Show that $\int\limits_0^2x^{-1}e^{-x} dx$ does not converge I'm trying to show that the integral

$$I:=\int\limits_0^2x^{-1}e^{-x}\mathrm dx$$

diverges. 

My attempt: We know that the problem is in the $x \to 0$ part so we have that $x \to 0$ implies $e^{-x} \to 1$ and $x^{-1} \to \infty$. We have that the function is not defined at zero so if we set this integral correctly we have that  
$$I= \lim_{\xi \to 0}\int_\xi^2x^{-1}e^{-x}\mathrm dx $$
I think is not possible to interchange the series of $e^x$ with the integral because it is a divergent integral of a series. (Is this correct? Thinking in the answer given HERE). 
We can use that $$E\mathrm i(-x) = \int x^{-1}e^{-x}\mathrm dx+\mathrm{Const.}$$
So $$I = E\mathrm i(-2) - \lim_{\xi \to 0}E\mathrm i(-\xi)$$
Can anyone set a hint for what can I do? 
 A: Since $x^{-1}e^{-x} \ge x^{-1}e^{-1}$ on $(0,1),$ and $\int_0^1 x^{-1}\,dx = \infty,$ we have $\int_0^1 x^{-1}e^{-x}\,dx=\infty.$
A: Let $g(x)=\frac{e^{-x}-1}{x}$ for $x\in (0,2]$. Then $g$ can be made continuous at $x=0$ by defining $g(0)=-1$.
But then $$\int_{\xi}^2 x^{-1}e^{-x}\,dx = \int_{\xi}^2 g(x)\,dx + \int_{\xi}^2 \frac{dx}{x}$$
Since $g$ is continuous on $[0,2]$ you get:
$$\lim_{\xi\to 0+}\int_{\xi}^2 x^{-1}e^{-x}\,dx = \int_{0}^{2}g(x)\,dx + \lim_{\xi\to 0+} \int_{\xi}^{2}\frac{dx}{x}$$
Basically, $x^{-1}e^{-x}$ behaves "like" $\frac{1}{x}$ when $x$ is near zero.
A: Note that 
\begin{align}
\lim_{\xi\to 0}\int_\xi^2x^{-1}e^{-x}\mathrm dx
=&
\lim_{\xi\to 0}\int_\xi^2x^{-1}D_x\big(-e^{-x}\big)\mathrm dx
\\
=&
\lim_{\xi\to 0}\left[
x^{-1}(-e^{-x})\Big|_{\xi}^{2}-\int_\xi^2D_x\big(x^{-1}\big)\big(-e^{-x}\big)\mathrm dx
\right]
\\
=&
\lim_{\xi\to 0}\left[
\frac{1}{x}(-e^{-x})\Big|_{\xi}^{2}
-
\int_\xi^2\left(\frac{1}{x^2}\right)(-e^{-x})\mathrm dx
\right]
\\
=&
\lim_{\xi\to 0}\left[
\underbrace{-\frac{1}{2}(e^{-2})+\frac{1}{\xi}(e^{-\xi})}_{\to \infty}
+
\underbrace{\int_\xi^2\left(\frac{1}{x^2}\right)(e^{-x})\mathrm dx}_{<\infty}
\right]
\end{align}
A: By convexity $e^{-x}\geq \max(1-x,0)$, hence
$$ \int_{0}^{2}\frac{e^{-x}}{x}\,dx \geq -1+\int_{0}^{1}\frac{dx}{x} $$
trivially implies divergence.
