Por favor, confirme que el $\int_{0}^{\infty}{\ln(x)\sin(x)\cos\left(x\over \sqrt{\phi}\right)\over x} dx={1\over 2}\pi \left(\ln\phi-\gamma\right)$

Question

Observar esta integral

Donde $\phi={\sqrt{5}+1\over 2}$

y $\gamma=0.5772156...$ es la Constante de Euler

$$\int_{0}^{\infty}{\ln(x)\sin(x)\cos\left(x\over \sqrt{\phi}\right)\over x}\mathrm dx={1\over 2}\pi \left(\ln\phi-\gamma\right)\tag1$$

$$I(a)=\int_{0}^{\infty}{\ln(x)\sin(x)\cos\left(ax\right)\over x}\mathrm dx\tag2$$

$$I^{'}(a)=-\int_{0}^{\infty}{\ln(x)\sin(x)\sin\left(ax\right)}\mathrm dx\tag3$$

El uso de

$2\sin A\sin B=\cos(A-B)-\cos(A+B)$

$\sin x\sin(ax)=\cos[(1-a)x]-\cos[(1+a)x]$

$$I^{'}(a)=-\int_{0}^{\infty}\cos[(1-a)x]\ln x \mathrm dx+\int_{0}^{\infty}\cos[(1+a)x]\ln x\mathrm dx\tag4$$

$$\int \cos[(1-a)x]\ln x \mathrm dx={\sin[(1-a)x]\ln x\over 1-a}-\int {\sin[(1-a)x]\over x}\tag 5$$

$$\int \cos[(1-a)x]\ln x \mathrm dx={\sin[(1+a)x]\ln x\over 1+a}-\int {\sin[(1+a)x]\over x}\tag 6$$

$(5)$ $(6)$ mostró divgergent integrales

Este enfoque que he probado no funciona, ¿qué otro método que podemos utilizar para verificar la $(1)?$

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\infty} {\ln\pars{x}\sin\pars{x}\cos\pars{x\over \root{\phi}} \over x}\,\dd x \\[5mm] = &\ {1 \over 2}\int_{0}^{\infty} {\ln\pars{x}\sin\pars{\bracks{1 + \phi^{-1/2}}x}\over x}\,\dd x + {1 \over 2}\int_{0}^{\infty} {\ln\pars{x}\sin\pars{\bracks{1 - \phi^{-1/2}}x}\over x}\,\dd x \\[5mm] = &\ {1 \over 2}\int_{0}^{\infty} {\ln\pars{x/\bracks{1 + \phi^{-1/2}}}\sin\pars{x}\over x}\,\dd x + {1 \over 2}\int_{0}^{\infty} {\ln\pars{x/\bracks{1 - \phi^{-1/2}}}\sin\pars{x}\over x}\,\dd x \\[5mm] = &\ -\,{1 \over 2}\,\ln\pars{1 - \phi^{-1}} \int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x + \int_{0}^{\infty}\ln\pars{x}\,{\sin\pars{x} \over x}\,\dd x \\[5mm] = &\ -\,{\pi \over 4}\,\ln\pars{1 - \phi^{-1}} + \bbox[#ffd,10px]{\ds{% \int_{0}^{\infty}\ln\pars{x}\,{\sin\pars{x} \over x}\,\dd x}} \end{align}

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\begin{align} &\bbox[#ffd,10px]{\ds{% \int_{0}^{\infty}\ln\pars{x}\,{\sin\pars{x} \over x}\,\dd x}} = \left.\partiald{}{\mu}\Im\int_{0}^{\infty}x^{\mu - 1}\expo{\ic x} \,\dd x\,\right\vert_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ \left.\partiald{}{\mu}\Im\int_{0}^{\infty\ic}x^{\mu - 1}\expo{\ic\pars{1 - \mu}\pi/2}\expo{-x} \pars{-\ic}\,\dd x\,\right\vert_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ \partiald{}{\mu}\bracks{\sin\pars{\mu\,{\pi \over 2}} \Gamma\pars{\mu}}_{\ \mu\ =\ 0^{+}} = \left.{1 \over 2}\,\pi\,\partiald{\Gamma\pars{\mu + 1}}{\mu} \right\vert_{\ \mu\ =\ 0^{+}} = -\,{1 \over 2}\,\pi\gamma \end{align}

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$$ \bbx{\ds{\int_{0}^{\infty} {\ln\pars{x}\sin\pars{x}\cos\pars{x\\raíz{\phi}} \over x}\,\dd x = {\pi \over 2}\bracks{-\,{1 \over 2}\,\ln\pars{1 - \phi^{-1}} - \gamma}}} $$

Tenga en cuenta que $\ds{-\ln\pars{1 - \phi^{-1}} = \ln\pars{3 + \root{5} \over 2}}$.

Desde $\ds{\phi^{2} - \phi - 1 = 0 \implica -\,{1 \over 2}\,\ln\pars{1 - \phi^{-1}} = \ln\pars{\phi}}$, una expresión alternativa es $$ \bbx{\ds{\int_{0}^{\infty} {\ln\pars{x}\sin\pars{x}\cos\pars{x\\raíz{\phi}} \over x}\,\dd x = \color{#f00}{+}\,{1 \over 2}\,\pi\bracks{\ln\pars{\phi} - \gamma}}} $$

Answer 2

Sugerencia: dejar $$I(a)=\int_{0}^{\infty}x^a\sin(x)\cos\left(x\over \sqrt{\phi}\right)dx$$ and then use Mellon transform to calculate it. Finally evaluate $I'(-1)$, que será la respuesta.