Evaluate this $\lim\limits_{n \to \infty}\int\limits_{1/n}^{n}\left(\cos x-\cos(x/2)\right)\frac{\ln x}{x}dx$ We seeking to evaluate this integral 
$$\lim_{n \to \infty}\int_{1/n}^{n}\left(\cos x-\cos(x/2)\right)\frac{\ln x}{x}\mathrm dx$$
using 
$\cos a -\cos b=-2\sin[(a+b)/2]\sin[(a-b)/2]$
$$\lim_{n \to \infty}-2\int_{1/n}^{n}\sin\left(\frac{3x}{4}\right)\sin\left(\frac{x}{4}\right)\frac{\ln x}{x}\mathrm dx$$
I stuck, not sure what to do next.

$v=\int \frac{\ln x}{x}=\frac{1}{2}\ln^2 x$
$u^{'}=-\sin x+\frac{1}{2}\sin(x/2)$
$$\int \left(\cos x-\cos(x/2)\right)\frac{\ln x}{x}\mathrm dx=\left[\cos x-\cos(x/2)\right]\frac{\ln^2 x}{2}-\frac{1}{2}\int \left[-\sin x+\frac{1}{2}\sin(x/2)\right]\ln^2 x$$
this integral it getting more complicated due to $\ln^2 x$
 A: By letting $t=x/2$, we have that
$$\int_{1/n}^{n}\cos(x/2)\frac{\ln (x)}{x}dx=\int_{1/(2n)}^{n/2}\cos(t)\frac{\ln(t) +\ln(2)}{t}dt$$
Hence
$$\begin{align}\int_{1/n}^{n}\left(\cos x-\cos(x/2)\right)\frac{\ln(x)}{x}\,dx&=
\int_{n/2}^{n}\frac{\cos(x)}{x}\ln(x)dx-\int_{1/(2n)}^{1/n}\frac{\cos(x)}{x}\ln(x)\,dx \\
&\qquad-\ln(2)\int_{1/(2n)}^{n/2}\frac{\cos(x)}{x}\,dx.\end{align}$$
Now show that
$$\begin{align}
&\lim_{n\to \infty}\int_{n/2}^{n}\frac{\cos(x)}{x}\ln(x)\,dx=0,\\
&\lim_{n\to \infty}\int_{1/(2n)}^{1/n}\frac{\cos(x)-1}{x}\ln(x)\,dx=0.
\end{align}$$
Morever
$$\lim_{n\to \infty}\left(\int_{1/(2n)}^{n/2}\frac{\cos(x)}{x}\,dx-\ln(n)\right)
=\lim_{n\to \infty} (-\text{Ci}(\frac{1}{2n})-\ln(n))=\ln(2)-\gamma.$$
where $\text{Ci}(x)$ is the cosine integral (recall that
$\text{Ci}(x)=\ln(x)+\gamma+o(1)$, as $x\to 0$).
Therefore 
$$\int_{1/(2n)}^{n/2}\frac{\cos(x)}{x}dx=\ln(n)+\ln(2)-\gamma+o(1),$$
and 
$$\int_{1/(2n)}^{1/n}\frac{\cos(x)}{x}\ln(x)\,dx
=-\ln(2)\ln(n)-\frac{\ln^2(2)}{2}+o(1).$$
Thus we may conclude that 
$$\lim_{n \to \infty}\int_{1/n}^{n}\left(\cos x-\cos(x/2)\right)\frac{\ln (x)}{x} \, dx=\boxed{\gamma  \ln (2)-\frac{\ln^2(2)}{2}}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\lim_{n \to \infty}
\int_{1/n}^{n}\bracks{\cos\pars{x} -\cos\pars{x \over 2}}{\ln\pars{x} \over x}\,\dd x}
\\[5mm] = &\
\lim_{\nu \to 0}\partiald{}{\nu}\,
\bbox[10px,#fee]{\Re\int_{0}^{\infty}\pars{\expo{\ic x} -
\expo{\ic x/2}}x^{\nu - 1}\,\dd x}\label{1}\tag{1}
\end{align}

I'll "close" the integration around a quarter circle in the first quadrant with
$\ds{\left. z^{\nu}\right\vert_{\
{\large z\ \not=\ 0} \atop
{\large\nu\ \in\ \pars{-1,1}}} =
\verts{z}^{\nu}\expo{\ic\nu\arg\pars{z}}}$ and
$\ds{\arg\pars{z} \in \pars{-1,1}}$:
\begin{align}
&\bbox[10px,#fee]{\Re\int_{0}^{\infty}\pars{\expo{\ic x} -
\expo{\ic x/2}}x^{\nu - 1}\,\dd x}
\\[5mm] = &\
-\lim_{R \to \infty}
\Re\int_{0}^{\pi/2}\bracks{\exp\pars{\ic R\expo{\ic\theta}} -
\exp\pars{\ic\,{R\expo{\ic\theta} \over 2}}}\times
\\[2mm] &\
R^{\nu - 1}\expo{\ic\pars{\nu - 1}\theta}
R\expo{\ic\theta}\ic\,\dd\theta
\\[5mm] &\
-\Re\int_{\infty}^{0}\pars{\expo{-y} - \expo{-y/2}}
y^{\nu - 1}\expo{\ic\pars{\nu - 1}\pi/2}\ic\,\dd y
\\[8mm] = &\
-\lim_{R \to \infty}
\Re\int_{0}^{\pi/2}\bracks{\exp\pars{\ic R\expo{\ic\theta}} -
\exp\pars{\ic\,{R\expo{\ic\theta} \over 2}}}\times
\\[2mm] &\
R^{\nu - 1}\expo{\ic\pars{\nu - 1}\theta}
R\expo{\ic\theta}\ic\,\dd\theta
\\[5mm] &\
+\cos\pars{\nu\pi \over 2}\int_{0}^{\infty}
\pars{\expo{-y} - \expo{-y/2}}y^{\nu - 1}\,\dd y
\label{2}\tag{2}
\end{align}

However,

\begin{align}
0 & < \verts{\int_{0}^{\pi/2}\exp\pars{\ic \Lambda\expo{\ic\theta}}
\Lambda^{\nu - 1}\expo{\ic\pars{\nu - 1}\theta}
\Lambda\expo{\ic\theta}\ic\,\dd\theta}
_{\ \nu\ \in\ \pars{-1,1}}
\\[5mm] & <
\Lambda^{\nu}\int_{0}^{\pi/2}
\exp\pars{-\Lambda\sin{\theta}}\,\dd\theta =
{\pi \over 2}\,
{1 - \exp\pars{-\Lambda} \over \Lambda^{1 - \nu}}
\,\,\,\stackrel{\mrm{as}\ \Lambda\ \to\ \infty}{\to}\,\,\, \color{red}{0}
\end{align}

such that the first term in the RHS of \eqref{2} vanishes out.

Then,
\begin{align}
&\bbox[10px,#fee]{\Re\int_{0}^{\infty}\pars{\expo{\ic x} -
\expo{\ic x/2}}x^{\nu - 1}\,\dd x} =
{1 - 2^{\nu} \over \nu}\cos\pars{\nu\pi \over 2}\Gamma\pars{\nu + 1}
\end{align}
With \eqref{1},
$$
\bbx{\bbox[10px,#ffd]{\lim_{n \to \infty}
\int_{1/n}^{n}\bracks{\cos\pars{x} -\cos\pars{x \over 2}}{\ln\pars{x} \over x}\,\dd x} =
\gamma\ln\pars{2} - {1 \over 2}\,\ln^{2}\pars{2}}
$$

Note that $\ds{\gamma = -\Psi\pars{1}}$.

