## anonymous one year ago For what values of r does the integral convege? $\int\limits_{0}^{+\infty}\frac{ 1 }{ x^r(x+2) }dx$ Well, i tried dividing the integral from 0 to 1 and from 1 to infinity and then compare with the integral of 1/x^r, but not sure how to compare the second one. Any ideas?

1. anonymous

So long as $$r$$ is sufficiently large, it should converge.

2. anonymous

If you think of it like a $$p$$ series, then $$r\geq 1$$ would be large enough, but that's just a speculation on my part.

3. anonymous

Let's say that: $\frac{1}{x^{r+1}+2x^r} <\frac{1}{x^{r+1}}$and perhaps use a comparison test.

4. anonymous

r is rational here? or possibly real? or is it just an integer?

5. anonymous

r is any real

6. anonymous

@wio but consider that $$\int_0^1 x^{-r} dx$$ fails to converge for $$r\ge1$$

7. anonymous

We're dealing with $$x^{-r-1}$$ though.

8. anonymous

I think the $$(0,1)$$ interval needs to be explored, but by comparison we can say the $$(1,\infty)$$ interval is convergent.

9. anonymous

so we must simultaneously bound both the $$(0,1)$$ and $$(1,\infty)$$ integrals: $$\int_0^{L>1}\frac1{x^r(x+2)} dx=\int_0^1\frac1{x^r (x+2)} dx+\int_1^L\frac1{x^r (x+2)} dx$$now we bound: $$3x^{r+1}<x^{r+1}+2x^r<3x^r$$ on $$(0,1)$$ so $$\int_0^1\frac1{3x^{r+1}}dx>\int_0^1\frac1{x^r(x+2)}dx>\int_0^1\frac1{3x^r}dx$$... which tells us that the $$(0,1)$$ integrals only if necessarily $$r<1$$ (but is this sufficient?)

10. anonymous

from playing around numerically we can see $$\int_0^1\frac1{x^{1-10^{-k} (x+2)}} dx\approx \frac12(10^k-1)$$ hmm so it does seem sufficient although this is without formal proof

11. anonymous

heh, even more interesting: $$\int_0^1\frac1{x^{1-10^{-k}}(x+n)} dx\approx\frac1n(10^k-1)$$ for positive integers $$n,k$$

12. anonymous

Actually, it seems like the real concern might be the $$0$$ limit.

13. anonymous

Since that is going to be a right hand limit, it will tend to $$\pm\infty$$.

14. anonymous

yes, in terms of $$x$$ -- but i'm talking about $$r$$ :)

15. anonymous

anyways the approximation stuff i have above is a special case of the fact that: $$\int_0^1 x^{-r} dx=\frac1{1-r}$$and when $$r=1-10^{-k}$$ that gives $$\int_0^1 \frac1{x^{1-10^{-k}}} dx=10^k$$ and then lastly we know that $$\frac1{x+n}=\frac1n\cdot\frac1{1+x/n}=\frac1n\left(1-\frac{x}n+\frac{x^2}{n^2}-\dots\right)\approx\frac1n\left(1-\frac{x}n\right)$$ so it seems that \begin{align*}\int_0^1 \frac1{x^{1-10^{-k}} (x+n)} dx&\approx\frac1n \int_0^1\left(\frac1{x^{1-10^{-k}}} -\frac1n\cdot\frac1{x^{-10^{-k}}}\right)dx\\&=\frac1n\left(10^k-\frac1n\cdot \frac1{1-10^{-k}}\right)\\&\approx\frac1n(10^k-\frac1n)\end{align*}

16. anonymous

cool stuff, not entirely relevant but it does seem to me that $$r<1$$ is necessary

17. anonymous

so i think the answer is $$0<r<1$$ taking into account both parts of the domain of integration but i think the argument i have for $$(0,1)$$ can be formalized (taking $$r\to1$$ using $$r=1-10^{-k},k\to\infty$$ shows the integral grows with $$10^k\to\infty$$ so it must be the cut-off for $$r$$)