We know that $$\int{e^{ax} }\cos bx\,\mathrm{d}x=\frac{e^{ax}}{a^2+b^2}\left(a\cos \left(bx\right)+b\sin \left(bx\right)\right)+C$$ for $a,b\in\mathbb{R}$. I'm interested in solving the following related improper integral, $$I=\int_{0}^{\infty}\frac{{e^{-tx} }}{x}(\cos ax-\cos bx)\,\mathrm{d}x$$ with $t\gt 0$ and $a,\,b,\,t\in\mathbb{R}$.
Where should I begin?
On
The usual Frullani-integral-type technique works in this case:
\begin{align*} \int_{0}^{\infty} e^{-tx} \frac{\cos(ax) - \cos(bx)}{x} \, dx &= \int_{0}^{\infty} e^{-tx} \bigg( \int_{a}^{b} \sin(xu) \, du \bigg) \, dx \\ &= \int_{a}^{b} \bigg( \int_{0}^{\infty} e^{-tx} \sin(xu) \, dx \bigg) \, du \\ &= \int_{a}^{b} \frac{u}{u^2 + t^2} \, du \\ &= \frac{1}{2}\log\left( \frac{b^2+t^2}{a^2+t^2} \right). \end{align*}
This technique can be also thought as the Laplace-transform technique in disguise.
On
Let's first assume $I$ as this (note that $I$ can be taken as a function of $t$) $$I(t)=\int_{0}^{\infty}\frac{{\mathrm{e}^{-tx} }}{x}(\cos ax-\cos bx)\,\mathrm{d}x$$ So, differentiating with respect to $t$ would give you \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}I(t)&=-\int_{0}^{\infty}\mathrm{e}^{-tx}(\cos ax-\cos bx)\,\mathrm{d}x\\ &=-\int_{0}^{\infty}\mathrm{e}^{-tx}\cos ax\,\mathrm{d}x +\int_{0}^{\infty}\mathrm{e}^{-tx}\cos bx\,\mathrm{d}x \end{align} But you know the closed form solution for the latter integrals (note that as $x\to\infty$ the result simplifies), hence \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}I(t)&=-\frac{t}{t^2+a^2}+\frac{t}{t^2+b^2}\\ \end{align} By integrating you'll get $$I(t)=\frac12 \log\left(\frac{t^2+b^2}{t^2+a^2}\right)+C$$ To determine $C$ you should note that $I(t)$ has to be zero for the special case where $a=b$. This implies $C=0$. Finally $$I(t)=\frac12 \log\left(\frac{t^2+b^2}{t^2+a^2}\right)$$
Hint- Take the Laplace transform of $\cos ax$ $$\mathcal{L}\{\cos ax\}=\frac s{s^2+a^2}$$ And also note that if $\mathcal{L}\{f(x)\}=F(s)$ then: $$\mathcal{L}\{\frac {f(x)}x\}=\int_s^\infty F(\sigma)d\sigma$$