0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.