Please confirm that $\int_{0}^{\infty}{\ln(x)\sin(x)\cos\left(x\over \sqrt{\phi}\right)\over x} dx={1\over 2}\pi \left(\ln\phi-\gamma\right)$ Observe this integral
Where $\phi={\sqrt{5}+1\over 2}$ 
and $\gamma=0.5772156...$ is Euler's Constant

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

$$I(a)=\int_{0}^{\infty}{\ln(x)\sin(x)\cos\left(ax\right)\over x}\mathrm dx\tag2$$
$$I^{'}(a)=-\int_{0}^{\infty}{\ln(x)\sin(x)\sin\left(ax\right)}\mathrm dx\tag3$$
Using
$2\sin A\sin B=\cos(A-B)-\cos(A+B)$
$\sin x\sin(ax)=\cos[(1-a)x]-\cos[(1+a)x]$
$$I^{'}(a)=-\int_{0}^{\infty}\cos[(1-a)x]\ln x \mathrm dx+\int_{0}^{\infty}\cos[(1+a)x]\ln x\mathrm dx\tag4$$
$$\int \cos[(1-a)x]\ln x \mathrm dx={\sin[(1-a)x]\ln x\over 1-a}-\int {\sin[(1-a)x]\over x}\tag 5$$
$$\int \cos[(1-a)x]\ln x \mathrm dx={\sin[(1+a)x]\ln x\over 1+a}-\int {\sin[(1+a)x]\over x}\tag 6$$
$(5)$ and $(6)$ showed divgergent integrals
This approach  I have tried is not working, what other method can we use to verify $(1)?$
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\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
$$
\bbx{\ds{\int_{0}^{\infty}
{\ln\pars{x}\sin\pars{x}\cos\pars{x\over \root{\phi}} \over x}\,\dd x =
\color{#f00}{+}\,{1 \over 2}\,\pi\bracks{\ln\pars{\phi} - \gamma}}}
$$
A: 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.
