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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.

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    @ncmathsadist no, it is $e^x$, I know this function is divergent, it is a matter of showing it using a certain method2012-07-18

4 Answers 4

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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.

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    @JosephSkelton I honestly don't find it worth thinking. In fact, you can use $\int_0^1 \frac{dx}{x}$ and $\int_1^\infty \frac{dx}{x}$ to prove both ends of the integral, which is as easy as $\pi$.2012-07-18
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Comparison to $e^{x/2}$ should work (taking $x \to \infty$). So should $1/x$ (either as $x \to 0+$ or as $x \to \infty$).

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    You're not going to find a simpler function to compare $e^x/x$ to as $x \to \infty$ if you insist on the limit of the ratio being finite and nonzero.2012-07-18
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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.

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    OK. It's your call.2012-07-18
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∫ 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.

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    You've been here a long time, but have you read the **big popup** that told you "Use MathJax to format equations. [MathJax reference.](http://meta.math.stackexchange.com/questions/5020/tex-latex-mathjax-basic-tutorial-and-quick-reference)"?2018-08-03