Proving of Integral $\int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}dx = \ln\left(\frac{a}{b}\right)$ 
Prove that
$$
\int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx = \ln\left(\frac{a}{b}\right)
$$

My Attempt:
Define the function $I(a,b)$ as
$$
I(a,b) = \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx
$$
Differentiate both side with respect to $a$ to get
$$
\begin{align}
\frac{dI(a,b)}{da} &= \int_{0}^{\infty}\frac{0-e^{-ax}(-x)}{x}\,dx\\
&= \int_{0}^{\infty}e^{-ax}\,dx\\
&= -\frac{1}{a}(0-1)\\
&= \frac{1}{a}
\end{align}
$$
How can I complete the proof from here?
 A: Note that the following is a general technique that can handle much harder problems. Recalling the Laplace transform 

$$ F(s) = \int_{0}^{\infty} f(x) e^{-sx} dx. $$

Consider the more general integral
$$ F(s) = \int_{0}^{\infty} \frac{e^{-bx}-e^{-ax}}{x} e^{-sx} dx \implies  F'(s) = -\int_{0}^{\infty} ({e^{-bx}-e^{-ax}}) e^{-sx} dx .$$
Now, it is just a matter of evaluating the last integral and integrating the answer with respect to $s$ and then taking the limit as $s\to 0$ to find the desired value.
Note: When you integrate with respect to $s$ do not forget the constant of integration. To find it use the fact that

$$ \lim_{s\to \infty} F(s) = 0. $$ 

A: $$
\begin{split}
\int_{0}^{\infty}\frac{\exp(-ax) - \exp(-bx)}{x}dx &= \lim_{\epsilon\to 0}\int_{\epsilon}^{\infty}\frac{\exp(-ax) - \exp(-bx)}{x}dx\\
&=\lim_{\epsilon\to 0}\left[\int_{\epsilon}^{\infty}\frac{\exp(-ax)}{x}dx - \int_{\epsilon}^{\infty}\frac{\exp(-bx)}{x}dx\right]\\
&=\lim_{\epsilon\to 0}\left[\int_{a\epsilon}^{\infty}\frac{\exp(-t)}{t}dt - \int_{b\epsilon}^{\infty}\frac{\exp(-t)}{t}dt\right]\\
&=\lim_{\epsilon\to 0}\int_{a\epsilon}^{b\epsilon}\frac{\exp(-t)}{t}dt=\lim_{\epsilon\to 0}\int_{a}^{b}\frac{\exp(-\epsilon u)}{u}du
\end{split}
$$
The integrand converges uniformly to $\frac{1}{u}$ within the finite integration limits, therefore we're allowed to move the limit inside the integral.
A: A problem-specific solution is as follows:
\begin{align*}
\int_{0}^{\infty} \frac{e^{-bx} - e^{-ax}}{x} \, dx
&= - \int_{0}^{\infty} \int_{a}^{b} e^{-xt} dt \, dx \\
&= - \int_{a}^{b} \int_{0}^{\infty} e^{-xt} dx \, dt \\
&= - \int_{a}^{b} \frac{dt}{t}
 = - \left[ \log x \right]_{a}^{b} = \log\left(\frac{a}{b}\right).
\end{align*}
Interchanging the order of integration is justified either by Fubini's theorem or Tonelli's theorem.
