Evaluar la Integral: $\int_0^\infty \frac{( \frac{1}{2} - \cos x )}{x} dx$

Question

Evaluar la Integral: $\int_0^\infty\left[\frac{1}{2} - \cos\left(x\right)\right]\,{\rm dx \over x}$

Me encontré con este límite: $\lim_{x\rightarrow\infty} -Ci(x) + Ci(1/x) +\ln(x)$, $\gamma$ ¿es?

Aquí $ Ci(x) = \gamma + \ln x + \int_0^x \frac{\cos t -1}{t} dt $ es el integral de coseno y $\gamma$ es la constante de Euler. El límite y el Integral parecen ser iguales.

Answer 1

$$\int_0^{+\infty}\left(\frac 1 2 - \cos x\right)\frac {dx} x = \gamma$$

\begin{align*}\int_{1/n}^n \left(\frac 1 2 - \cos x\right)\frac {dx} x &= \int_{1/n}^1 \left(1 - \cos x\right)\frac {dx} x -\int_1^n \cos x\frac {dx} x+ \frac 1 2 \left(\int_1^n \frac {dx} x -\int_{1/n}^1 \frac {dx} x\right) \\ &=\int_{1/n}^1 (1-\cos x) \frac {dx} x - \int_1^n \cos x \frac {dx} x \end{align*} $$\bbox[0.2 ex,borde:0.5 pt solid black]{\int_0^1 \frac {1-\cos(y)}y\,dy -\int_1^{+\infty} \frac {\cos(y)} y\,dy=\gamma}$$

He aquí la prueba de Gronwall, 1918. Para $n$ un entero positivo $$\begin{align} A(n) &= \int_0^{n\pi} \frac{1-\cos (y)}{y} dy - \ln ( n\pi ) \\ &= \int_0^1 \frac{1-\cos (y)}{y} dy + \int_1^{n\pi} \frac{1-\cos (y)}{y} dy - \ln ( n\pi ) \\ &= \int_0^1 \frac{1-\cos(y)}{t} dy - \int_1^{n\pi} \frac{\cos(y)}{y} dy \end{align}$$

Con un cambio de variable $x = 2 \pi ny$:

$$\begin{align} \int_0^{n\pi} \frac{1-\cos(x)}{x} xy &= \int_0^{\frac{1}{2}} \frac{1 - \cos (2\pi ny)}{y} dy \\ &= \pi \int_0^{\frac{1}{2}} \frac{1 - \cos (2 \pi ny)}{\sin(\pi y)} dy + \int_0^{\frac{1}{2}} g(y) dy - \int_0^{\frac{1}{2}} g(y) \cos ( 2 \pi ny) dy \end{align}$$

$g(y) = \frac{1}{y} - \frac{\pi}{\sin (\pi y)}$ que es continua en a$[0,\frac{1}{2}]$$g(0) = 0$.

Con Riemann-Lebesgue teorema de $$\int_0^\frac{1}{2} g(y) \cos ( 2 \pi n y ) dy \rightarrow 0 \text{ for } n \rightarrow + \infty$$

$$\begin{align} \int_0^\frac12 g(y) dy &= \lim_{s \to 0^+} \int_s^\frac12 \left( \frac 1y - \frac \pi{\sin(\pi y)} \right) dy \\ &= \lim_{s \to 0^+} \left[ \ln \left( \frac{y}{\tan ( \pi/2\;y )} \right) \right]_s^{\frac{1}{2}} = \ln ( \pi ) - 2 \ln (2) \end{align}$$

En el otro lado $$\begin{align} \pi \int_{0}^{\frac{1}{2}} \frac{1 - \cos ( 2 \pi n y)}{\sin ( \pi y )} dy &= 2\pi \int_{0}^{\frac{1}{2}} \sum_{k=1}^{n} \sin ( (2k-1) \pi y ) dy \\ &= \sum_{k=1}^{n} \frac{2}{2k-1} ={\ln (n) + 2 \ln (2) + \gamma + o(1)} \end{align}$$

Finalmente $$A(n) = \gamma + o(1)$$

Muchas gracias a Sladjan Stankovik para facilitar la comprensión de este problema para mí.

Answer 2

$\newcommand{\+}{^{\daga}} \newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, nº 1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\a la derecha\vert\,} \newcommand{\cy}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{1/\epsilon} \bracks{\media \cos\pars{x}}\,{\dd x \sobre x} = \gamma.\quad}$ $\ds{\quad\gamma}$ es la De Euler-Mascheroni Constante.

\begin{align} &\color{#c00000}{ \lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{1/\epsilon}\bracks{ \half - \cos\pars{x}}\,{\dd x \over x}} =\lim_{\epsilon \to 0^{+}}\bracks{-\int_{\epsilon}^{1/\epsilon}{ \cos\pars{x} - 1 \over x}\,\dd x - \half\int_{\epsilon}^{1/\epsilon}{\dd x \over x}} \\[3mm]&=\lim_{\epsilon \to 0^{+}}\braces{ -\bracks{{\rm Ci}\pars{1 \over \epsilon} - \ln\pars{1 \over \epsilon} - \gamma} + \bracks{{\rm Ci}\pars{\epsilon} - \ln\pars{\epsilon} - \gamma} -\half\ln\pars{1/\epsilon \over \epsilon}}\tag{1} \end{align} donde $\ds{{\rm Ci}\pars{x}}$ es uno de los Coseno Funciones Integrales.

La expresión de $\pars{1}$ puede escribirse como \begin{align} &\color{#c00000}{ \lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{1/\epsilon}\bracks{ \half - \cos\pars{x}}\,{\dd x \over x}} =\gamma + \lim_{\epsilon \to 0^{+}}\braces{ -{\rm Ci}\pars{1 \over \epsilon} + \bracks{{\rm Ci}\pars{\epsilon} - \ln\pars{\epsilon} - \gamma}}\tag{2} \end{align}

Sin embargo $$ \lim_{\epsilon \to 0^{+}}{\rm Ci}\pars{1 \over \epsilon} = 0\,,\qquad \lim_{\epsilon \to 0^{+}}\bracks{ {\rm Ci}\pars{\epsilon} - \gamma \ln\pars{\epsilon}} = 0 $$

tal que $\pars{2}$ conduce a $$\color{#00f}{\large \lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{1/\epsilon}\bracks{ \media \cos\pars{x}}\,{\dd x \sobre x} = \gamma} $$