1
$\begingroup$

I'm trying to show

$$\lim_{x \to \infty} \int_x^\infty (y \log y)^{-1} dy = 0$$

In order to finish a proof. The problem I'm having is that without the limit, I know the integral diverges, and hence when I use substitution I end up with indeterminate form.

I think rather than using a substitution like $v = \log x$, I need to re-write the integral in the form of $e^{-t}$ so that the integral can be expressed in a proper form.

Any hints/advice is appreciated thank you!

Edit:

Using the substitution $t=y \log y$ I end up at an integral of the form $\int \frac{1}{1+e^t}dt$ which does not seem to help as I again end up with an indeterminate form

$\endgroup$
1
  • 1
    $\begingroup$ We have $\int(y\log y)^{-1}dy=\log(\log y).$ I am afraid the indefinite integral doesn't converge. $\endgroup$
    – user376343
    Oct 6, 2018 at 11:01

1 Answer 1

0
$\begingroup$

$\int \limits_{x}^{\infty} \frac{1}{y \ln y} dy > \int \limits_{x}^{x^2} \frac{1}{y \ln y} dy = \ln \ln x^2 -\ln \ln x = \ln 2 \approx 0.693 > 0$ for all $0 < x \in \mathbb{R}$, so if $ \lim \limits_{x \to \infty} \int \limits_{x}^{\infty } \frac{1}{y \ln y} dy = 0$ then we would have that for all $ x > x_0$ that $ \int \limits_{ x}^{x^2} \frac{1}{y \ln y }d y < \epsilon$ which contradict the fact that $\int \limits_{x}^{x^2} \frac{1}{y \ln y} dy = \ln \ln x^2 -\ln \ln x = \ln 2 \approx 0.693 $ and $\epsilon = 0.1$

$\endgroup$
4
  • $\begingroup$ Thanks for your answer. That makes a lot of sense to me. Do you have any advice for how else I can show $ \lim{ x \to \infty} x\sum_{k=x}^\infty (k^2logk)^{-1}=0$? $\endgroup$
    – Xiaomi
    Oct 6, 2018 at 11:10
  • 2
    $\begingroup$ for positive $x$ , every term is positive so we know for sure that the limits is $\geq 0$ , and also $x \sum \limits_{k=x}^{\infty} \frac{1}{ k^2 \ln k} <x \int \limits_{x}^{\infty } \frac{1}{t^2 \ln t} dt <x \frac{1}{\ln x} \int \limits_{x}^{\infty} \frac{1}{t^2} dt = x \frac{1}{\ln x} \frac{1}{x} = \lim \limits_{x \to \infty } \frac{1}{\ln x} = 0$ and by squeeze theorem the limit is $0$ . $\endgroup$
    – Ahmad
    Oct 6, 2018 at 11:16
  • $\begingroup$ The question does not make sense because $\int_x^{\infty} (y log \, y)^{-1}\, dy$ does not exist. $\endgroup$ Oct 6, 2018 at 11:45
  • $\begingroup$ @KaviRamaMurthy because of that i gave a proof by contradiction, which is essentially to say that the integral is divergent. $\endgroup$
    – Ahmad
    Oct 6, 2018 at 11:47

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.