anonymous
  • anonymous
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?
Mathematics
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anonymous
  • anonymous
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?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
So long as \(r\) is sufficiently large, it should converge.
anonymous
  • 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.
anonymous
  • anonymous
Let's say that: \[ \frac{1}{x^{r+1}+2x^r} <\frac{1}{x^{r+1}} \]and perhaps use a comparison test.

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anonymous
  • anonymous
r is rational here? or possibly real? or is it just an integer?
anonymous
  • anonymous
r is any real
anonymous
  • anonymous
@wio but consider that \(\int_0^1 x^{-r} dx\) fails to converge for \(r\ge1\)
anonymous
  • anonymous
We're dealing with \(x^{-r-1}\) though.
anonymous
  • anonymous
I think the \((0,1)\) interval needs to be explored, but by comparison we can say the \((1,\infty)\) interval is convergent.
anonymous
  • 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: $$3x^{r+1}\int_0^1\frac1{x^r(x+2)}dx>\int_0^1\frac1{3x^r}dx$$... which tells us that the \((0,1)\) integrals only if necessarily \(r<1\) (but is this sufficient?)
anonymous
  • 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
anonymous
  • anonymous
heh, even more interesting: $$\int_0^1\frac1{x^{1-10^{-k}}(x+n)} dx\approx\frac1n(10^k-1)$$ for positive integers \(n,k\)
anonymous
  • anonymous
Actually, it seems like the real concern might be the \(0\) limit.
anonymous
  • anonymous
Since that is going to be a right hand limit, it will tend to \(\pm\infty\).
anonymous
  • anonymous
yes, in terms of \(x\) -- but i'm talking about \(r\) :)
anonymous
  • 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 $$\begin{align*}\int_0^1 \frac1{x^{1-10^{-k}} (x+n)} dx&\approx\frac1n \int_0^1\left(\frac1{x^{1-10^{-k}}} -\frac1n\cdot\frac1{x^{-10^{-k}}}\right)dx\\&=\frac1n\left(10^k-\frac1n\cdot \frac1{1-10^{-k}}\right)\\&\approx\frac1n(10^k-\frac1n)\end{align*}$$
anonymous
  • anonymous
cool stuff, not entirely relevant but it does seem to me that \(r<1\) is necessary
anonymous
  • anonymous
so i think the answer is \(0

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