How do you determine if the improper integral converges or diverges $\int \frac{1}{x} \cdot {(ln x)}^{p} d x$ $int 1/x*(lnx)^p dx$ from 2 to infinity?

Question

8061 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-15T15:13:05

Integrate by substitution and evaluate the limit.

For $p \neq - 1$ $p != -1$ , we get

$lim_(brarroo) int_2^b (lnx)^p 1/x dx = lim_(brarroo) (lnx)^(p+1)/(p+1)]_2^b$

$= 1/(p+1) lim_(brarroo)[(lnb)^(p+1) - (ln2)^(p+1)]$

For $p > -1$ , we have $lim_(brarroo) (lnb)^(p+1) = oo$ so the integral diverges.

For $p < -1$ , we have $lim_(brarroo) (lnb)^(p+1) = 0$ so the integral converges.

For $p = -1$ , we get $int 1/(xlnx) dx = ln(lnx)$ , so the integral diverges.

How do you determine if the improper integral converges or diverges int 1/x*(lnx)^p dx∫1x⋅(lnx)pdxint 1/x*(lnx)^p dx from 2 to infinity?