If $f(t)=\frac{e^{bt}-e^{at}}{t}$ how to compute $\mathcal{L} \{ f(t) \}$ I have a problem trying to solve 
$$
\int_0^{+\infty}\frac{e^{(b-s)t}}{t}
$$
I know that I can use the Ei (exponential integral function) but after that I don't know what's exactly this means.
I begin with the definition 
$
\mathcal{L} \{ f(t) \}:=\int_0^{+\infty}e^{-st}f(t)\,dt
$
In my laplace transform table says that $\mathcal{L} \{ f(t) \}=\ln \frac{s-a}{s-b}$
How I know that is true?
 A: Don't know what Frullani's integral is? No problem. Your integral can instead be converted to a double integral first by noting that
$$\int_{s - b}^{s - a} e^{-xt} \, dx = \frac{e^{-(s - b)t} - e^{-(s - a) t}}{t}.$$
Then 
\begin{align}
\mathcal{L} \{f(t)\} &= \int_0^\infty \frac{e^{-(s - b)t} - e^{-(s - a) t}}{t} \, dt\\
&= \int_0^\infty \int_{s - b}^{s - a} e^{-xt} \, dx \, dt\\
&= \int_{s - b}^{s - a} \int_0^\infty e^{-xt} \, dt \, dx\\
&= \int_{s - b}^{s - a} \frac{dx}{x}\\
&= \ln \left (\frac{s - a}{s - b} \right ).
\end{align}
Note when $a \neq b$, for convergence we require $s > \max \{a,b\}$.
A: Here, we will use Feynman's Trick for differentiating under the integral.  
Let $F(s)$ be defined by the integral 
$$\begin{align}
F(s)&=\int_0^\infty \frac{e^{-(s-b)t}-e^{-(s-a)t}}{t}\,dt\tag1
\end{align}$$
Differentiating under the integral in $(1)$ reveals
$$\begin{align}
F'(s)&=\int_0^\infty \left(e^{-(s-a)t}-e^{-(s-b)t}\right)\,dt\\\\
&=\frac1{s-a}-\frac1{s-b}\tag 2
\end{align}$$
Next, integrating both sides of $(2)$ we obtain
$$\int_{s}^\infty F'(u)\,du=\int_s^\infty \left(\frac1{u-a}-\frac1{u-b}\right)\,du$$
from which we see that 
$$F(s)=\log\left(\frac{s-a}{s-b}\right)$$
A: You are trying to compute the following integral:
$$\int_0^\infty \frac{e^{-t(s-b)}-e^{-t(s-a)}}{t}\,dt$$
By Frullani's integral, your integral is just
$$\ln\frac{s-b}{s-a}$$
