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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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$\frac{e^x}{x} \geq \frac{1}{x}$ when $x >0$. –  copper.hat Jul 17 '12 at 22:38
    
that's good for direct comparison, which I know can be done easily with this. But I wanted to do it with limit comparison –  Joseph Skelton Jul 17 '12 at 22:39
    
Do you mean $e^{x}$ or $e^{-x}$? If it's $e^x$, you know that $e^x/x\rightarrow\infty$ as $x\to\infty$ and you are dead right there. –  ncmathsadist Jul 17 '12 at 23:27
    
@ncmathsadist no, it is $e^x$, I know this function is divergent, it is a matter of showing it using a certain method –  Joseph Skelton Jul 18 '12 at 0:11

4 Answers 4

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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but this again, is direct comparison. You are chosing divergent functions that are less than the original and comparing them to show that the original is divergent, I know that this can easily be done here. What I am asking is if this can be done doing limit comparison. Which is finding a $g(x)$ such that $lim_{x\rightarrow \infty} \frac{f(x)}{g(x)} = L$ for a positive finite number L. –  Joseph Skelton Jul 18 '12 at 0:08
    
@JosephSkelton OK, let me think about it. I think, however, that that is a little more farfetched than direct comparison which works swiftly. –  Pedro Tamaroff Jul 18 '12 at 0:12
    
I know, which is part of the reason it has intrigued me, because it is a method that wouldn't necessarily be used, but seemingly is still possible. –  Joseph Skelton Jul 18 '12 at 0:14
    
@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 $\pi$. –  Pedro Tamaroff Jul 18 '12 at 0:41

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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but if you do $e^{x/2}$ then you just get $\frac{e^{x/2}}{x} \longrightarrow \infty$ and the same with $1/x$ with limit comparison we need $\frac{f(x)}{g(x)} = L$ for $0<L<\infty$ so doesn't getting infinity give us an inconclusive answer? –  Joseph Skelton Jul 17 '12 at 22:46
    
@JosephSkelton You're not understanding what Robert is saying. See my answer below. –  Pedro Tamaroff Jul 18 '12 at 0:01
    
@PeterTamaroff ok, I understand what he is saying now, but it is still a direct comparison, not a limit comparison. –  Joseph Skelton Jul 18 '12 at 0:12
    
A better version of Limit Comparison is this (let's say for an integral that is improper at $+\infty$). Suppose $f(x)$ and $g(x)$ are continuous and strictly positive on $[a,\infty)$. If $\lim_{x \to \infty} f(x)/g(x) = L < \infty$ and $\int_a^\infty g(x)\ dx$ converges then $\int_a^\infty f(x)\ dx$ converges. Under the same conditions if $\int_a^\infty f(x)\ dx$ diverges then $\int_a^\infty g(x)\ dx$ diverges. This includes the case $L=0$. –  Robert Israel Jul 18 '12 at 1:06
    
Similarly if $\lim_{x \to \infty} f(x)/g(x) = L > 0$ (including the case $L=\infty$) and $\int_a^\infty g(x)\ dx$ diverges then $\int_a^\infty f(x)\ dx$ diverges. Under the same conditions if $\int_a^\infty f(x)\ dx$ converges then $\int_a^\infty g(x)\ dx$ converges. –  Robert Israel Jul 18 '12 at 1:10

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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What is the upper limit of the integral?. –  Pedro Tamaroff Jul 18 '12 at 0:43
    
It does not matter; for any fixed positive value it will be $+\infty$. –  ncmathsadist Jul 18 '12 at 0:45
    
True, but maybe it is worth mentioning. –  Pedro Tamaroff Jul 18 '12 at 1:00
    
You are done at the bottom. It is not. –  ncmathsadist Jul 18 '12 at 1:01
    
OK. It's your call. –  Pedro Tamaroff Jul 18 '12 at 1:04

∫ 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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