It seems that $\displaystyle\int_0^{+\infty}\frac{\cos x}{x}$ is divergent, so how to solve this problem?
$$\int_0^\infty \frac{\cos bx -\cos ax}{x}\, dx\quad,\quad\mbox{where}\,a,b>0$$
It's really a great challenge for me.
If there's any advice of hint I would be grateful!
On
Denote $$I(y)=\int_0^{+\infty}\frac{\cos bx-\cos ax}{x}e^{-yx}dx, y\in[0,+\infty)$$ Then $$I'(y)=-\int_0^{+\infty}(\cos bx-\cos ax)e^{-yx}dx=\frac{y}{y^2+a^2}-\frac{y}{y^2+b^2}$$ and $$I(+\infty)=0$$ So $$I(0)=-\int_0^{+\infty}(\frac{y}{y^2+a^2}-\frac{y}{y^2+b^2})dy=\ln\frac{a}{b}$$
On
\begin{align} \int_0^\infty \frac{\cos bx -\cos ax}{x}\, dx&=\int_0^\infty \bigg[\cos bx -\cos ax\bigg]\int_0^\infty e^{-xy}\,dy\,\, dx\\ &=\int_0^\infty\int_0^\infty \bigg[e^{-xy}\cos bx -e^{-xy}\cos ax\bigg] \,dx\, dy\\ &=\lim_{s\to\infty}\int_0^s \bigg[\frac{y}{y^2+b^2} -\frac{y}{y^2+a^2}\bigg] \,dy\\ &=\ln\left(\frac{a}{b}\right) \end{align}
On
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$\ds{a\,,\ b\ >\ 0}$.
\begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{\cos\pars{bx} - \cos\pars{ax} \over x}\,\dd x} =\Re\ \overbrace{\int_{0}^{\infty}{\expo{\ic bx} - \expo{\ic ax} \over x}\,\dd x} ^{\ds{\dsc{x}\ =\ \dsc{\ic t}\ \imp\ \dsc{t}\ =\ \dsc{-\ic x}}} \\[5mm]&=\Re\int_{0}^{-\infty\ic}{\expo{-bt} - \expo{-at} \over \ic t}\,\ic\,\dd t\\[5mm]&=\Re\lim_{R\ \to\ \infty}\braces{% \left.-\int_{-\pi/2}^{0}{\expo{-bt} - \expo{-at} \over t}\,\dd t\, \right\vert_{\, t\ =\ R\expo{\ic\theta}} -\int_{R}^{0}{\expo{-bt} - \expo{-at} \over t}\,\dd t} \end{align}
Indeed $\ds{\pars{~\dsc{\tt\mbox{there is a proof at the very end}}~}}$, $$ \lim_{R\ \to\ \infty}\Re\braces{\left.-\int_{-\pi/2}^{0}{\expo{-bt} - \expo{-at} \over t}\,\dd t\, \right\vert_{\, t\ =\ R\expo{\ic\theta}}} = 0 $$ such that
\begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{\cos\pars{bx} - \cos\pars{ax} \over x}\,\dd x} =\int_{0}^{\infty}{\expo{-bt} - \expo{-at} \over t}\,\dd t \\[5mm]&=-\int_{0}^{\infty}\ln\pars{t}\bracks{-b\expo{-bt} + a\expo{-at}}\,\dd t =\int_{0}^{\infty}\ln\pars{t \over b}\expo{-t}\,\dd t -\int_{0}^{\infty}\ln\pars{t \over a}\expo{-t}\,\dd t \\[5mm]&=\ln\pars{a \over b}\int_{0}^{\infty}\expo{-t}\,\dd t =\color{#66f}{\large\ln\pars{a \over b}} \end{align}
Note that
\begin{align} 0&<\verts{% \Re\braces{\left.-\int_{-\pi/2}^{0}{\expo{-bt} - \expo{-at} \over t}\,\dd t\, \right\vert_{\, t\ =\ R\expo{\ic\theta}}}} =\verts{\Re\int_{-\pi/2}^{0} \pars{\expo{-bR\expo{\ic\theta}} - \expo{-aR\expo{\ic\theta}}}\,\dd\theta} \\[5mm]&<\int_{-\pi/2}^{0} \pars{\expo{-bR\cos\pars{\theta}} + \expo{-aR\cos\pars{\theta}}}\,\dd\theta =\int_{0}^{\pi/2} \pars{\expo{-bR\sin\pars{\theta}} + \expo{-aR\sin\pars{\theta}}}\,\dd\theta \\[5mm]&<\int_{0}^{\pi/2} \pars{\expo{-2bR\theta/\pi} + \expo{-2aR\theta/\pi}}\,\dd\theta \\[5mm]&={\pi \over 2b}\,{1 - \expo{-bR} \over R} +{\pi \over 2a}\,{1 - \expo{-aR} \over R}\color{#66f}{\large\ \to\ 0\quad \mbox{when}\quad R\ \to\ \infty} \end{align}
On
The cosine integral is defined by $$ \text{Ci}(x) = - \int_{x}^{\infty} \frac{\cos t}{t} \ dt. $$
Using this definition, $$ \int_{x}^{\infty} \frac{\cos at}{t} \ dt = \int_{ax}^{\infty} \frac{\cos u}{u} \ du = - \text{Ci}(ax). $$
Therefore,
$$ \begin{align} \int_{0}^{\infty} \frac{\cos bx - \cos ax}{x} \ dx &= \lim_{\epsilon \to 0^{+}} \left( \int_{\epsilon}^{\infty} \frac{\cos bx}{x} \ dx - \int_{\epsilon}^{\infty}\frac{\cos ax}{x} \ dx \right) \\ &= \lim_{\epsilon \to 0^{+}} \left(-\text{Ci}(b \epsilon) + \text{Ci}(a \epsilon) \right) \\ &= \lim_{\epsilon \to 0^{+}} \left( - \ln(b \epsilon) + \ln(a \epsilon) +\mathcal{O}(\epsilon^{2})\right) \tag{1} \\ &= \lim_{\epsilon \to 0^{+}} \left( \ln \left( \frac{a}{b}\right) + \mathcal{O}(\epsilon^{2})\right) \\ &= \ln \left( \frac{a}{b} \right). \end{align}$$
$(1)$ http://en.wikipedia.org/wiki/Trigonometric_integral#Convergent_series
In general, if $f(x)$ is a continuous function over $(0, \infty)$ such that following two limits exist
$$\begin{cases} f_0 &= \lim\limits_{x\to 0} f(x),\\ f_\infty &= \lim\limits_{x\to\infty} f(x) \end{cases}$$ then $$\int_0^\infty \frac{f(bx) - f(ax)}{x} dx = (f_0 - f_\infty)\log\frac{a}{b}\tag{*1}$$
This is known as Frullani integral.
In your cases $f_0$ exists but $f_\infty$ doesn't. However, there is a generalization which only require the existence of following limit: $$f_\infty^{alt} = \lim_{L\to\infty}\frac1L\int_1^L f(t) dt$$
For $f(t) = \cos(t)$, $f_0 = 1, f_\infty^{alt} = 0$. $(*1)$ tells us
$$\int_0^\infty \frac{\cos(bx)-\cos(ax)}{x} dx = (1 - 0) \log\frac{a}{b} = \log\frac{a}{b}$$
For a proof of Frullani's theorem, look at answers in this question. In particular, this answer which cover the generalized version you need.