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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?
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So long as \(r\) is sufficiently large, it should converge.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0Let's say that: \[ \frac{1}{x^{r+1}+2x^r} <\frac{1}{x^{r+1}} \]and perhaps use a comparison test.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0r is rational here? or possibly real? or is it just an integer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@wio but consider that \(\int_0^1 x^{r} dx\) fails to converge for \(r\ge1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We're dealing with \(x^{r1}\) though.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think the \((0,1)\) interval needs to be explored, but by comparison we can say the \((1,\infty)\) interval is convergent.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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?)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from playing around numerically we can see $$\int_0^1\frac1{x^{110^{k} (x+2)}} dx\approx \frac12(10^k1)$$ hmm so it does seem sufficient although this is without formal proof

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0heh, even more interesting: $$\int_0^1\frac1{x^{110^{k}}(x+n)} dx\approx\frac1n(10^k1)$$ for positive integers \(n,k\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually, it seems like the real concern might be the \(0\) limit.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since that is going to be a right hand limit, it will tend to \(\pm\infty\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, in terms of \(x\)  but i'm talking about \(r\) :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anyways the approximation stuff i have above is a special case of the fact that: $$\int_0^1 x^{r} dx=\frac1{1r}$$and when \(r=110^{k}\) that gives $$\int_0^1 \frac1{x^{110^{k}}} dx=10^k$$ 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 $$\begin{align*}\int_0^1 \frac1{x^{110^{k}} (x+n)} dx&\approx\frac1n \int_0^1\left(\frac1{x^{110^{k}}} \frac1n\cdot\frac1{x^{10^{k}}}\right)dx\\&=\frac1n\left(10^k\frac1n\cdot \frac1{110^{k}}\right)\\&\approx\frac1n(10^k\frac1n)\end{align*}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cool stuff, not entirely relevant but it does seem to me that \(r<1\) is necessary

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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 \(r=110^{k},k\to\infty\) shows the integral grows with \(10^k\to\infty\) so it must be the cutoff for \(r\))
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