Hay otros métodos para evaluar el $\int_1^{\infty}\frac{\{x\}-\frac{1}{2}}{x}\ \mathrm dx$?

Question

$$\displaystyle\int_1^{\infty}\frac{\{x\}-\frac{1}{2}}{x}\ \mathrm dx$$

donde $\{x\}$ denota la parte fraccionaria de x (por ejemplo, - $\{3.141\}=0.141)$

Estoy buscando métodos alternativos (tal vez también los que son más cortos) a lo que ya conoce (método que he utilizado más adelante), para Evaluar esto.

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El único método que conozco:

$$\ln(n!)=\sum_{k=2}^{n} \ln k$$

$$\ln(n!)=\sum_{k=2}^{n} \color{brown}{\int_1^k \frac{dx}{x}}$$

Desde: $$\color{brown}{\int_1^k \frac{dx}{x}}=\sum_{j=1}^{k-1} \int_j^{j+1} \frac{dx}{x}$$

$$\ln(n!)=\sum_{k=2}^{n} {\Bigg\{\sum_{j=1}^{k-1} \int_j^{j+1} \frac{dx}{x}}\Bigg\}$$

$$\ln(n!)=(n-1)\int_1^2 \frac{dx}{x}+(n-2)\int_2^3 \frac{dx}{x}+\cdots+\int_{n-1}^n \frac{dx}{x}$$

$$ln(n!)=\int_1^2 \frac{(n-1) dx}{x}+\int_2^3 \frac{(n-2)dx}{x}+\cdots+\int_{n-1}^n \frac{dx}{x}$$

La forma general de los términos en la última suma es, con $1\leq j\leq n-1$

$$\int_j^{j+1} \frac{(n-j)}{x}dx=\int_j^{j+1} \frac{(n-[x])}{x} dx$$

donde la notación $[x]$ significa que la parte entera de la $x$

porque, para la integración intervalo de $j\leq x <j+1$

$$\ln(n!)=\int_1^n \frac{n-[x]}{x} dx$$

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Ahora :

$$x=[x]+\{x\}\implies[x]=x-\{x\}$$

$$\Large{\color{orange}{\ln(n!)}} = \int_1^n \frac{n-x+\{x\}}{x}dx=n\ln(n)-n+1+\int_1^n \frac{\{x\}}{x}dx$$

$$=n\ln(n)-n+1+\frac{1}{2}\ln(n)+\int_1^n \frac{\{x\}-\frac{1}{2}}{x}dx$$ $$\left(n+\frac{1}{2}\right)\ln(n)-n+1+\int_1^n \frac{\{x\}-\frac{1}{2}}{x} dx$$

$$=\ln\left(n^{n+\frac{1}{2}}\right)+\ln(e^{-n})+1+\int_1^n \frac{\{x\}-\frac{1}{2}}{x} dx$$

$$=\ln\left(e^{-n}n^{n+\frac{1}{2}}\right)+\ln\left(e^{1+\int_1^n \frac{\{x\}-\frac{1}{2}}{x}dx}\right)$$

$$=\color{crimson}{\Large{\ln\left(e^{-n}n^{n+\frac{1}{2}}e^{1+\int_1^n \frac{\{x\}-\frac{1}{2}}{x}dx}\right)}}$$

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$$\Large{\color{orange}{n!}=\color{Crimson}{e^{-n}n^{n+\frac{1}{2}}e^{1+\int_1^n \frac{\{x\}-\frac{1}{2}}{x}dx}}}$$

$$\Large{e^{1+\int_1^n \frac{\{x\}-\frac{1}{2}}{x}dx}=\frac{n!}{e^{-n}n^{n+\frac{1}{2}}}}$$

Si dejamos $n\to \infty$ Y el uso de Stirling de la fórmula, tenemos:

$$\Large{e^{1+\int_1^{\infty} \frac{\{x\}-\frac{1}{2}}{x}dx}}=\sqrt{2\pi}$$

Por lo tanto, $$1+\int_1^{\infty} \frac{\{x\}-\frac{1}{2}}{x}dx=\ln(\sqrt{2\pi})$$

$$\bbox[8pt,border:2px #0099FF solid]{\int_1^{\infty}\frac{\{x\}-\frac{1}{2}}{x} dx=-1+\ln(\sqrt{2\pi})}$$

Answer 1

\begin{align} \int_1^{\infty}\frac{\{x\}-\frac{1}{2}}{x}\ \mathrm dx &=\sum_{n=1}^\infty\int_n^{n+1}\frac{x-n-\frac{1}{2}}{x}\ \mathrm dx\\[9pt] &=\sum_{n=1}^\infty\left[1-\left(n+\frac{1}{2}\right)\ln\left(\frac{n+1}{n}\right)\right]\\[9pt] &=\sum_{n=1}^\infty\left[\ln e+\left(n+\frac{1}{2}\right)\ln\left(\frac{n}{n+1}\right)\right]\\[9pt] &=\sum_{n=1}^\infty \ln\left(e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}\right)\\ &= \ln\left(\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}\right)\\[9pt] &= \ln\left(\frac{\sqrt{2\pi}}{e}\right)\tag{%#%#%}\\[9pt] &=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\ln\left(\,\sqrt{2\pi}\,\right)-1}} \end{align}

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$\spadesuit$$

Answer 2

El uso de $\color{#00A000}{\text{telescoping series}}$$\color{#C00000}{\text{Stirling's Formula}}$, obtenemos $$ \begin{align} \int_1^\infty\frac{\{x\}-\frac12}{x}\mathrm{d}x &=\sum_{k=1}^\infty\int_0^1\frac{t-\frac12}{k+t}\mathrm{d}t\\ &=\lim_{n\to\infty}\sum_{k=1}^n\left(1\color{#00A000}{-(k+\tfrac12)\log\left(\frac{k+1}{k}\right)}\right)\\ &=\lim_{n\to\infty}\left(n\color{#00A000}{+\sum_{k=1}^n(k+\tfrac12)\log(k)-\sum_{k=2}^{n+1}(k-\tfrac12)\log(k)}\right)\\ &=\lim_{n\to\infty}\left(n+\color{#C00000}{\sum_{k=2}^n\log(k)}-(n+\tfrac12)\log(n+1)\right)\\ &=\lim_{n\to\infty}\left(n+\color{#C00000}{\left[(n+\tfrac12)\log(n)-n+\tfrac12\log(2\pi)\right]}-(n+\tfrac12)\log(n+1)\right)\\ &=\tfrac12\log(2\pi)+\lim_{n\to\infty}\left((n+\tfrac12)\log\left(\frac{n}{n+1}\right)\right)\\[4pt] &=\tfrac12\log(2\pi)-1 \end{align} $$

Answer 3

$\newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, nº 1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert}$ 'Un buen subproducto" de la siguiente evaluación es la identidad: $$ \color{#66f}{\large% \int_{1}^{\infty}{\llaves{x} - 1/2 \sobre x}\,\dd x} =\color{#66f}{\large\int_{0}^{1}\ln\pars{\Gamma\pars{x + 1}}\,\dd x} $$

\begin{align}&\color{#66f}{\large% \int_{1}^{\infty}{\braces{x} - 1/2 \over x}\,\dd x} =\int_{1}^{\infty}{x - \floor{x} - 1/2 \over x}\,\dd x \\[5mm]&=\int_{1}^{2}{x - 1 - 1/2 \over x}\,\dd x +\int_{2}^{3}{x - 2 - 1/2 \over x}\,\dd x + \cdots \\[5mm]&=\int_{0}^{1}{x - 1/2 \over x + 1}\,\dd x +\int_{0}^{1}{x - 1/2 \over x + 2}\,\dd x + \cdots =\lim_{N\ \to\ \infty}\int_{0}^{1} \sum_{n\ =\ 0}^{N}{x - 1/2 \over x + n + 1}\,\dd x \\[5mm]&=\lim_{N\ \to\ \infty}\int_{0}^{1}\pars{x - \half} \sum_{n\ =\ 0}^{\infty}\pars{{1 \over n + x + 1} - {1 \over n + x + N + 2}}\,\dd x \\[5mm]&=\lim_{N\ \to\ \infty}\int_{0}^{1}\pars{x - \half}\bracks{% \Psi\pars{x + N + 2} - \Psi\pars{x + 1}}\,\dd x \end{align} donde $\ds{\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}}$ es la Función Digamma .

A continuación, \begin{align}&\color{#66f}{\large% \int_{1}^{\infty}{\braces{x} - 1/2 \over x}\,\dd x} =\lim_{N\ \to\ \infty}\int_{0}^{1}\pars{x - \half} \bracks{\ln\pars{x + N + 2} - \Psi\pars{x + 1}}\,\dd x \\[5mm]&=\ \overbrace{% \lim_{N\ \to\ \infty}\int_{0}^{1}\pars{x - \half}\ln\pars{x + N + 2}\,\dd x} ^{\ds{=}\ \dsc{0}}\ -\ \int_{0}^{1}\pars{x - \half}\,\totald{\ln\pars{\Gamma\pars{x + 1}}}{x}\,\dd x \\[5mm]&=-\left.\ln\pars{\Gamma\pars{x + 1}}\pars{x - \half} \right\vert_{x\ =\ 0}^{x\ =\ 1} +\int_{0}^{1}\ln\pars{\Gamma\pars{x + 1}}\,\dd x \\[5mm]&=\underbrace{% -\ln\pars{\Gamma\pars{2}}\half + \ln\pars{\Gamma\pars{1}}\pars{-\,\half}} _{\ds{=}\ \dsc{0}}\ +\ \int_{0}^{1}\ln\pars{\Gamma\pars{x + 1}}\,\dd x \end{align}

$$ \color{#66f}{\large% \int_{1}^{\infty}{\llaves{x} - 1/2 \sobre x}\,\dd x} =\color{#66f}{\large\int_{0}^{1}\ln\pars{\Gamma\pars{x + 1}}\,\dd x} $$

Sin embargo, \begin{align} \int_{0}^{1}\ln\pars{\Gamma\pars{x + 1}}\,\dd x& =\int_{0}^{1}\ln\pars{x}\,\dd x +\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x =-1+\int_{0}^{1}\ln\pars{\Gamma\pars{1 - x}}\,\dd x \\[5mm]&=-1+\int_{0}^{1}\ln\pars{\pi \over \sin\pars{\pi x}\Gamma\pars{x}}\,\dd x \\[5mm]&=-1 + \ln\pars{\pi} -{1 \over\pi}\ \underbrace{\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\,\dd x} _{\ds{=}\ \dsc{-\pi\ln\pars{2}}} -\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x \\[5mm]&=\ln\pars{2\pi} - 2 - \int_{0}^{1}\ln\pars{\Gamma\pars{x + 1}}\,\dd x \end{align}

La integral de la $\ds{\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\,\dd x=-\pi\ln\pars{2}}$ aparece con frecuencia en M. SE .

Finalmente, $$ \color{#66f}{\large% \int_{1}^{\infty}{\llaves{x} - 1/2 \sobre x}\,\dd x} =\color{#66f}{\large\media\,\ln\pars{2\pi} - 1} $$