2

Suppose $f$ is a continuous decreasing function on $[0, \infty)$ and $\int^{\infty}_0 f(t)\ \mathrm {d} t$ converges. Prove that $\lim\limits_{x\to\infty} xf(x) = 0\ $.

I am currently preparing for my final calculus examination and this question appeared in last semester's paper. I have been staring at it for hours but have no idea how to even approach it. Any hints/intuitions on how to work this will be greatly appreciated :)

P.S. I am only taking an introductory calculus module in college, so please do not provide solutions which require "heavy machinery" :)

Ethan Mark
  • 2,167
  • Another one: https://math.stackexchange.com/q/328045. – Martin R Nov 26 '20 at 16:15
  • Well, at least, "staring at it for hours" is honest. It doesn't work, unless you're better at hypnotizing than most people, but the latter will most likely phantasize about "struggling for hours" or "trying 23 methods I know" (meaning exactly the same, though). –  Nov 26 '20 at 16:21
  • @MartinR Perhaps I should have added that I did see a few solutions to this, but I do not understand them fully. For example, in https://math.stackexchange.com/questions/119409/if-monotone-decreasing-and-int-0-infty-fxdx-infty-then-lim-x-to-infty, the top voted answer uses $\frac x 2$ and $x$ as the limits of integration, but how did he arrive at those values? Are they arbitrary? If so, what is the intuition behind it? – Ethan Mark Nov 26 '20 at 16:48
  • The idea is that the convergence of $\int^{\infty}_0 f(t) dt $ implies that $\int_a^b f(t) dt$ becomes small for large $a, b$, and $\int_a^b f(t) dt > (b-a) f(b)$ due to the monotony. Then find “suitable” values to get the desired conclusion, e.g. $a=x$ and $b=2x$. – Btw, if you have seen previous solutions to the same problem (but not understood) then it is a good idea to add that information to the question. – Martin R Nov 26 '20 at 17:04
  • @MartinR Yes. On hindsight, I should have done so, but my post has already been closed, so it is a little too late for that. Haha. Regarding $\int^b_a f(t)\ \mathrm{d} t > (b - a)f(b)\ $, in layman terms, means that the area under the graph of $f(t)$ is always greater than the average area (a rectangle) under the same graph right? However, I am unable to see why that is true. – Ethan Mark Nov 26 '20 at 17:18
  • $f(t) \ge f(b)$ for all $t \in [a, b]$, therefore $\int_a^b f(t) dt \ge \int_a^b f(b) dt = (b-a) f(b)$. – Martin R Nov 26 '20 at 17:19
  • @MartinR I see. In that case I understand the accepted answer in https://math.stackexchange.com/questions/119409/if-monotone-decreasing-and-int-0-infty-fxdx-infty-then-lim-x-to-infty more or less, but could you expound on why the convergence of $\int^{\infty}_0 f(t) \mathrm {d} t\ $ implies that $\int^b_a f(t) \mathrm {d} t\ $ tends to $0$ for large $a$ and $b$? I thought that convergence just means the integral tends to some constant value (not necessarily $0$)? In particular, I do not understand the "added" portion of the accepted answer. Unless... my understanding of convergence is flawed. – Ethan Mark Nov 26 '20 at 17:24
  • $\int_a^\infty f(t) dt \to 0$ for $a \to \infty$. – Martin R Nov 26 '20 at 17:33
  • @MartinR Oh. Okay the approach to this question seems quite trivial now. Thank you for your help and sorry for bothering :) – Ethan Mark Nov 26 '20 at 17:35

0 Answers0