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Given the integral:

$$\int_a^\infty \frac{1}{x^α}\,\text{d}x$$

and knowing that it converges when $α>1$ and it diverges when $α\le1$

I would like to know how can I get to write the integral in the next way by more or less a couple simple steps I can't really get to figure out.

$$\int_0^b \frac{1}{x^α}\,\text{d}x$$

and knowing that it converges when $α<1$ and it diverges when $α\ge1$.

Thank you very much.

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What do you mean by "how can I get to write the integral in the next way..."? –  Thomas Andrews Nov 13 '12 at 19:27
    
Oh, it looks line one of the editors deleted the second integral from your question. –  Thomas Andrews Nov 13 '12 at 19:28
2  
@Cameron - apologies, your edit was correct. –  amWhy Nov 13 '12 at 19:34
    
@amWhy: Quite alright. I've goofed on edits, myself. –  Cameron Buie Nov 13 '12 at 19:48
    
Hint: try the change of variable $x=1/t$. –  Did Nov 13 '12 at 20:05

1 Answer 1

up vote 1 down vote accepted

Note that $$\int_0^b\frac{\mathrm dx}{x^{\beta}}\stackrel{(x=1/t)}{=}\int_{1/b}^{+\infty}\frac1{t^{-\beta}}\frac{\mathrm dt}{t^2}=\int_{1/b}^{+\infty}\frac{\mathrm dx}{x^{2-\beta}}, $$ hence $$ (\alpha+\beta=2\quad\&\quad ab=1)\implies\int_0^b\frac{\mathrm dx}{x^{\beta}}=\int_{a}^{+\infty}\frac{\mathrm dx}{x^{\alpha}}. $$

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Could you please explain to me why did you do this -> $$(\alpha+\beta=2\quad\&\quad ab=1) I don't understand it sorry. Thank's again. –  Ana Lopez Nov 14 '12 at 11:11
    
If the last integral in the first displayed line of my answer is to be equal to the last integral in the second displayed line, then one should have $1/b=a$ and $2-\beta=\alpha$. These conditions are equivalent to the identities at the beginning of the second displayed line. –  Did Nov 14 '12 at 11:20
    
And why does it turn over and from being convergent when \alfa>1 and divergent when \alfa≤1 becomes convergent when \alfa<1 and divergent when \alfa≥1? Thank you very much. –  Ana Lopez Nov 14 '12 at 11:39

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