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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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Split the integral up first as $\int_0^1{1\over x^p}\, dx+\int_1^\infty{1\over x^p}\, dx$. Then consider three cases: $p<1$, $p=1$, and $p>1$. For each case determine if the original integral converges (both $\int_0^1{1\over x^p}\, dx$ and $\int_1^\infty{1\over x^p}\, dx$ have to converge in order for the original integral to converge; if one of the two diverges, so does the original integral). –  David Mitra Feb 23 '12 at 1:23
    
Oh, if you have the $p$-test in hand, you could certainly appeal to it... –  David Mitra Feb 23 '12 at 1:32

1 Answer 1

up vote 4 down vote accepted

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 !! –  Ryu Feb 23 '12 at 4:48

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