Question

Determine the convergence or divergence (posting problem)

Answer 1

\[\int\limits_{-\infty}^{\infty}\frac{ 1 }{ x^{2}(e^{x}+1) }dx\]
I know it diverges, but I have no idea how to integrate this in order to show that.

Answer 2

There are three improper portions of this integral, \[- \infty, 0, \infty \], so you must break up the integral and make it proper by taking the limits of each value of these

Answer 3

I have no problem with the improper portion of it, its the actual integrating.

Answer 4

I couldn't find some sort of brute force u-sub and I don't think it can be done by parts. So kind of at a dead end.

Answer 5

I'm kind of thinking that you're not supposed to integrate it, but make some sort of argument as to why it converges or diverges. But even then, I wouldn't know how to go about it.

Answer 6

You can use the Direct Comparison test since

\[ e^x+1 \geq 1 \] \[ -\infty \leq x \geq \infty \]

So then
\[ \frac{1}{x^2(e^x+1)} \leq \frac{1}{x^2} \]
for all real numbers. And 1/x^2 has a closed form anti-derivative

Answer 7

Disregard the incorrect sign for the x interval

Answer 8

So even though this is an integral, I can treat it like a series problem? The integral itself diverges, but if this is treated as two series, 1/x^2 converges by p-series meaning my original problem converges. But is that how this is treated, like a series and not an area under a curve?

Answer 9

DCT also works for integrals, the logic still follows right? If f(x) is always bigger than g(x) and the area under g(x) is infinite than so to must f(x) be.

Answer 10

Yeah, but isn't this backwards then? The integral of 1/x^2 diverges, but if we're going to use a comparison that diverges, doesn't our original f(x) have to be greater than 1/x^2?

Answer 11

Correct, I apologize for that, one moment.

Answer 12

Is this for just calculus based course, calc 2?

Answer 13

Yes, this is calculus 2.

Answer 14

Since you can do a direct comparison argument, could you maybe make another argument, limit comparison for example? Use 1/x^2 as your g(x) and show that f(x)/g(x) has a finite limit as x goes to infinity? Then say that since g(x) diverges so must f(x)?

Answer 15

I like the way you are thinking, but I believe the limit will tend to 0 and we have an inconclusive case as before, although if not there may have been a way to justify it.

An explanation, though not rigorous, could be found in the fact that \[ lim_{x\to 0^-} f(x) = \infty\] and since \[ f(x) \geq 0 \] then we can show that \[\int_\infty^0 f(x) dx \text{ diverges}\] but I haven't come up with a satisfactory explanation yet.

Answer 16

should be \[ - \infty\]

Answer 17

Ah, I forgot, limit comparison excludes 0. So how do each of those two facts combine to showing that integral diverges? I guess I don't see the connection right off the bat.

Answer 18

Well, I just to say that on the interval \[(-\infty,0)\] x is unbounded and monotonic increasing, the integral over that interval should be divergent

Answer 19

*f(x)

Answer 20

Ah, alright. I have to go unfortunately, but thanks a lot. I never knew the sequence and series arguments could be applied to integrals as well. Very very much appreciated ^_^

Answer 21

Not a problem, sorry I couldn't be of more help. If I find a more satisfactory answer, I'll message you it.

Answer 22

Thank you.