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$\newcommand\Ga\Gamma \newcommand{\R}{\mathbb{R}} \newcommand{\de}{\delta} \newcommand{\ga}{\gamma} \newcommand{\Si}{\Sigma}$ We have to show that for $a:=-\ln(2^{1/b-1/2}/v)\in\R$ and $b:=\beta\in(1, …

We want to upper-bound $|I_{n,k}|$, where $$I_{n,k}:=\int_0^{\pi/3} e^{-itn} (1-e^{it})^k\,dt.$$ Integrating by parts, we have $$I_{n,k}=\frac{e^{-itn}}{-in}(1-e^{it})^k\Big|_0^{\pi/3} -\int_0^{\pi/3} …

A counterexample: $$f(z,t):=e^{tz}[1+(e^{|z|^2}-1)h(t)],$$ where $h(t):=e^{-1/|t|}$ for $t\ne0$, with $h(0):=0$. Then all the assumptions on $f$ hold, but the conclusion $$|f(z,t)|\le e^{c|z|}\ \;\for …

Note that $g(1)=g'(1)=1$ and for real $x\in(-1,1)$ \begin{equation*} g''(x)=\frac1{\pi\sqrt{1-x^2}}. \end{equation*} The map $z\mapsto1-z^2$ maps the set \begin{equation*} R:=\mathbb C\setminu …

$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No, there is no uniqueness here. Indeed, let $\hat f$ denote the Fourier transform of a (say integrable) function $f\colon\R\to\C$, so that $\hat f(t) …

This is to complement the inequality $$|(1+it)^n-1|\ge b_1(n,t):=|e^{int}-1|,\tag{10}\label{10}$$ proved by Terry Tao for real $t$ and natural $n$, by the following inequality: $$|(1+it)^n-1|\ge b_2(n …

This is to detail Carlo Beenakker's assertion about the poles of the integrand. Suppose that $t=x+iy$ is such a pole, where $x$ and $y$ are real. Then $$1-y=e^{-uy}\cos ux,\quad x=e^{-uy}\sin ux.$$ Su …

This is a slight modification of Alexander Kalmynin's answer. For $n\ge3$. let $z$ be a multiple root of $p_n$. Then $0=p_n'(z)=1+2z-nz^{n-1}$ and, as shown by Alexander Kalmynin, $|z|>1$. So, $n|z|^{ …

$\newcommand{\R}{\mathbb{R}} \newcommand{\si}{\sigma} \newcommand{\supp}{\operatorname{\mathrm supp}} \newcommand{\cch}{\operatorname{\mathrm cch}} $ If $F\in L^2$, then the condition \begin{equation …

Using the formula for the sum of the first $n$ terms of a geometric series, we have $$|f(x)|^2=\frac{\sin^2(Nx/2)}{\sin^2(x/2)}$$ and hence for $t\downarrow 0$ $$g(t)=2\int_0^t |f(x)|^2\,dx =2\int_0^t …

Such a function does not exist, because the constant value $1$ of $|f(z)|^2$ on the circle $\{z\in\mathbb C\colon|z|^2=1/2\}$ is less than the value $e^{1/4}$ of $|f(z)|^2$ at the center $z=0$ of the …

The key is the following integral representation: \begin{equation} \begin{aligned} s_n(z)&:= \sum_{j>n} \frac{z^j e^{-j/n}}j \\ &=\sum_{j>n} z^j e^{-j/n} \int_0^\infty du\,e^{-ju} \\ & …

$\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$As shown in the post by Alexandre Eremenko, the answer to this question is yes if $C=1$ (and hence if $C\le1$). On the othe …

$\newcommand\R{\mathbb R}$Assume that the integral $$I_n:=\oint_{\mathbb{S}^1} \frac{1-\hat{f}(\xi)}{1-\xi}\,\frac{d\xi}{\xi^{n+1}} =\int_0^{2\pi} \frac{1-\hat{f}(e^{it})}{1-e^{it}}\,\frac{e^{it}\,i\, …

Let $L$ denote the left-hand side of your identity \eqref{2}. Then, using the identity $$J_a(x)=\sum_{m\ge0}\frac{(-1)^m}{m!(m+a)!}(x/2)^{2m+a} \tag{$\dagger$}\label{3},$$ we get $$ \begin{split} L&=2 …