Integration with a little bit of Leibnitz rule The problem is simple. But the solution is where I get my doubts. Also I would appreciate if anyone could fill me in with all the necessary points where I need to check for convergation for the procedure to correct.
$$\int_0^\infty \frac{\cos(bx)-\cos(cx)}x e^{-ax} \,dx $$ for all $a,b,c \in \rm I\!R  ; a>0$
From here it's obvious that we can rewrite this as
$$\frac d{dt} \left.\int_0^\infty \frac{cos(tx)}x e^{-ax}\right|_c^b \,dx =\int_0^\infty \int_c^b {\sin(tx)} e^{-ax} \,dx \,dt= \int_c^b I\,dt$$
Then via double per partes 
$$I=\int_0^\infty {\sin(tx)} e^{-ax}\,dx = \left.\frac{-1}{a}\sin(tx)e^{-ax}\right|_{t=0}^\infty + \frac ta \int_0^\infty {cos(tx)} e^{-ax}dx =$$
$$=\left. \frac{-\sin(tx)e^{-ax}}a \right|_{t=0}^\infty - \left. \frac t a {\cos(tx)e^{-ax}}\right|_{t=0}^\infty - \frac{t^2}{a^2}\int_0^\infty {\sin(tx)} e^{-ax}\,dx$$
$$I=\frac ta-\frac{t^2}{a^2}I$$
$$I=\frac{\frac ta}{1+\frac{t^2}{a^2}}$$
$$\int_c^b\frac{\frac ta}{1+\frac{t^2}{a^2}}=\int_{c/a}^{b/a}\frac{u}{1+u^2}=\frac12\ln(\frac dc)$$
I find it odd that my final result is not dependant at all of parameter $a$ Just does not sound right.
And I would appreciate filling out where covnergation or any other possible breaking points need to be checked.
EDIT: A typo...probably more to come.
 A: In the OP, there was a flaw in the beginning of the development.  The integral of interest is $I(a)=\int_0^\infty \frac{\cos(bx)-\cos(cx)}{x}e^{-ax}\,dx$.  
But, $I(a)$ is not equal to $\left.\left(\frac{d}{dt}\int_0^\infty \frac{\cos(tx)}{x}e^{-ax}\,dx\right)\right|_{c}^{b}$.  In fact, this latter integral fails to exist due to the $\frac1x$ singularity at $x=0$.
Herein, we present two methodologies that we can use to evaluate $I(a)$.


NOTE: Since the integral is even in $b$ and $c$, we may assume without loss of generality that $b\ge 0$ and $c \ge 0$.



METHODOLOGY $(1)$:  Differentiating Under the Integral Sign

Here is a straightforward approach.  Let $I(a)$ be given by
$$I(a)=\int_0^\infty \frac{\cos(bx)-\cos(cx)}{x}e^{-ax}\,dx$$
Differentiating reveals
$$I'(a)=\int_0^\infty e^{-ax}\left(\cos(cx)-\cos(bx)\right)\,dx=\frac{a}{a^2+c^2}-\frac{a}{a^2+b^2}$$
Integrating $I'(a)$, we obtain
$$I(a)-I(0)=\int_0^a \left(\frac{t}{t^2+c^2}-\frac{t}{t^2+b^2}\right)\,dt=\frac12\log\left(\frac{a^2+c^2}{a^2+b^2}\right)-\log(c/b)$$
Using a slightly modified version of Frullani's Theorem, we find that $I(0)=\log(c/b)$. Therefore,

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{\cos(bx)-\cos(cx)}{x}e^{-ax}\,dx=\frac12\log\left(\frac{a^2+c^2}{a^2+b^2}\right)}$$



METHODOLOGY $(2)$:  Applying Frullani's Theorem for Complex Parameters

In THIS ANSWER, I showed that if $f(z)$ is analytic, then the following generalization of Frullani's Theorem holds for complex parameters $a$ and $b$:
$$\begin{align}
\int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx&=f(0)\log(|b/a|)\\\\
&+if(0)\left(\arctan\left(\frac{\text{Re}(a\bar b)-|a|^2}{\text{Im}(a\bar b)}\right)-\arctan\left(\frac{|b|^2-\text{Re}(a\bar b)}{\text{Im}(a\bar b)}\right)\right) \tag 1
\end{align}$$
Then, we can write
$$I(a)=\text{Re}\left(\int_0^\infty \frac{e^{-(a-ib)x}-e^{-(a-ic)x}}{x}\,dx\right)$$
Applying $(1)$ with $f(z)=e^{-z}$, $a\to a-ib$ and $b\to a-ic$ yields

$$\begin{align}
I(a)&=e^{-(0)}\log\left(\frac{\sqrt{a^2+c^2}}{\sqrt{a^2+b^2}}\right)\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac12\log\left(\frac{a^2+c^2}{a^2+b^2}\right)}
\end{align}$$

as expected!
A: I get:
$\frac {d}{dt}\int_0^\infty \frac{\cos(tx)}x e^{-ax}|_c^b dx \\
\int_0^\infty \int_c^b {\sin(tx)} e^{-ax} dt dx\\
 \int_c^b\int_0^\infty  {\sin(tx)} e^{-ax} dxdt\\
 \int_c^b \frac {t}{a^2+t^2} dt\\
 \frac 12 (\ln(a^2 + b^2) - \ln(a^2 + c^2))$
