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

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?

• 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? – 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 $$L=0$$.