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

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Answer 1 · 2016-03-15T15:13:05

Integrate by substitution and evaluate the limit.

For #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.