ℂ, ∃ n : ℕ,
      iteratedDeriv n f z₀ / (n.factorial : ℂ) = 0) :
    ∃ (p : Polynomial ℂ), f = fun z => Polynomial.aeval z p := by
  sorry

Say that at least one of the coefficients of the Taylor series vanishes is the same as saying that for every $a \in \mathbb{C}$ there is $m \in \mathbb{N}$ such that $f^{(m)}(a)=0$.
Consider $A_n:=\left\{z \in \mathbb{C}: f^{(n)}(z)=0\right\}$ for each $n \in \mathbb{N}$. Note that:
$f$ is polynomial iff $A_n$ is not countable for some $n \in \mathbb{N}$.
Indeed, if $f$ is polynomial of degree $n$, then $f^{(n+1)}(z)=0$ for all $z \in \mathbb{C}$, then $A_{n+1}=\mathbb{C}$, so, $A_{n+1}$ is not countable. Conversely, if there is $n \in \mathbb{C}$ such that $A_n$ is not countable, then $A_n$ has a limit point, then by Identity principle we have $f^{(n)}(z)=0$ for all $z \in \mathbb{C}$, so, $f$ is a polynomial of degree at most $n-1$.

Therefore, tt suffices to show that there is $n \in \mathbb{N}$ such that $A_n$ is not countable. Indeed, consider $\bigcup_{n \in \mathbb{N}} A_n$, by hypothesis for each $a \in \mathbb{C}$ there is $m \in \mathbb{N}$ such that $f^{(m)}(a)=0$, then $\mathbb{C} \subseteq \bigcup_{n \in \mathbb{N}} A_n$. Therefore, $\bigcup_{n \in \mathbb{N}} A_n$ is not countable, then there is $n \in \mathbb{N}$ such that $A_n$ is not countable.

to avoid data contamination, the formal proof is in proofnet_verified_gt.zip


import Mathlib

open Complex Filter Function Metric Finset MeasureTheory Real Set
open scoped BigOperators Topology FourierTransform RealInnerProductSpace Complex Interval

Show that $ \int_{-\infty}^{\infty} \frac{x \sin x}{x^2 + a^2} dx = \pi e^{-a}$ for $a > 0$.

theorem Shakarchi_exercise_3_4 (a : ℝ) (ha : 0 < a) :
    Tendsto (λ y => ∫ x in -y..y, x * Real.sin x / (x ^ 2 + a ^ 2))
    atTop (𝓝 (Real.pi * (Real.exp (-a)))) := by
  sorry


import Mathlib

open Complex Filter Function Metric Finset MeasureTheory
open scoped BigOperators Topology Polynomial

Prove that all entire functions that are also injective take the form $f(z) = az + b$, $a, b \in \mathbb{C}$ and $a \neq 0$.

theorem Shakarchi_exercise_3_14 {f : ℂ → ℂ} (hf : Differentiable ℂ f)
   (hf_inj : Function.Injective f) :
   ∃ (a b : ℂ), f = (λ z => a * z + b) ∧ a ≠ 0 := by
  sorry

Look at $f(1 / z)$. If it has an essential singularity at 0 , then pick any $z_0 \neq 0$. Now we know that the range of $f$ is dense as $z \rightarrow 0$. We also know that the image of $f$ in some small ball around $z_0$ contains a ball around $f\left(z_0\right)$. But this means that the image of $f$ around this ball intersects the image of $f$ in any arbitrarily small ball around 0 (because of the denseness). Thus, $f$ cannot be injective. So the singularity at 0 is not essential, so $f(1 / z)$ is some polynomial of $1 / z$, so $f$ is some polynomial of $z$. If its degree is more than 1 it is not injective (fundamental theorem of algebra), so the degree of $f$ is 1 .


Prove that if $f$ is holomorphic in the unit disc, bounded and not identically zero, and $z_{1}, z_{2}, \ldots, z_{n}, \ldots$ are its zeros $\left(\left|z_{k}\right|<1\right)$, then $\sum_{n}\left(1-\left|z_{n}\right|\right)<\infty$.

theorem Shakarchi_exercise_5_1
    (f : ℂ → ℂ) (hf : DifferentiableOn ℂ f (ball 0 1))
    (hb : Bornology.IsBounded (f '' (ball (0 : ℂ) 1)))
    (h0 : ¬ Set.EqOn f (0 : ℂ → ℂ) (ball (0 : ℂ) 1)) :
    Summable (fun z : {z : ℂ // f z = 0 ∧ z ∈ ball (0 : ℂ) 1} =>
      ((analyticOrderAt f (z : ℂ)).toNat : ℝ) * (1 - ‖(z : ℂ)‖)) := by
  sorry

Fix $\mathrm{N}$ and let $D(0, R)$ contains the first $\mathrm{N}$ zeroes of f. Let $S_N=\sum_{k=1}^N\left(1-\left|z_k\right|\right)=$ $\sum_{k=1}^N \int_{\left|z_k\right|}^1 1 d r$. Let $\eta_k$ be the characteristic function of the interval $\left.\| z_k \mid, 1\right]$. We have $S_N=\sum_{k=1}^N \int_0^1 \eta(r) d r=\int_0^1\left(\sum_{k=1}^N \eta_k(r)\right) d r \leq \int_0^1 n(r) d r$, where $n(r)$ is the number of zeroes of $f$ at the disk $D(0, r)$. For $r \leq 1$ we have $n(r) \leq \frac{n(r)}{r}$. This means that $S_N \leq \int_0^1 n(r) \frac{d r}{r}$. If $f(0)=0$ then we have $f(z)=z^m g(z)$ for some integer $\mathrm{m}$ and some holomorphic $g$ with $g(0) \neq 0$. The other zeroes of $\mathrm{f}$ are precisely the zeroes of $g$. Thus we have reduced the problem to $f(0) \neq 0$. By the Corollary of the Jensen's equality we get $S_N \leq \int_0^1 n(r) \frac{d r}{r}=\frac{1}{2 \pi} \int_0^{2 \pi} \log \left|f\left(R e^{i \pi}\right)\right| d \phi-\log |f(0)|<M$ since $f$ is bounded. The partial sums of the series are bounded and therefore the series converges.