# Thread: Can someone please check my work here? (Improper integrals, converge/diverge)

1. ## Can someone please check my work here? (Improper integrals, converge/diverge)

I have the following two problems I am working on:

$\int_{2}^{\\infty}dx/sqrt[x^3-1]{}$

For this one I am using the limit comparison test, I did

f(x)= 1/(x^3-1)^1/2 / g(x) = 1/x

which ended up at x/(x^3-1)^1/2 which I can determine as x->infty is going to be greater than 0 but less than infinity (thus both either converge or diverge). I then integrated my g(x) which I chose as 1/x and got as a limit infty, thus the functions diverge. Is this correct?

The second one is:

$\int_{2}^{\infty}x^2/sqrt[x^8-1]{}$

For this one I took a similiar approach, except g(x) = x^2/sqroot(x^8) and determined f(x)/g(x) = 1, thus greater than zero but less than infty. So I integrated my g(x) down to -1/x, and as x->infty the limit is zero. No positive finite limit, thus it diverges.

Hope I'm doing these right. Any help would be great thanks in advance.

2. Originally Posted by joseph_
I have the following two problems I am working on:

$\int_{2}^{\\infty}dx/sqrt[x^3-1]{}$

For this one I am using the limit comparison test, I did

f(x)= 1/(x^3-1)^1/2 / g(x) = 1/x

which ended up at x/(x^3-1)^1/2 which I can determine as x->infty is going to be greater than 0 but less than infinity

Nop. The asymptotic behaviour of the denominator here is like that of $\displaystyle x^{3\slash 2}$, so the whole thing behaves like $\displaystyle x^{-1\slash 2}\xrightarrow [x\to\infty] {} 0$ ...But you can try the same test with $\displaystyle \frac{1}{x^{3\slash 2}}$ , getting something that converges to 1. BTW, your integral converges.

(thus both either converge or diverge). I then integrated my g(x) which I chose as 1/x and got as a limit infty, thus the functions diverge. Is this correct?

The second one is:

$\int_{2}^{\infty}x^2/sqrt[x^8-1]{}$

For this one I took a similiar approach, except g(x) = x^2/sqroot(x^8) and determined f(x)/g(x) = 1

You mean the limit, of course.

, thus greater than zero but less than infty. So I integrated my g(x) down to -1/x, and as x->infty the limit is zero. No positive finite limit, thus it diverges.

Nop, again: the limit OF THE IMPROPER INTEGRAL exists and it's finite and thus THAT is the value of the improper integral. Again, this integral converges.

Tonio

Hope I'm doing these right. Any help would be great thanks in advance.
.

3. Originally Posted by tonio
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So for the first one g(x) is 1/x^3/2 and when compared with the original f(x) lim as x->infty is equal to one because infinity / infinity is equal to 1. 1 is a pos finite limit so therefore f(x) converges because g(x) converges. So if I get a one from doing the LCT my integral converges? My book just says 0 < L < infty of lim->x f(x)/g(x) then both either converge or diverge. But you're saying if I get one then I have convergence? Sorry if this is confusing.

As far as the second one, so I had the correct comparison g(x) but because this turns out to be 1 the integral converges as well?

4. note that for $\displaystyle x\ge2$ the following estimation holds: $\displaystyle \frac{1}{\sqrt{x^{3}-1}}<\frac{\sqrt{2}}{x^{3/2}}.$

hence, the first integral converges, and you can apply this for the second one.