Divergent or convergent integral

I have the next integral:

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

I would like to know how to determine whether it's convergent or divergent according to the values given to $\mu$.

I can't really figure out how to proceed and how the procedure would look like. Thank you very much in advanced.

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That you don't know the answer seems contradicted by this other question. – Did Nov 14 '12 at 9:54
Sorry, I didn't want to be missunderstood, what I aim in this case is to know wether the integral is divergent or convergent for the different possible values of #μ not how to get the other expression that appeared in the question you have just mentioned. Thank you very much for your time @did. – Ana Lopez Nov 14 '12 at 10:00
But you stated exactly w[h]ether the integral is divergent or convergent for the different possible values of #μ in this other question... Well, whatever. – Did Nov 14 '12 at 11:17

If $\mu = 1$, obviously integral diverges. $\displaystyle \int_a^{+\infty} \frac 1x dx = \lim_{x \rightarrow + \infty} \ln x - \ln a \rightarrow \infty$. If $\mu \neq 1$ then $\displaystyle \int_a^{+\infty} x^{-\mu} dx = \frac 1{-\mu + 1} \left(\lim_{x \rightarrow +\infty} x^{-\mu + 1} - a^{-\mu + 1}\right)$. In this case, integral diverges or converges depending on the value of limit. If $-\mu + 1 < 0 \Rightarrow \mu > 1$ limit exists and is finite, so integral converges. For all other values of $\mu$ integral diverges. So
$\mu > 1$ converges
$\mu \le 1$ diverges