Evaluating $\int_{0}^{\infty}\frac{\cos(x/a)-1}{x}\cdot\cos(x)\ln(x) dx$ 
$$I=\large \int_{0}^{\infty}\frac{\cos(x/a)-1}{x}\cdot\cos(x)\ln(x)\mathrm dx$$

Attempt: 
Split them into
$$\large \int_{0}^{\infty}\frac{\cos(x/a)}{x}\cdot\cos(x)\ln(x)\mathrm dx-\int_{0}^{\infty}\frac{1}{x}\cdot\cos(x)\ln(x)\mathrm dx=J-K$$
Applying integration by parts, $\large u=\cos(x)$
$\large \mathrm du=-\sin(x)$
$\large \mathrm dv=\frac{\ln(x)}{x}\mathrm dx$
$\large v=\frac{\ln^2(x)}{2}$
$$K=\frac{\cos(x)\ln^2(x)}{2}+\frac{1}{2}\int\sin(x)\ln^2(x)\mathrm dx$$
Try another integration by parts,
$\large u=\sin(x)\ln(x)$
$\large \mathrm du=\frac{}\sin(x){x}+\cos(x)\ln(x)$
$\large \mathrm dv=\sin(x)\mathrm dx$
$\large v=\sin(x)$
$$\int\sin(x)\ln^2(x)\mathrm dx=-\cos(x)\sin(x)\ln(x)+\int \left(\frac{\sin(2x)}{2x}+\ln(x)-\sin^2(x)\ln(x)\right)\mathrm dx$$
$$\int\sin(x)\ln^2(x)\mathrm dx=-\cos(x)\sin(x)\ln(x)+x\ln(x)-x+\int \left(\frac{\sin(2x)}{2x}-\sin^2(x)\ln(x)\right)\mathrm dx$$
I have try doing integration by parts to reduce it but, it is not working.
I have try and look up standard table of integrals but can't find much to help me.
I am unable to proceed. Can anyone please lead the way? 
 A: 
Lemma. For $a > 0$, we have
  $$ f(a) := \int_{0}^{\infty} \frac{\cos (ax) - \cos x}{x}\log x \, dx
= \frac{(2\gamma + \log a)\log a}{2}. $$

Proof. Consider the function $I(s) = \int_{0}^{\infty} \frac{\cos(ax) - \cos x}{x^{1+s}} \, dx$. Then $I(s)$ defines an analytic function on the strip $\operatorname{Re}(s) \in (-1, 1)$. Also, for $s \in (-1, 0)$ we have
\begin{align*}
I(s)
&= \int_{0}^{\infty} \left( \frac{1}{\Gamma(1+s)} \int_{0}^{\infty} u^s e^{-xu} \, du \right) (\cos(ax) - \cos x) \, dx \\
&= \frac{1}{\Gamma(1+s)} \int_{0}^{\infty} u^s \left( \int_{0}^{\infty} e^{-xu} (\cos(ax) - \cos x) \, dx \right) \, du \\
&= \frac{1}{\Gamma(1+s)} \int_{0}^{\infty} \left( \frac{u^{s+1}}{u^2+a^2} - \frac{u^{s+1}}{u^2 + 1} \right) \, du \\
&= \frac{(a^s - 1)\mathrm{B}(\frac{s}{2}+1,-\frac{s}{2})}{2\Gamma(1+s)} \\
&= -\frac{\pi (a^s - 1)}{2\sin(\frac{\pi s}{2})\Gamma(1+s)}.
\end{align*}
This formula continues to hold on all of the strip by the principle of analytic continuation. So it follows that
$$ f(a)
= -I'(0) = -\lim_{s \to 0} I'(s) = \frac{(2\gamma + \log a)\log a}{2}. $$

Using the lemma and $\cos(x/a)\cos x = \frac{1}{2}\left( \cos (\frac{a+1}{a}x) + \cos(|\frac{a-1}{a}|x) \right)$, we obtain
$$ \int_{0}^{\infty} \frac{\cos (ax) - 1}{x}\cos x\log x \, dx
= \frac{f\left(\frac{a+1}{a}\right) + f\left(\left|\frac{a-1}{a}\right|\right)}{2}. $$
A: Let $b=\frac{1}{a} > 0$ to avoid fractions. Using trigonometric identities we have
$$ [\cos(b x) - 1]\cos(x) = - \sin\left(\frac{b}{2}x\right) \left\{\sin\left[\left(\frac{b}{2}-1\right)x\right] + \sin\left[\left(\frac{b}{2}+1\right)x\right]\right\} \, . $$
In my answer to this question I have shown that
$$ \int \limits_0^\infty \frac{\sin(\alpha x) \sin(\beta x)}{x} \ln(x) \, \mathrm{d} x = \frac{1}{4} \ln \left(\frac{\lvert \alpha - \beta\rvert}{\alpha + \beta}\right)[\ln(\lvert \alpha^2-\beta^2\rvert) + 2 \gamma]$$
holds for $\alpha, \beta > 0$ if $\alpha \neq \beta$ . Therefore we obtain for $a \neq 1 \Leftrightarrow b \neq 1$
\begin{align}
I &= \int \limits_0^\infty \frac{[\cos(b x)-1]\cos(x)}{x} \ln(x) \, \mathrm{d} x \\
&= - \int \limits_0^\infty \frac{\sin\left(\frac{b}{2}x\right) \left\{\sin\left[\left(\frac{b}{2}-1\right)x\right] + \sin\left[\left(\frac{b}{2}+1\right)x\right]\right\}}{x} \ln(x) \, \mathrm{d} x \\
&= \frac{1}{4} \ln(\lvert 1-b\rvert) [\ln(\lvert 1-b\rvert) + 2 \gamma] + \frac{1}{4} \ln( 1+b) [\ln( 1+b) + 2 \gamma] \\
&= \frac{1}{4} [\ln^2 (\lvert 1-b\rvert) + \ln^2  (1+b) + 2 \gamma \ln (\lvert 1-b^2 \rvert)] \, .
\end{align}
A: This is not an answer but it is too long for a comment.
As I wrote in a comment, I suppose that there is a trick but I suppose that we must not split as $I=J−K$ since there is a serious problem with $K$ around $x=0$.
Close to zero
$$\frac{\cos \left(\frac{x}{a}\right)-1}{x}\,\cos (x)\,\log (x)=-\frac{x \log (x)}{2 a^2}+O\left(x^3\right) $$
$$\int \frac{\cos \left(\frac{x}{a}\right)-1}{x}\,\cos (x)\,\log (x)\,dx=\frac{x^2 (1-2 \log (x))}{8 a^2}+O\left(x^4\right)$$ does not make any problem.
Being short of ideas and lazy, just by curiosity, I gave it to a CAS and obtained the result
$$I=\frac{1}{16} \left(\log ^2\left(\frac{(a-1)^2}{a^2}\right)+4 \gamma  \log
   \left(\frac{(a-1)^2}{a^2}\right)+4 \log \left(\frac{1+a}{a}\right) \left(\log
   \left(\frac{1+a}{a}\right)+2 \gamma \right)\right)$$ How to get it, this is the question !
