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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,

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    Oh, if you have the $p$-test in hand, you could certainly appeal to it...2012-02-23

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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.

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    I got it !! Thank you !!2012-02-23