# Transformation of $\int_a^\infty \frac{1}{x^\alpha}\,\text{d}x$ into $\int_0^b \frac{1}{x^\beta}\,\text{d}x$

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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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
@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

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