Calculate $\int_0^\infty\frac{\sin(x)\log(x)}{x}\mathrm dx$. 
Calculate $\displaystyle\int_0^\infty\dfrac{\sin(x)\log(x)}{x}\mathrm dx$.

I tried to expand $\sin(x)$ at zero, or use SI(SinIntegral) function, but it did not work. Besides, I searched the question on math.stackexchange, nothing found.
Mathematica tells me the answer is $-\dfrac{\gamma\pi}{2}$, I have no idea how to get it.
Thanks for your help!
 A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\color{#f00}{\int_{0}^{\infty}{\sin\pars{x}\ln\pars{x} \over x}\,\dd x} =
\int_{0}^{\infty}\sin\pars{x}\ln\pars{x}\int_{0}^{\infty}\expo{-xt}\,\dd t\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}\
\overbrace{\Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x}
^{\ds{\equiv\ \mc{J}}}\ \,\dd t\label{1}\tag{1}
\end{align}

Hereafter the $\ds{\ln}$-branch-cut runs along $\ds{\left(-\infty,0\right]}$ with $\ds{\ln\pars{z}\ \,\mrm{arg}}$ given by
  $\ds{-\pi < \mrm{arg}\pars{z} < \pi}$.


\begin{align}
\mc{J} & \equiv
\Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x =
\Im\bracks{{1 \over t - \ic}\int_{0}^{\pars{t - \ic}\infty}
\ln\pars{x \over t - \ic}\expo{-x}\,\dd x}
\\[5mm] & =
-\,\Im\braces{{t + \ic \over t^{2} + 1}\int_{\infty}^{0}\bracks{\ln\pars{x \over \root{t^{2} + 1}} + \arctan\pars{1 \over t}\ic}\expo{-x}\,\dd x}
\\[5mm] & = 
\Im\braces{{t + \ic \over t^{2} + 1}\bracks{%
\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x -
{1 \over 2}\ln\pars{t^{2} + 1}+ \arctan\pars{1 \over t}\ic}}
\end{align}


Note that
  
  $\ds{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x =
\left.\partiald{}{\mu}\int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x\,
\right\vert_{\ \mu\ =\ 0} = \Gamma\,'\pars{1}\ =\ \overbrace{\Gamma\pars{1}}^{\ds{=\ 1}}\
\overbrace{\Psi\pars{1}}^{\ds{-\gamma}}\ =\ -\gamma\quad}$ where $\ds{\Gamma}$ 
  
  and $\ds{\Psi}$ are the Gamma and Digamma Functions, respectively. Then,

\begin{align}
\mc{J} & \equiv \bbox[8px,#ffe,border:0.1em groove navy]{%
\Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x} =
-\,{\gamma \over t^{2} + 1} -
{1 \over 2}\,{\ln\pars{t^{2} + 1} \over t^{2} + 1} +
{t\arctan\pars{1/t} \over t^{2} + 1}
\\[5mm] & =
\bbox[8px,#ffe,border:0.1em groove navy]{-\,{\gamma \over t^{2} + 1} +
\totald{}{t}\bracks{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}}
\label{2}\tag{2}
\end{align}

Note that$\ds{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}$ vanishes out when $\ds{t \to \infty}$ and $\ds{t \to 0^{+}}$.

By replacing \eqref{2} in \eqref{1}:
\begin{align}
&\color{#f00}{\int_{0}^{\infty}{\sin\pars{x}\ln\pars{x} \over x}\,\dd x} =
-\gamma\int_{0}^{\infty}{\dd t \over t^{2} + 1} =
\color{#f00}{-\,{1 \over 2}\,\gamma\pi}
\end{align}
A: From the integral representation of the beta function and Euler's reflection formula, we get
\begin{align*}
\int_{0}^{\infty} \frac{\sin x}{x^s} \, \mathrm{d}x
&= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} t^{s-1}e^{-xt} \, \mathrm{d}t \right) \sin x \, \mathrm{d}x \\
&= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-tx} \sin x \, \mathrm{d}x \right) t^{s-1} \, \mathrm{d}t \\
&= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1}}{1+t^2} \, \mathrm{d}t \\
&= \frac{1}{2\Gamma(s)} \beta\left(\frac{s}{2}, 1-\frac{s}{2}\right) \\
&= \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)}.
\end{align*}
Here, interchanging the two integral signs is a priori valid on the strip $1 < \Re(s) <2$ by Fubini's theorem, and then this identity extends to the larger strip $0 < \Re(s) < 2$ by analytic continuation.
Now differentiating both sides, we get
$$ \int_{0}^{\infty} \frac{\sin x \log x}{x^s} \, \mathrm{d}x
\stackrel{(*)}{=} -\frac{\mathrm{d}}{\mathrm{d}s} \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)}
= \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)} \left( \psi(s) + \frac{\pi}{2}\cot\left(\frac{\pi s}{2}\right) \right), $$
where $\psi$ is the digamma function. So by plugging $s = 1$, we conclude:
$$ \int_{0}^{\infty} \frac{\sin x \log x}{x} \, \mathrm{d}x = -\frac{\gamma \pi}{2}. $$

Justification of $\text{(*)}$. Let us prove that
$$F(s) := \int_{0}^{\infty} \frac{\sin x}{x^s} \, \mathrm{d}x \tag{1}$$
is analytic on the open strip $\mathcal{S} = \{ s \in \Bbb{C} : 0 < \Re(s) < 2 \}$ and its derivative can be computed by differentiation under the integral sign (Leibniz's integral rule). A major issue of $\text{(1)}$ is that the integral converges only conditionally for $0 < \Re(s) \leq 1$ and hence poses technical difficulty in adapting the usual proof of Leibniz's integral rule. In order to circumvent this, we accelerate the speed of convergence using integration by parts:
$$ F(s) = \underbrace{\left[ \frac{1-\cos x}{x^s} \right]_{0}^{\infty}}_{=0} + s \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, \mathrm{d}x. $$
Note that the resulting integral is absolutely convergent on the strip $\mathcal{S}$. Now we claim that
$$ g(s) := \frac{F(s)}{s} = \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, \mathrm{d}x $$
is differentiable on $\mathcal{S}$ and its derivative can be computed by the Leibniz's integral rule. Fix $s \in \mathcal{S}$ and choose $\varepsilon$ so that $0 < \varepsilon < \Re(s) < 2-\varepsilon $. Then whenever $0 < |h| < \varepsilon$,
\begin{align*}
&\frac{g(s+h) - g(s)}{h} - \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, \mathrm{d}x \\
&\qquad = \int_{0}^{\infty} \frac{1-\cos x}{x^{1+s}} \left( \frac{x^{-h} - 1}{h} + \log x \right) \, \mathrm{d}x \\
&\qquad = \int_{0}^{\infty} \frac{1-\cos x}{x^{1+s}} \left( \int_{0}^{1} (1 - x^{-ht}) \log x \, \mathrm{d}t \right) \, \mathrm{d}x.
\end{align*}
Using the inequality $\cos x \geq 1 - x^2/2$, we find that the integrand is dominated by
$$ \left| \frac{1-\cos x}{x^{1+s}} \left( \int_{0}^{1} (1 - x^{-ht}) \log x \, \mathrm{d}t \right) \right|
\leq \frac{\min\{2, x^2/2\}}{x^{1+\Re(s)}} (1 + \max\{ x^{\varepsilon}, x^{-\varepsilon} \}) \left|\log x\right| $$
This dominating function is integrable. So by the dominated convergence theorem, as $h \to 0$, we have
$$ g'(s) = \lim_{h \to 0} \frac{g(s+h) - g(s)}{h} = \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, \mathrm{d}x. $$
Plugging this back, we know that $F(s)$ is differentiable on $\mathcal{S}$ . Moreover, by integration by parts,
\begin{align*}
F'(s)
&= g(s) + sg'(s) \\
&= g(s) + \underbrace{\left[ \frac{(1-\cos x)\log x}{x^s} \right]_{0}^{\infty}}_{=0} - \int_{0}^{\infty} \left( \frac{\sin x \log x}{x^s} + \frac{1-\cos x}{x^{1+s}} \right) \, \mathrm{d}x \\
&= - \int_{0}^{\infty} \frac{\sin x \log x}{x^s} \, \mathrm{d}x.
\end{align*}
Therefore $F'(s)$ coincides with the integral obtained by differentiation under the integral sign.
