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Given the integral:

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

and knowing that it converges when $\alpha >1$ and it diverges when $\alpha\le1$, I would like to know how I can transform the integral into

$\int_0^b \frac{1}{x^\beta}\,\text{d}x$

which converges when $\beta<1$ and diverges when $\beta\ge1$ by a couple of more or less simple steps. I can't really figure it out.

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    See the answer below.2012-11-13

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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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    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.2012-11-14