Evaluación de un tipo particular de integral que involucra logaritmos y función trigonométrica

Question

¿Existe alguna forma cerrada para $$ \int _0 ^{\infty}\int _0 ^{\infty}\int _0 ^{\infty} \log(x)\log(y)\log(z)\cos(x^2+y^2+z^2)dzdydx$$ Si es así, ¿cómo se puede demostrar?

Answer 1

Una forma de atacar esto es explotar la simetría de la integral. Empieza por expandir el término del coseno en trozos individuales, es decir

$$\begin{align}\cos{(x^2+y^2+z^2)} &= \cos{x^2} \cos{(y^2+z^2)} - \sin{x^2} \sin{(y^2+z^2)}\\ &= \cos{x^2} \cos{y^2} \cos{z^2} - \cos{x^2} \sin{y^2} \sin{z^2} \\ &\quad -\sin{x^2} \sin{y^2} \cos{z^2} - \sin{x^2} \cos{y^2} \sin{z^2} \end{align}$$

Siguiente definición

$$C = \int_0^{\infty} dx \, \cos{x^2} \: \log{x} $$ $$S = \int_0^{\infty} dx \, \sin{x^2} \: \log{x} $$

Entonces la integral triple es igual a $C^3-3 C S^2$ .

Obtenemos $C$ y $S$ utilizando el teorema de Cauchy para encontrar $C+i S$ . Para ello, consideramos la integral

$$\oint_{\eta} dz \, e^{i z^2} \log{z}$$

donde $\eta$ es una cuña circular de radio $R$ y el ángulo $\pi/4$ en el semiplano superior con base en el eje real positivo. Se puede demostrar que la integral sobre el arco circular desaparece como $\pi \log{R}/(4 R)$ como $R \to \infty$ . Así, tenemos

$$C+i S = e^{i \pi/4} \int_0^{\infty} dt \, e^{-t^2} \left (\log{t} + i \frac{\pi}{4} \right ) $$

Se puede derivar el valor de esta integral observando que

$$\int_0^{\infty} dt \, e^{-t^2} \log{t} = -\frac{\sqrt{\pi}}{4} (\gamma + 2 \log{2})$$

Este valor se deriva de utilizar el valor de

$$\frac12 \left [\frac{d}{d\alpha} \Gamma \left (\frac{\alpha+1}{2} \right ) \right ]_{\alpha=0} = \frac14 \Gamma \left (\frac12 \right ) \psi \left ( \frac12 \right )$$

También utilizamos

$$\int_0^{\infty} dt \, e^{-t^2} = \frac{\sqrt{\pi}}{2}$$

para obtener

$$C = -\frac18 \sqrt{\frac{\pi}{2}} (\pi + 2 \gamma + 4 \log{2})$$ $$S = \frac18 \sqrt{\frac{\pi}{2}} (\pi - 2 \gamma - 4 \log{2})$$

y por lo tanto la integral triple es

$$C^3-3 C S^2 = \frac{\pi ^{3/2} \left(4 \gamma ^2-8 \gamma \pi +\pi ^2+16 \log{2} \,(\gamma -\pi +\log{2})\right) (2 \gamma +\pi +4 \log{2})}{512 \sqrt{2}}$$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \ln\pars{x}\ln\pars{y}\ln\pars{z}\cos\pars{x^{2} + y^{2} + z^{2}} \,\dd z\,\dd y\,\dd x:\ {\large ?}}$

\begin{align}&\color{#c00000}{\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \ln\pars{x}\ln\pars{y}\ln\pars{z}\cos\pars{x^{2} + y^{2} + z^{2}} \,\dd z\,\dd y\,\dd x} \\[3mm]&=\Re\pars{\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \ln\pars{x}\ln\pars{y}\ln\pars{z}\expo{\ic\pars{x^{2} + y^{2} + z^{2}}} \,\dd z\,\dd y\,\dd x} \\[3mm]&=\Re\pars{\bracks{\int_{0}^{\infty}\ln\pars{x}\expo{\ic x^{2}}\,\dd x}^{3}} ={1 \over 64}\,\Re\pars{\bracks{% \int_{0}^{\infty}x^{-1/2}\ln\pars{x}\expo{\ic x}\,\dd x}^{3}} \\[3mm]&={1 \over 64}\,\Re\pars{\bracks{\lim_{\mu \to -1/2}\partiald{}{\mu} \color{#00f}{\int_{0}^{\infty}x^{\mu}\expo{\ic x}\,\dd x}}^{3}} \quad\mbox{where}\quad -1 < \mu < 0\tag{1} \end{align}

Con $\ds{\ic x \equiv -t\quad\imp\quad x = \ic t}$ \begin{align}&\color{#00f}{\int_{0}^{\infty}x^{\mu}\expo{\ic x}\,\dd x} =\expo{\pi\mu\ic/2}\int_{0}^{-\infty\ic}t^{\mu}\expo{-t}\ic\,\dd x \\[3mm]&=\ic\expo{\pi\mu\ic/2}\pars{\int_{0}^{\infty}t^{\mu}\expo{-t}\,\dd x -\lim_{R \to \infty} \int_{\verts{z}\ =\ R\atop {\vphantom{\Huge A} -\pi/2\ <\ {\rm Arg}\pars{z}\ <\ 0}} z^{\mu}\expo{-z}\,\dd z}\tag{2} \end{align}

Sin embargo, \begin{align}&0 < \verts{\int_{\verts{z}\ =\ R\atop {\vphantom{\Huge A} -\pi/2\ <\ {\rm Arg}\pars{z}\ <\ 0}}z^{\mu}\expo{-z}\,\dd z} <\int_{-\pi/2}^{0}R^{\mu}\exp\pars{-R\cos\pars{\theta}}R\,\dd\theta \\[3mm]&=R^{\mu + 1}\int_{0}^{\pi/2}\expo{-R\sin\pars{\theta}}\,\dd\theta <R^{\mu + 1}\int_{0}^{\pi/2}\expo{-2R\theta/\pi}\,\dd\theta ={\pi \over 2}\,R^{\mu}\pars{1 - \expo{-R}} \end{align}

Entonces, $$ \lim_{R \to \infty} \int_{\verts{z}\ =\ R\atop {\vphantom{\Huge A} -\pi/2\ <\ {\rm Arg}\pars{z}\ <\ 0}} z^{\mu}\expo{-z}\,\dd z = 0 \quad\mbox{since}\quad \mu <0. $$ y, a partir de la expresión $\pars{2}$ , $$ \color{#00f}{\int_{0}^{\infty}x^{\mu}\expo{\ic x}\,\dd x} =\ic\expo{\pi\mu\ic/2}\Gamma\pars{\mu + 1} $$ donde $\ds{\Gamma\pars{z}}$ es el Función gamma ${\bf\mbox{6.1.1}}$ . También $$\lim_{\mu \to -1/2}\partiald{}{\mu} \color{#00f}{\int_{0}^{\infty}x^{\mu}\expo{\ic x}\,\dd x} =-\,{\pi \over 2}\,\expo{-\pi\ic/4}\Gamma\pars{\half} +\ic\expo{-\pi\ic/4}\Gamma\pars{\half}\Psi\pars{\half}\tag{3} $$ $\ds{\Psi\pars{z}}$ es el Función Digamma ${\bf\mbox{6.3.1}}$ .

Con $\ds{\Gamma\pars{\half} = \root{\pi}}$ ( véase ${\bf\mbox{6.1.8}}$ ) y $\ds{\Psi\pars{\half} = -\gamma - 2\ln\pars{2}}$ ( véase ${\bf\mbox{6.3.3}}$ ), expresión $\pars{3}$ se reduce a: $$ \lim_{\mu \to -1/2}\partiald{}{\mu} \color{#00f}{\int_{0}^{\infty}x^{\mu}\expo{\ic x}\,\dd x} =-\expo{-\pi\ic/4}\root{\pi}\braces{{\pi \over 2} +\bracks{\gamma + 2\ln\pars{2}}\ic}\tag{4} $$ $\ds{\gamma}$ es el Constante de Euler-Mascheroni ${\bf\mbox{6.1.3}}$ .

Con $\pars{1}$ y $\pars{4}$ encontramos: \begin{align}&\color{#c00000}{\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \ln\pars{x}\ln\pars{y}\ln\pars{z}\cos\pars{x^{2} + y^{2} + z^{2}} \,\dd z\,\dd y\,\dd x} \\[3mm]&={\pi^{3/2} \over 64}\,\Re\pars{\pars{% -\expo{-\pi\ic/4}\braces{{\pi \over 2} +\bracks{\gamma + 2\ln\pars{2}}\ic}}^{3}} \\[3mm]&=-\,{\root{2}\pi^{3/2} \over 2048} \Re\pars{\braces{\bracks{1 - \ic}\bracks{\pi + \varphi\ic}}^{3}} \quad\mbox{where}\quad\varphi\equiv 2\bracks{\gamma + 2\ln\pars{2}} \end{align}

\begin{align}&\color{#c00000}{\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \ln\pars{x}\ln\pars{y}\ln\pars{z}\cos\pars{x^{2} + y^{2} + z^{2}} \,\dd z\,\dd y\,\dd x} \\[3mm]&=-\,{\root{2}\pi^{3/2} \over 1024} \Re\pars{\bracks{-1 - \ic}\bracks{\pi + \varphi\ic}^{3}} \\[3mm]&=-\,{\root{2}\pi^{3/2} \over 1024} \Re\pars{\bracks{-1 - \ic} \braces{\pi\bracks{\pi^{2} - 3\varphi^{2}}+ \bracks{3\pi^{2} - \varphi^{2}}\varphi\ic}} \\[3mm]&=-\,{\root{2}\pi^{3/2} \over 1024}\braces{% \pi\bracks{3\varphi^{2} - \pi^{2}} + \bracks{3\pi^{2} - \varphi^{2}}\varphi} \\[3mm]&=-\,{\root{2}\pi^{3/2} \over 1024}\pars{% -\varphi^{3} + 3\pi\varphi^{2} + 3\pi^{2}\varphi - \pi^{3}} \end{align}

\begin{align}&\color{#44f}{\large\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \ln\pars{x}\ln\pars{y}\ln\pars{z}\cos\pars{x^{2} + y^{2} + z^{2}} \,\dd z\,\dd y\,\dd x} \\[3mm]&=\color{#44f}{\large-\,{\root{2}\pi^{3/2} \over 1024} \pars{\pi + \varphi}\bracks{\pars{2 - \root{3}}\pi - \varphi} \bracks{\pars{-2 - \root{3}}\pi + \varphi}} \\[3mm]&\approx -1.3077\,,\qquad\qquad \varphi \equiv 2\bracks{\gamma + 2\ln\pars{2}} \end{align}

Answer 3

En realidad, resolver $$J(a):=\int_0^\infty \log(x) \cos(x^2 + a) dx$$ con los rendimientos de Maple $$J(a) = -\frac{\sqrt{2\pi}}{16} (4\log 2 \cos(a) + 2\cos(a) \gamma + \cos(a) \pi - 4\log 2 \sin(a) - 2\sin(a) \gamma + \sin(a) \pi)$$ Wich da esperanza hacia la existencia de la integral, ya que $$I = \int_0^\infty \int_0^\infty \log x \log y J(x^2 + y^2)\ dy\ dx$$ Y $J$ puede expresarse como una combinación lineal de $\cos(a)$ y $\cos(a + \frac{\pi}2) = \sin(a)$ Así que $$\int \log x J(x^2+y^2) dx = \sum_i \lambda_i J(?_i)$$

Answer 4

Desde $\log x \rightarrow \infty$ como $x\rightarrow\infty$ , intente probar que esta integral no existe (debido a la oscilación el $\cos$ factor trae consigo.