You are presented with $I=\int_0^\infty \frac{e^x}{x}dx$
It is clear the function is bounded at any point inside $(0,\infty)$ so we're worried about the extrema of the interval. Split the integral at, say $1$, we have
$I=\int_0^1 \frac{e^x}{x}dx+\int_1^\infty \frac{e^x}{x}dx$
We need to analyze, then
$\lim_{\epsilon \to 0}\int_\epsilon^1 \frac{e^x}{x}dx$
and
$\lim_{m \to \infty}\int_1^m \frac{e^x}{x}dx$
But note that for $x\in(0,1)$, we have
$\frac{1}{x}<\frac{e^x}{x}$
so that for $\epsilon >0$
$\int_\epsilon^1\frac{dx}{x}<\int_\epsilon^1\frac{e^x}{x}dx$
If we let $\epsilon \to 0$ we see that
$\lim_{\epsilon \to 0}\int_\epsilon^1\frac{dx}{x}<\lim_{\epsilon \to 0}\int_\epsilon^1\frac{e^x}{x}dx$
But $\displaystyle \lim_{\epsilon \to 0}\int_\epsilon^1\frac{dx}{x}$ diverges, so that $\displaystyle \lim_{\epsilon \to 0}\int_\epsilon^1 \frac{e^x}{x}dx$ forcedfully, diverges too.
Now consider $e^{x/2}$ in $(1,\infty)$. You can check that
$e^{x/2}<\frac{e^x}{x}$
so
$\int_1^me^{x/2}dx<\int_1^m\frac{e^x}{x}dx$
for $m>1$. But now if we let $m\to \infty$ we see that
$\lim_{m \to \infty}\int_1^me^{x/2}dx$
diverges, so $\lim_{m \to \infty}\int_1^m\frac{e^x}{x}dx$ diverges forcedfully, too.
In conclusion, you integral diverges.