Convergence of Integral: Splitting and Comparing Functions

[Dell]

i am given the following, and asked if it converges or diverges

\intdx/ln(x) (from 0-1)

what i did was look for a similar function
i took 1/x which is bigger than 1/lnx,
but 1/x diverges from 0-1

can i split my function up and compare
\intdx/ln(x) (from 0-0.5) + \intdx/ln(x) (from 0.5-1)
now i know that 1/x converges from 0.5-1 so 1/lnx does as well

now i need to find a fuction to compare 1/lnx with from 0-0.5
1/lnx < 1/(x+1)
1/(x+1) converges
therefore 1/lnx does as well

therefore 1/lnx is made up of two converging parts from 0-1 so the whole also converges


is this correct?

 

[phreak]

The integral is finite for any interval (k,1), where 0 < k < 1, because continuous, bounded functions are always integrable. You can't compare 1/ln(x) to 1/x or 1/(x+1) for that matter because the technique you're working with only works for positive functions. We're trying to check that the integral isn't NEGATIVE infinity, so finding a function that's larger than it on an interval doesn't work. You need to find a function that decreases to -infinity as x goes to 0 SLOWER than 1/ln(x) on an arbitrary small interval around 0. Considering that ln(x) goes to -infinity as x goes to 0 at the same rate that e^x goes to infinity as x goes to infinity, what could you use?

 

[Dell]

i don't know, 1/e^x ??
i understand what you are saying but don't understand how to put it into practice

 

[Dell]

i tried 1/(e^1/x) but i found that this converges and is smaller than 1/lnx so it doesn't help, please give some idea of what to do

 

[tnutty]

use the integral test, by using integration by parts.

 

[Dell]

how do i integrate in parts, i tried and i keep on going in cirlces!

i made t=lnx, dt=dx/x

then got

integral of ((e^t)/t)dt, made 1/t=u e^tdt=dv, and even tried the other way round, -nothing!