Please confirm that $\int_{0}^{\infty}{\ln(x)\sin(x)\cos\left(x\over \sqrt{\phi}\right)\over x} dx={1\over 2}\pi \left(\ln\phi-\gamma\right)$
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
\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}
$$ \bbx{\ds{\int_{0}^{\infty} {\ln\pars{x}\sin\pars{x}\cos\pars{x\over \root{\phi}} \over x}\,\dd x = {\pi \over 2}\bracks{-\,{1 \over 2}\,\ln\pars{1 - \phi^{-1}} - \gamma}}} $$
Note that $\ds{-\ln\pars{1 - \phi^{-1}} = \ln\pars{3 + \root{5} \over 2}}$.
Since $\ds{\phi^{2} - \phi - 1 = 0 \implies -\,{1 \over 2}\,\ln\pars{1 - \phi^{-1}} = \ln\pars{\phi}}$, an alternative expression is $$ \bbx{\ds{\int_{0}^{\infty} {\ln\pars{x}\sin\pars{x}\cos\pars{x\over \root{\phi}} \over x}\,\dd x = \color{#f00}{+}\,{1 \over 2}\,\pi\bracks{\ln\pars{\phi} - \gamma}}} $$
Hint: let $$I(a)=\int_{0}^{\infty}x^a\sin(x)\cos\left(x\over \sqrt{\phi}\right)dx$$ and then use Mellon transform to calculate it. Finally evaluate $I'(-1)$ which will be the answer.