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!


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

  • 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


$\int \limits_{x}^{\infty} \frac{1}{y \ln y} dy > \int \limits_{x}^{x^2} \frac{1}{y \ln y} dy = \ln \ln x^2 -\ln \ln x = \ln 2 \approx 0.693 > 0$ 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 $\epsilon = 0.1$

  • $\begingroup$ 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

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