is this true

Question

is this true

anonymous · Answer

$$x \ln ^{p} x  = x( \ln x )^{p}$$

anonymous · Answer

i havent learned that yet, im sorry :(

e.mccormick · Answer

Ummm.... what exactly is going to a power?  The ln?  If so, how?

e.mccormick · Answer

$\ln x^a=a\ln x$

anonymous · Answer

the original question is the integration 
$$\int\limits_{1}^{2} \frac{ dx }{ x \ln ^{p}x }$$

anonymous · Answer

i was thinking of changing the look

anonymous · Answer

this is for finding the value of p for improper integral converges

e.mccormick · Answer

Hmm.  Well, I have no clue how they are doing $\ln^p$ or what that means...  I have never heard of raising the log function itself to a power.  So it must be some symbolism I am just not familiar with.

anonymous · Answer

i have no idea approaching to the answer the value of p

anonymous · Answer

same here

anonymous · Answer

both links solve it but in series http://math.stackexchange.com/questions/641310/find-the-values-of-p-for-which-sum-n-2-infty-frac1n-ln-np-is-conve http://www.math.toronto.edu/canghel/mat135Q/Q11_3.pdf

anonymous · Answer

im stuck

e.mccormick · Answer

well, $\ln^p x$ and $(\ln x)^p$ are different.  The second makes sense because you would evaluate and then take it to a power.  I mean, it would be no different than saying:

$\ln x\cdot  \ln x \cdot \ln x \cdot ... $ for p values of $\ln x$

anonymous · Answer

so its true

e.mccormick · Answer

Not that I know of.  Everything I see in the documentation is like the second notation.  That first notation just makes no sense to me.

anonymous · Answer

aha

anonymous · Answer

found this online : https://www.physicsforums.com/threads/i-dont-get-this-convergent-divergent-problem.109830/

e.mccormick · Answer

Yes, which is still the second notation.

anonymous · Answer

still cant solve it