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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?

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  1. anonymous
    • one year ago
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    So long as \(r\) is sufficiently large, it should converge.

  2. anonymous
    • one year ago
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    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
    • one year ago
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    Let's say that: \[ \frac{1}{x^{r+1}+2x^r} <\frac{1}{x^{r+1}} \]and perhaps use a comparison test.

  4. anonymous
    • one year ago
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    r is rational here? or possibly real? or is it just an integer?

  5. anonymous
    • one year ago
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    r is any real

  6. anonymous
    • one year ago
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    @wio but consider that \(\int_0^1 x^{-r} dx\) fails to converge for \(r\ge1\)

  7. anonymous
    • one year ago
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    We're dealing with \(x^{-r-1}\) though.

  8. anonymous
    • one year ago
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    I think the \((0,1)\) interval needs to be explored, but by comparison we can say the \((1,\infty)\) interval is convergent.

  9. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Actually, it seems like the real concern might be the \(0\) limit.

  13. anonymous
    • one year ago
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    Since that is going to be a right hand limit, it will tend to \(\pm\infty\).

  14. anonymous
    • one year ago
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    yes, in terms of \(x\) -- but i'm talking about \(r\) :)

  15. anonymous
    • one year ago
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    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
    • one year ago
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    cool stuff, not entirely relevant but it does seem to me that \(r<1\) is necessary

  17. anonymous
    • one year ago
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    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 \(r\))

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