Ejercicio de revisión de ayuda en la integración de Apostol Vol 1

Question

5.11.9

Muestran que

$$\displaystyle \int_0^x{\frac{\sin(t)}{t+1}\ dt} \geq 0$$ $\forall x\geq 0$

Mi intento de-

Por el 2 º Teorema de valor medio,

$\displaystyle \int_0^x{\frac{\sin(t)}{t+1}\ dt} = \frac{1}{0+1} (1 - \cos(c)) + \frac{1}{x+1}(\cos(c) - \cos(x)) = 1 - \frac{x}{x+1}\cos(c) - \frac{\cos(x)}{x+1}$

Pero ahora la forma de demostrar lo anterior.

Answer 1

Por integración parcial, $$\int^x_0\frac{\sin t}{t+1}\,dt=\frac{1-\cos x}{x+1}+\int^x_0\frac{1-\cos t}{(t+1)^2}\,dt.$$ Everything on the RHS is $\ge0$ for $x\ge0.$
Nota: este dispositivo funciona para cada integral $\int^x_0f(t)\,\sin t\,dt,$ donde $f$ es positivo, diferenciable y monótona decreciente. El dispositivo de @robjohn no necesita differentiability, por lo que es (potencialmente) más general.

Answer 2

Sugerencia: Tenga en cuenta que $$\begin{align} &\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(t)}{t+1}\,\mathrm{d}t+\int_{(2k+1)\pi}^{(2k+2)\pi}\frac{\sin(t)}{t+1}\,\mathrm{d}t\\ &=\int_0^{\pi}\frac{\sin(t)}{t+2k\pi+1}\,\mathrm{d}t-\int_0^\pi\frac{\sin(t)}{t+(2k+1)\pi+1}\,\mathrm{d}t\\ &=\pi\int_0^{\pi}\frac{\sin(t)}{(t+2k\pi+1)(t+(2k+1)\pi+1)}\,\mathrm{d}t \end {alinee el} $, al final de cada intervalo de disminución, la integral es positiva.

Answer 3

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

Esta prueba es válida $\ds{{\Huge\forall}\ x\ \in\ \mathbb{R}_{\ \geq\ 0}}$ !!! que "$\underline{includes}$", como un caso particular, "$\ds{x \over 2\pi}$ es un $\ds{\underline{integer}}$" !!!.

$$\begin{align} \left.\int_{0}^{x}{\sin\pars{t} \over t + 1}\,\dd t\, \right\vert_{\ x\ \geq\ 0} & = \int_{0} ^{\left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + \left\{x/\pars{2\pi}\right\}2\pi} {\sin\pars{t} \over t + 1}\,\dd t \\[5mm] & = \int_{0}^{\left\lfloor x/\pars{2\pi}\right\rfloor 2\pi} {\sin\pars{t} \over t + 1}\,\dd t + \int_{0}^{\left\{x/\pars{2\pi}\right\}2\pi} {\sin\pars{t} \over t + \left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + 1}\,\dd t \\[5mm] & = \left\lfloor{x \over 2\pi}\right\rfloor\ \overbrace{\int_{0}^{2\pi}\sin\pars{t} \sum_{n = 0}^{\left\lfloor{x/\pars{2\pi}}\right\rfloor - 1} {1 \over t + 2\pi n + 1}\,\dd t}^{\ds{\mc{I}_{1}}} \\[2mm] & +\ \underbrace{\int_{0}^{\left\{x/\pars{2\pi}\right\}2\pi} {\sin\pars{t} \over t + \left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + 1} \,\dd t} _{\ds{\equiv\ \mc{I}_{2}}}\label{1}\tag{1} \end{align}$$

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$\ds{\Large\mc{I}_{1} \geq 0:\ ?}$ $$\begin{align} \mc{I}_{1} & = -\int_{-\pi}^{\pi}\sin\pars{t} \sum_{n = 0}^{\left\lfloor{x/\pars{2\pi}}\right\rfloor - 1} {1 \over t + \pars{2n + 1}\pi + 1}\,\dd t \\[5mm] & = -\int_{0}^{\pi}\sin\pars{t}\bracks{% \sum_{n = 0}^{\left\lfloor{x/\pars{2\pi}}\right\rfloor - 1} {1 \over t + \pars{2n + 1}\pi + 1} - \sum_{n = 0}^{\left\lfloor{x/\pars{2\pi}}\right\rfloor - 1} {1 \over -t + \pars{2n + 1}\pi + 1}}\,\dd t \\[5mm] & = 2\int_{0}^{\pi}\sin\pars{t} \sum_{n = 0}^{\left\lfloor{x/\pars{2\pi}}\right\rfloor - 1} {t \over \bracks{\pars{2n + 1}\pi + 1}^{\,2} - t^{2}}\,\dd t\ \bbox[10px,#ffe,border:1px dotted navy]{\Large\geq 0} \end{align}$$

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$\ds{\Large\mc{I}_{2} \geq 0:\ ?}$

1. Es claro que $\ds{I_{2} \geq 0}$ siempre $\ds{\bbx{0 \leq \braces{x \over 2\pi}2\pi \leq \pi}}$.
2. Al $\ds{\bbx{\pi < \braces{x \over 2\pi}2\pi < 2\pi}}$: $$\begin{align} \mc{I}_{1} & = \int_{0}^{\pi}{\sin\pars{t} \over t + \left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + 1}\,\dd t + \int_{\pi}^{\left\{x/\pars{2\pi}\right\}2\pi}{\sin\pars{t} \over t + \left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + 1}\,\dd t \\[5mm] & = \int_{0}^{\pi}{\sin\pars{t} \over t + \left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + 1}\,\dd t - \int_{0}^{\left\{x/\pars{2\pi}\right\}2\pi - \pi}{\sin\pars{t} \over t + \pars{2\left\lfloor x/\pars{2\pi}\right\rfloor + 1}\pi + 1}\,\dd t \\[1cm] & = \overbrace{% \int_{0}^{\left\{x/\pars{2\pi}\right\}2\pi - \pi}\sin\pars{t}\bracks{{1 \over t + \pars{\left\lfloor x/\pars{2\pi}\right\rfloor + \color{#f00}{0}}2\pi + 1} - {1 \over t + \pars{\left\lfloor x/\pars{2\pi}\right\rfloor + \color{#f00}{1/2}}2\pi + 1}}\,\dd t}^{\large\geq\ 0} \\[2mm] & +\ \underbrace{% \int_{\left\{x/\pars{2\pi}\right\}2\pi - \pi}^{\pi}{\sin\pars{t} \over t + \pars{2\left\lfloor x/\pars{2\pi}\right\rfloor + 1}\pi + 1}\,\dd t} _{{\large \geq\ 0}.\ \mbox{See}\ \ds{1.}}\quad \bbox[#ffe,10px,border:1px dotted navy]{\Large \geq 0} \end{align}$$