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 $\log{t} \le t$.


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

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

  • $\begingroup$ 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

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