0
$\begingroup$

Show, using a comparison test, that $\displaystyle \int_2^\infty{\frac{\log{t}}{t^{\frac32}}}\mathrm{d}t$ converges.

All the answers I've tried shows it diverges, taking $\log{t} \le t^{1/2}$ and $\log{t} \le t$.

Cheers

$\endgroup$
  • $\begingroup$ Show us one of those answers. $\endgroup$ – Soham Chowdhury May 17 '13 at 12:42
1
$\begingroup$

Solving $\log t = t^{1/4}$ we get $t_1=4.177, t_2=5503.66$ from W|A.

Differentiating $(\log t)' = \frac{1}{x}$ and $(t^{1/4})' = \frac 1 4 t^{-3/4}$. And $t>t^{3/4}\implies 1/t<t^{-3/4}$.

for $t \ge t_2$, $\log t \le t^{1/4}$.

$$\int_2^\infty \frac{\log t}{t^{3/2}}dt = \int_2^{t_2} \frac{\log t}{t^{3/2}}dt + \int_{t_2}^\infty \frac{\log t}{t^{3/2}}dt \le \int_2^{t_2} \frac{\log t}{t^{3/2}}dt + \int_{t_2}^\infty \frac{t^{1/4}}{t^{3/2}}dt$$.

We know that $\displaystyle \int_{t_2}^\infty \frac{t^{1/4}}{t^{3/2}}dt$ converges. So the integral converges by comparison.

Integrating by parts:: $$\int_2^\infty \frac{\log x}{x^{3/2}}dx = -\frac{2 \log x}{\sqrt x} \big |_2^\infty - \int_2^{\infty} \frac 1 x \frac{-2}{\sqrt x}dx = \sqrt 2 \log 2 + 2 \sqrt 2 $$

$\endgroup$
  • $\begingroup$ Why do you need all that rigmarole about a comparison test? If we're evaluating it, just do it all the way, it's not hard. Then take a limit. $\endgroup$ – leeabarnett May 17 '13 at 12:48
  • $\begingroup$ And that way you don't have to use Wolfram, the calculus student's succubus $\endgroup$ – leeabarnett May 17 '13 at 12:51
  • $\begingroup$ @leeabarnett yes I see your point, since it is asked to show using comparison test, I showed it ... but can't find nice way for nice expression of $t_2$. $\endgroup$ – Santosh Linkha May 17 '13 at 12:55
  • $\begingroup$ Oh I didn't see that about having to use comparison tests. I think there must be a better way then. $\endgroup$ – leeabarnett May 17 '13 at 12:58
  • $\begingroup$ @leeabarnett i am hoping so ${{}}$. Well it was my try :(( $\endgroup$ – Santosh Linkha May 17 '13 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.