How to calculate the integral $\int_{0}^{\infty} \frac{e^{-ax}-e^{-bx}}{x}\ dx $ given $ b>a>0$? $ b>a>0$. Calculate the integral $\displaystyle \int_{0}^{\infty} \frac{e^{-ax}-e^{-bx}}{x}\ dx $
Integration by parts doesn't work, nor can I find a proper substituion.
 A: This is a special case of the question already posted in the comments, but a quick direct way to proceed is the following:
$$
F(a,b)=\int_0^\infty \frac{e^{-ax}-e^{-bx}}{x}dx
$$
satisfies
$$
\partial_a F(a,b)=-\int_0^\infty e^{-ax}dx=-\frac{1}{a}
$$
and similarly
$$
\partial_b F(a,b)=\frac{1}{b}.
$$
Therefore,
$$
F(a,b)=\log\frac{b}{a}.
$$
You can verify that the integration constants are correct as well.
A: \begin{align*}
  \int_0^\infty \frac{e^{-ax}-e^{-bx}}{x} dx &=
  \int_0^\infty \int _a^b e^{-xy} \, dy \, dx \\
  &= \int_a^b \int_0^\infty e^{-xy} \, dx \, dy \\
  &= \int_a^b \left[ -\frac{e^{-xy}}{y} \right]_{x=0}^\infty \, dy \\
  &= \int_a^b \frac{1}{y} \, dy \\
  &= \ln \frac{b}{a}
\end{align*}
A: Another Way
Also you can use Laplace transform. Let $\mathcal{L}[f(t)]=F(s)$, We have
$$\int_{0}^{\infty}F(s)ds=\int_{0}^{\infty}\int_{0}^{\infty}e^{-st}f(t) \mathrm{dt}\mathrm{ds}=\int_{0}^{\infty}\left(\int_{0}^{\infty}e^{-st}\mathrm{ds}\right)f(t)dt=\int_{0}^{\infty}\frac{f(t)}{t}\mathrm{dt}$$
As a result
$$\int_{0}^{\infty}F(s)ds=\int_{0}^{\infty}\frac{f(t)}{t}\mathrm{dt}$$
Now apply the result for $f(x)=e^{-ax}$ , and $f(x)=e^{-bx}$
$$\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}dx=\int_{0}^{\infty}\left(\frac{1}{s+a}-\frac{1}{s+b}\right)ds=\ln\left(\frac{s+a}{s+b}\right)\Big{|}_{0}^{\infty}=-\ln\left(\frac{a}{b}\right)=\ln\left(\frac{b}{a}\right)$$
