I have the following statement which I have to prove.

If $\int_a^{+\infty} f(x)dx $ converges, and $\lim_{x\to + \infty} f(x) = L$. Then prove that $L=0$

The idea is to prove it only using the improper integral definition, but I don’t know how to do it. Thanks in advance!

Edit: I wrote down the limit definition for f, and then integrate both sides. Doing that I got: $\int_a^{+\infty} L-\varepsilon \ dx \leq L’ $ (Being $L’= \int_a^{+\infty} f(x)dx $ (Since it converges) That is the hint that the professor showed us in a drawing. But how can I use this information to conclude that L must be equal to 0?

  • $\begingroup$ WLOG let $a=0$. Since $\int_0^\infty f(x)\,dx = \sum_{n=0}^\infty\int_n^{n+1}f(x)\,dx$ converges, we have $\int_{n}^{n+1}f(x)\,dx\to 0$ as $n\to\infty$. Can you go on? $\endgroup$ – amsmath Sep 2 at 18:14

Hint: Assume that $L \neq 0$, and prove that the integral diverges.


Note that for any $m\ge a$ we have $\lim_{n \to \infty} {1 \over n-m} \int_m^n f(x)dx =0$.

Let $\epsilon>0$ and choose $m \ge a$ such that $|f(x)-L| < \epsilon$. Then we see that $L-\epsilon \le \lim_{n \to \infty} {1 \over n-m} \int_m^n f(x)dx \le L+\epsilon$ and so $0 \in [L-\epsilon, L+\epsilon]$ for all $\epsilon>0$. Hence $L=0$.

  • $\begingroup$ Why the downvote? $\endgroup$ – copper.hat Sep 2 at 19:08
  • 1
    $\begingroup$ Nice solution, I like it more than my hint! (+1) $\endgroup$ – Botond Sep 2 at 20:13

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