Improper Integral

[akoaysigod]

int(1 to infinity) 1/x^p dx

Find the values of p so that the following integral converges.

I've gotten it down to

limit t->infinity of -1/(p+1) x^(-p+1) from 1 to t

t^(-p+1)/(-p+1) + 1^(p+1)/p+1

I think that's right up to there but I'm not sure how to interpret that to get an answer for the values of p that I would need. I'm pretty sure I don't need to do anything with the right hand side, I think. Which leaves just the left half.

 

[DrLee]

int(1 to infinity) 1/x^p dx

Find the values of p so that the following integral converges.

I've gotten it down to

limit t->infinity of -1/(p+1) x^(-p+1) from 1 to t

t^(-p+1)/(-p+1) + 1^(p+1)/p+1

I think that's right up to there but I'm not sure how to interpret that to get an answer for the values of p that I would need. I'm pretty sure I don't need to do anything with the right hand side, I think. Which leaves just the left half.

I compute the expression as 1/(1-p) [t^(1-p) - 1^(1-p)] for which we need to evaluate lim t-> infinity.

For p = 1, the expression blows up, but for any p>1, the first term in brackets goes to zero as t-> infinity. Therefore it looks like the integral converges for all p>1.

 

[BigGlenntheHeavy]

\(\displaystyle \int_{1}^{\infty}\frac{dx}{x^{p}} \ = \ \lim_{b\to\infty} \int_{1}^{b}x^{-p}dx\)

\(\displaystyle = \ \lim_{b\to\infty} \frac{x^{1-p}}{1-p}\bigg]_{1}^{b} \ = \ \lim_{b\to\infty}\frac{b^{1-p}-1}{1-p}\)

\(\displaystyle If \ p \ = \ 1, \ we \ have \ indeterminate \ form \ \frac{0}{0} \ which \ diverges\)

\(\displaystyle If \ p \ < \ 1, \ we \ have \ divergence \ (try \ example).\)

\(\displaystyle Hence, \ for \ all \ p \ > \ 1, \ we \ have \ convergence \ (try \ example).\)

\(\displaystyle Note: \ For \ p \ a \ positive \ constant, \ \int_{1}^{\infty}\frac{1}{x^{p}}dx \ converges \ if \ p \ > \ 1, \ and \ diverges \ if \ 0 \ < \ p \ \le \ 1.\)

 

[akoaysigod]

Oh wow, I should have got that, thank you.