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.

  • $\begingroup$ Hint: try the change of variable $x=1/t$. $\endgroup$ – Did Nov 13 '12 at 20:05
  • $\begingroup$ @did could you please specify how to do the variable change? I don`t get it, sorry. Thank you very much for your time. $\endgroup$ – Ana Lopez Nov 13 '12 at 20:27
  • $\begingroup$ See the answer below. $\endgroup$ – Did Nov 13 '12 at 20:38

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 $$ (\alpha+\beta=2\quad\&\quad ab=1)\implies\int_0^b\frac{\mathrm dx}{x^{\beta}}=\int_{a}^{+\infty}\frac{\mathrm dx}{x^{\alpha}}. $$

  • $\begingroup$ 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. $\endgroup$ – Ana Lopez Nov 14 '12 at 11:11
  • $\begingroup$ 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. $\endgroup$ – Did Nov 14 '12 at 11:20
  • $\begingroup$ 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. $\endgroup$ – Ana Lopez Nov 14 '12 at 11:39

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