Is ${\int_{0}^{+\infty}\frac{e^{-x} - e^{-2x}}{x}dx}$ finite? I want to know if $\displaystyle{\int_{0}^{+\infty}\frac{e^{-x} - e^{-2x}}{x}dx}$ is finite, or in the other words, if the function  $\displaystyle{\frac{e^{-x} - e^{-2x}}{x}}$ is integrable in the neighborhood of zero.
 A: $$
\frac1x\,(e^{-x}-e^{-2x})=\frac1x\,(1-x+O(x^2)-(1-2x+O(x^2)))=\frac1x\,(x+O(x^2))=1+O(x).
$$
So the function can be extended to $x=0$ in a continuous way, and it thus integrable on any interval $[0,k]$. 
A: Let $f(x) = \frac{e^{-x} - e^{-2x}}{x}$.
L'Hopital gives $\lim_{x \to 0} f(x)= 1$. Hence in some neighborhood $B(0,\epsilon)$ , $|f(x)| <2$. For $x\geq \epsilon$, we have $\frac{1}{x} \leq \frac{1}{\epsilon}$, and the function $x \mapsto e^{-x} - e^{-2x}$ is clearly integrable.
Hence $\int_0^\infty |f(x)| dx \leq 2 \epsilon + \frac{1}{\epsilon}\int_{\epsilon}^\infty |e^{-x} - e^{-2x}| dx $, and it follows that $f$ is integrable.
A: In general, when $f$ is "well-behaved" at zero and infinity:
$$\int_0^{\infty} dx \frac{f(a x) - f(b x)}{x} = (f(\infty)-f(0)) \log{\frac{a}{b}}$$
You can see this from this (rough) "proof":
$$\begin{align}\int_0^{\infty} dx \frac{f(a x) - f(b x)}{x} &= \int_0^{\infty} dx \: \int_b^a du \, \frac{d}{du} f(u x) \\ &= \int_b^a du \: \int_0^{\infty} dx \, \frac{d}{dx} f(u x)\\ &= \int_b^a \frac{du}{u} (f(\infty)-f(0)) \end{align}$$
The result follows.  In this case, $f(x) = e^{-x}$, $a=1$, and $b=2$; the integral is then
$$(0-1)\log{\frac{1}{2}} = \log{2}$$
A: It suffice to expand the function locally:$$e^{-x}-e^{-2x}=(1-x+x^{2}/2)+..-(1-2x+4x^{2}/2)-...=x-3x^{3}/2+...$$ where $...$ are terms of power at least cubic. It is not difficult to see the above expression divided by $x$ should be locally integrable around 0. 
A: Claim: $$\int_0^{\infty} \frac{e^{-u}-e^{-2u}}{u} du = \ln(2).$$
Proof: Let
\begin{align}
C &\equiv \int_0^{\infty} \frac{e^{-u}-e^{-2u}}{u} du\\\ \\
&=\lim_{x=0}\left[ \operatorname{Ei}(1,x) - \operatorname{Ei}(1,2x)\right],
\end{align}
where
$$
\operatorname{Ei}(1,x) \equiv \int_x^\infty \frac{e^{-u}}{u} du.
$$
Now, let $$f(x) \equiv \int_1^x \frac{e^{-u}}{u} du.$$
Note:$$\frac{d\operatorname{Ei}(1,x)}{dx} = - \frac{df}{dx},$$
so $$f(x) = -\operatorname{Ei}(1,x) + c,$$ where $c\in \mathbb{R}$. Then $f(1) = -\operatorname{Ei}(1,1)+c$. However,
$$
f(1) = \int_1^1 \frac{e^{-u}}{u} du = 0.
$$
$\therefore c=\operatorname{Ei}(1,1)$, i.e. 
$$
\operatorname{Ei}(1,x) = \operatorname{Ei}(1,1) - \int_1^x \frac{e^{-u}}{u} du
$$
Considering that
$$
\ln(x) = \int_1^x \frac{1}{u} du,
$$
we have
$$
\operatorname{Ei}(1,x) = -\ln(x) + \operatorname{Ei}(1,1) + \int_1^x\frac{1-e^{-u}}{u} du \tag{$\star$}.
$$
$(\star)$ applied to the definition of $C$ gives:
\begin{align}
\int_0^{\infty} \frac{e^{-u}-e^{-2u}}{u} du &=\lim_{x=0}\left[ \operatorname{Ei}(1,x) - \operatorname{Ei}(1,2x)\right]\\
&=\lim_{x=0}\left[ \ln(2)-\ln(1) - \int_x^{2x} \frac{1-e^{-u}}{u}du \right]\\
&=\ln(2).
\end{align}
Q.E.D.
