Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\displaystyle \int \frac{1}{t} dt = \ln t$ diverges.

How do I show that $\displaystyle \int_2^\infty \frac{1}{t ~\log^2 t} dt$ is convergent?

share|improve this question
    
Hint: let $u=\log t$. (For the integral from $2$ to $\infty$, say.) –  David Mitra Aug 3 '12 at 2:37
1  
And, you do need limits on those integrals in order for your question to make sense... –  David Mitra Aug 3 '12 at 2:46
    
If you don't write down any limits for your integral there is no meaning to "divergent", "convergent" –  DonAntonio Aug 3 '12 at 2:58
    
On what interval? –  ncmathsadist Aug 3 '12 at 2:59

1 Answer 1

up vote 3 down vote accepted

To help yourself figure out the appropriate substitution, reorganize the integral into: $$ \int \frac{1}{\log^2 t} \frac{dt}{t} $$ Does this ring a bell? $\dfrac{dt}{t}$? If we take $u = \log t,$ then we have $du = \dfrac{dt}{t}$ and $\dfrac{1}{\log^2 t} = \dfrac{1}{u^2}.$ So the integral is: $$ \int \frac{1}{t\log^2 t} dt = \int \frac{1}{u^2} du = - \frac{1}{u} + \text{const} = -\frac{1}{\log t} + \text{const}.$$ Now $$ \int_2^{\infty} \frac{1}{t\log^2 t} dt = - \lim_{n \to \infty} \frac{1}{\log n} + \frac{1}{\log 2} = 0 + \frac{1}{\log 2}.$$

share|improve this answer
1  
Did you guess the integral's limits? –  DonAntonio Aug 3 '12 at 3:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.