Determine divergence of $\int_0^\infty\frac{ e^x}{x}$ with limit comparison test. Today I was asked if you can determine the divergence of $$\int_0^\infty \frac{e^x}{x}dx$$ using the limit comparison test.
I've tried things like $e^x$, $\frac{1}{x}$, I even tried changing bounds by picking $x=\ln u$, then $dx=\frac{1}{u}du$. Then the integral, with bounds changed becomes $\int_1^\infty \frac{1}{\ln u}du$ This didn't help either.
This problem intrigued me, so any helpful pointers would be greatly appreciated.
 A: 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.
A: Comparison to $e^{x/2}$ should work (taking  $x \to \infty$).  So should $1/x$ (either as $x \to 0+$ or as $x \to \infty$).  
A: In that case, do this
You have $$e^x/x\sim {1\over x}$$
as $x\to 0$ and 
$$\int _{0^+} {dx\over x} = +\infty.$$
Now you are done.
A: ∫  ε to 1  e^x / x dx > ∫  ε to 1 1/x dx,  → ∞  as ε → 0; and
∫  1 to N  e^x / x dx > ∫  1 to N  1/x dx,  → ∞  as  N → ∞;
and so we are done.
