Work out the value of $\int_{0}^{\infty}{\cos{x}\over x}[1-\cos(nx)]\mathrm dx$ How to show that $(1)=\ln(n^2-1)?$

$$2\int_{0}^{\infty}{\cos{x}\over x}[1-\cos(nx)]\mathrm dx=\ln(n^2-1)\tag1$$

$n>1$
$\cos(nx)=2\cos{x}\cos[(n-1)x]-\cos[(n-2)x]$
$$2\int_{0}^{\infty}\left({\cos{x}\over x}-{{2\cos^2{x}\cos[(n-1)x]\over x}}+{\cos{x}\cos[(n-2)x]\over x}\right)\mathrm dx\tag2$$
 A: \begin{align}
2\int_{0}^{\infty}{\cos{x}\over x}[1-\cos(nx)]\mathrm dx 
&= \int_0^\infty\dfrac{2\cos x-\cos(n+1)x-\cos(n-1)x}{x}\mathrm dx \\
&= \int_0^\infty\dfrac{2s}{s^2+1}-\dfrac{s}{s^2+(n+1)^2}-\dfrac{s}{s^2+(n-1)^2}\mathrm ds \\
&= \ln\dfrac{s^2+1}{\sqrt{(s^2+(n+1)^2)(s^2+(n-1)^2)}}\Big|_0^\infty \\
&= \color{blue}{\ln(n^2-1)}
\end{align}
A: You may just use the complex version of Frullani's theorem, or the Laplace transform.
$$ \int_{0}^{+\infty}\frac{\cos(x)-\cos(x)\cos(nx)}{x}\,dx =\\= \int_{0}^{+\infty}\left[\frac{s}{1+s^2}-\frac{s}{2}\left(\frac{1}{(n+1)^2+s^2}+\frac{1}{(n-1)^2+s^2}\right)\right]\,ds $$
The latter is an elementary integral, equal to $\frac{1}{2}\left(\log(n-1)+\log(n+1)\right)$ for any $n>1$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\left.2\int_{0}^{\infty}{\cos\pars{x} \over x}\,\bracks{1 -\cos\pars{nx}}\dd x
\,\right\vert_{\ n\ >\ 1}
\\[5mm] = &\
\int_{0}^{\infty}{2\cos\pars{x} - \cos\pars{\bracks{n + 1}x} - \cos\pars{\bracks{n - 1}x}\over x}\,\,\dd x
\\[5mm] = &\
\Re\int_{0}^{\infty}\bracks{{2\expo{\ic x} -
\expo{\ic\pars{n + 1}x} - \expo{\ic\pars{n - 1}x}}}
\pars{\int_{0}^{\infty}\expo{-xt}\,\dd t}\dd x
\\[5mm] = &\
\Re\int_{0}^{\infty}\bracks{{2 \over t - \ic} -
{1 \over t - \pars{-n - 1}\ic} - {1 \over t - \pars{-n + 1}\ic}}\dd t
\\[5mm] = &\
\lim_{R \to \infty}\int_{0}^{R}\bracks{{2t \over t^{2} + 1} -
{t \over t^{2} + \pars{n + 1}^{2}} - {t \over t^{2} + \pars{n - 1}^{2}}}\,\dd x
\\[5mm] = &\
\lim_{R \to \infty}\braces{\ln\pars{R^{2} + 1} -
{1 \over 2}\ln\pars{R^{2} + \bracks{n + 1}^{2} \over \pars{n + 1}^{2}} -
{1 \over 2}\ln\pars{R^{2} + \bracks{n - 1}^{2} \over \pars{n - 1}^{2}}}
\\[5mm] = &\
\bbx{\ln\pars{n^{2} - 1}\,,\qquad n > 1}
\end{align}
A: $\displaystyle \int\limits_0^{a\geq 0} \frac{\cos x}{x}(1-\cos(nx))dx = \int\limits_0^a \int\limits_0^n \sin(tx)dt \cos x dx =  \int\limits_0^n \int\limits_0^a \sin(xt) \cos x dx dt $ 
$\displaystyle =  \int\limits_0^n \frac{t-t\cos(a)\cos(at) - \sin(a)\sin(at) }{t^2-1} dt $
$\displaystyle =\frac{1}{2} (\ln|n^2-1|-\text{Ci}(a|1+n|)-\text{Ci}(a|1-n|)+2\text{Ci}(a))\,\, \to \,\,\frac{1}{2} \ln|n^2-1|$ 
for $\,\,a\to\infty\,\,$ where $\,\,|n|\neq 1\,\,$ ; $\enspace\text{Ci}(x)\,\,$ is the Cosine Integral 
(e.g. http://mathworld.wolfram.com/CosineIntegral.html)
A: If you are not comfortable or familiar with the Laplace transform method suggested by @MyGlasses and @Jack D'Aurizio you could try Feynman's trick for differentiating under the integral sign. This, as I will show, is essentially the Laplace transform method in disguise.  
Since 
$$2 \cos x \cos nx = \cos (n - 1) x + \cos (n + 1) x,$$
the integral can be rewritten as
$$2 \int^\infty_0 \frac{\cos x(1 - \cos nx)}{x} \, dx = \int^\infty_0 \frac{2 \cos x - \cos (n - 1)x - \cos (n + 1) x}{x} \, dx, \quad n > 1.$$
Now consider
$$I(a) = \int^\infty_0 \frac{e^{-ax} \{2 \cos x - \cos (n - 1)x - \cos (n + 1)x \}}{x} \, dx, \quad a \geqslant 0.$$
Differentiating under the integral sign with respect to $a$ we have
$$I'(a) = -\int^\infty_0 \left (2 e^{-ax} \cos x - e^{-ax} \cos (n - 1)x - e^{-ax} \cos (n + 1) x \right ) \, dx.$$
Noting that
$$\int^\infty_0 e^{-ax} \cos (kx) \, dx = \frac{a}{a^2 + k^2},$$
a result that can be found using integration by parts twice, one has
$$I'(a) = - \left [\frac{2a}{a^2 + 1} - \frac{a}{a^2 + (n - 1)^2} - \frac{a}{a^2 + (n + 1)^2} \right ].$$
Note here that this is essentially the Laplace transform method in disguise. What we in effect have done is found the Laplace transform for the cosine function.
We require $I(0)$. Since $I(\infty) =0$, we have
$$\int^\infty_0 I'(a) \, da = I(\infty) - I(0) = -I(0),$$
or
$$I(0) = \int^\infty_0 \left [\frac{2a}{a^2 + 1} - \frac{a}{a^2 + (n - 1)^2} - \frac{a}{a^2 + (n + 1)^2} \right ] \, da,$$
and is exactly the same point reached by both @MyGlasses and @Jack D'Aurizio using either a method based on Laplace transforms or the complex version of Frullani's theorem. 
