# Showing the limit of $\int_x^\infty (y \log y)^{-1}dy$ is zero

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

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

$$\int \limits_{x}^{\infty} \frac{1}{y \ln y} dy > \int \limits_{x}^{x^2} \frac{1}{y \ln y} dy = \ln \ln x^2 -\ln \ln x = \ln 2 \approx 0.693 > 0$$ 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 $$\epsilon = 0.1$$
• Thanks for your answer. That makes a lot of sense to me. Do you have any advice for how else I can show $\lim{ x \to \infty} x\sum_{k=x}^\infty (k^2logk)^{-1}=0$? Oct 6, 2018 at 11:10
• 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 $0$ . Oct 6, 2018 at 11:16
• The question does not make sense because $\int_x^{\infty} (y log \, y)^{-1}\, dy$ does not exist. Oct 6, 2018 at 11:45