Evaluate $\int_0^{\infty}\frac{e^{-x}-e^{-2x}}{x}dx$ using a double integral I was given the following problem:

Evaluate the following integrate using a double integral: $\int_0^{\infty}\frac{e^{-x}-e^{-2x}}{x}dx$.

The professor told us off the bat the answer was $\ln(2)$. He wants us to show our work and prove this is true. My attempt is below. 


*

*$I(x)$=$\int_0^{\infty}\frac{e^{-x}-e^{-2x}}{x}dx$ = $\int_0^{\infty}\frac{e^{-y}-e^{-2y}}{y}dy$ =$I(y)$.

*Thus I can say $=I(x)=\sqrt{I(x)I(y)}$ 


\begin{eqnarray}
\int_0^{\infty}\frac{e^{-x}-e^{-2x}}{x}dx &=&\sqrt{\int_0^{\infty}\frac{e^{-x}-e^{-2x}}{x}dx\int_0^{\infty}\frac{e^{-y}-e^{-2y}}{y}dy}\\
& = & \sqrt{\int_0^{\infty}\int_0^{\infty}\frac{(e^{-x}-e^{-2x})(e^{-y}-e^{-2y})}{xy}dxdy}
\end{eqnarray}
This is where things get a little nasty. I can decided to make a change of variables. Letting $x=r\cos{\theta}$, $y=r\sin{\theta}$, thus $dA=dxdy=rdrd\theta$ by the Jacobian. Thus:
\begin{eqnarray}
\int_0^{\infty}\frac{e^{-x}-e^{-2x}}{x}dx & = & \sqrt{\int_0^{2\pi}\int_0^{\infty}\frac{(e^{-r\cos{\theta}}-e^{-2r\cos{\theta}})(e^{-r\sin{\theta}}-e^{-2r\sin{\theta}})}{r^2\cos{\theta}\sin{\theta}}rdrd\theta}\\
& = & \sqrt{\int_0^{2\pi}\int_0^{\infty}\frac{(e^{-r(\cos{\theta}+\sin{\theta})}-e^{-r(2\cos{\theta}+\sin{\theta})}-e^{-r(\cos{\theta}+2\sin{\theta})}-e^{-2r(\cos{\theta}+\sin{\theta})}}{r\cos{\theta}\sin{\theta}}drd\theta}
\end{eqnarray}
This is where I am stuck. From here I did a lot of trial and error trying to solve for the problem. From changing the order of integration, integration by parts, and subsitution. Is there something I am missing? Maybe a trig identity that will help simplify the expression. 
Thank You for your time and I greatly appreciate any feedback you give me.
 A: \begin{eqnarray}
\int_0^\infty\frac{e^{-x}-e^{-2x}}{x}\,dx&=&\int_0^\infty\left[\int_1^2e^{-yx}\,dy\right]\,dx=\int_1^2\left[\int_0^\infty e^{-yx}\,dx\right]\,dy\\
&=&\int_1^2\left[-\frac{e^{-yx}}{y}\right]_0^\infty\,dy=\int_1^2\frac1y\,dy=\ln2.
\end{eqnarray}
A: This is called a Frullani integral. The idea is to write
$$ \frac{e^{-x}-e^{-2x}}{x} = \int_1^2 e^{-tx} \, dt, $$
and then change the order of integration, which gives you
$$ \int_1^2 \int_{0}^{\infty} e^{-tx} \, dx \, dt = \int_1^2 \frac{dt}{t} = \log{2}. $$
A: Assume the function $\phi$ has a continuous first derivative.
Let's look at the integral 
$$\int_a^b \int_c^d \varphi' (xy) dy dx$$
Note that $\frac{1}{x}\frac{\partial \varphi}{\partial y}  = \frac{1}{y}\frac{\partial \varphi}{\partial x}$.  Thus, we have 
$$\begin{align}
\int_a^b \int_c^d \varphi' (xy) dy dx &=\int_a^b \int_c^d \frac{1}{x}\frac{\partial \varphi (xy)}{\partial y} dy dx \\
&=\int_a^b \frac{1}{x} \left( \varphi (dx)-\varphi (cx)\right) dx 
\end{align}$$
We also have
$$\begin{align}
\int_a^b \int_c^d \varphi' (xy) dy dx &=\int_a^b \int_c^d \frac{1}{y}\frac{\partial \varphi (xy)}{\partial x} dy dx \\
&=\int_c^d \frac{1}{y} \left( \varphi (by)-\varphi (ay)\right) dy 
\end{align}$$
Assume that $\lim_{z \to \infty} \varphi (z) =0$.  Then, as $a \to 0$ and $b \to \infty$, we see that
$$\begin{align}
\int_0^{\infty} \int_c^d \varphi' (xy) dy dx &=- \varphi (0) \int_c^d \frac{1}{y}  dx \\
&=-\varphi (0) \log (d/c)
\end{align}$$
Putting it all together, we have 
$$\int_0^{\infty} \frac{\left( \varphi (cx)-\varphi (dx)\right)}{x}  dx =\varphi (0) \log (d/c)$$
Here, let $\phi(x)=e^{-x}$, $c=1$, and $d=2$.  Then, the result is $\log2$ as expected.
A: Here is another approach.
$$
\begin{align}
\int_0^\infty\frac{e^{-x}-e^{-2x}}{x}\,\mathrm{d}x
&=\int_0^\infty(e^{-x}-e^{-2x})\int_0^\infty e^{-tx}\,\mathrm{d}t\,\mathrm{d}x\\
&=\int_0^\infty\int_0^\infty(e^{-(1+t)x}-e^{-(2+t)x})\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_0^\infty\left(\frac1{1+t}-\frac1{2+t}\right)\,\mathrm{d}t\\
&=\lim_{L\to\infty}\left(\int_0^L\frac1{1+t}\,\mathrm{d}t-\int_1^{L+1}\frac1{1+t}\,\mathrm{d}t\right)\\
&=\int_0^1\frac1{1+t}\,\mathrm{d}t-\lim_{L\to\infty}\int_L^{L+1}\frac1{1+t}\,\mathrm{d}t\\[9pt]
&=\log(2)
\end{align}
$$
