How can I determine convergence or divergence? I have stacked about expression of the text book. 
The question was that 

Find the value of $p$ for which the integral converges and evaluate the integral for those values of $p$
  $$\int_0^\infty \frac{1}{x^p} \; dx$$

Do I need to use $p$ test to determine if it were conversion or diversion?
OR Do I still use $\displaystyle\lim\limits_{t \to \infty}\int_0^t \frac{1}{x^p} \; dx$?
If you have any idea, please post it on the wall 
thank you,
 A: The integral $\int_0^\infty {1\over x^p}\, dx$ is improper in two ways: the interval of integration is infinite and  the integrand "blows up" at 0.  Thus, you need to split it up:
$$
I=\int_0^\infty {1\over x^p}\, dx=
\underbrace{\int_0^1 {1\over x^p}\, dx}_{I_0}\ +\ 
\underbrace{\int_1^\infty {1\over x^p}\, dx}_{I_\infty}
$$
Then the integral $I$ converges if and only if both of the integrals $I_o$ and $I_\infty$ converge. If $I$ converges, it converges to the value of $I_0+I_\infty$.  Note that to show $I$ diverges (if it does) it suffices to show that one of $I_o$ or $I_\infty$ diverges.
If you have the $p$-test in hand, this should be an easy problem. Consider the integral $I_\infty$ for $p\le1$, and consider the integral $I_o$ for $p>1$.
If you don't have the $p$-test in hand, you'd compute:
$$\tag{1}
I_0=\int_0^1 {1\over x^p}\, dx=\lim_{a\rightarrow0^+} \int_a^1 {1\over x^p}\, dx
$$
and
$$\tag{2}
I_\infty=\int_1^\infty {1\over x^p}\, dx
=\lim_{b\rightarrow\infty} \int_1^b {1\over x^p}\, dx
$$
The integral  $I_o$ or $I_\infty$ converges if and only if the respective limit above converges. 
It would be best here to consider three cases: $p>1$, $p<1$, and $p=1$. A hint here (as above) is to consider $I_o$ for $p>1$ and $I_\infty$ for the other cases.
