Converging improper integral have sequence with limit of zero

Question

I've the following statement:

Let $ f : \mathbb{R} \rightarrow \mathbb{R} $ integrable function. If $ \int_0^\infty f(t)dt $ converge, does sequence $ (x_n)\in \mathbb{R} $ exist such that: $ \lim_{n\rightarrow 0} x_n = \infty $ and $ \lim_{n\rightarrow 0} f(x_n) = 0 $

I think it is true, because in order to the integral to converge it must be near zero infinite number of times so it possible to build sequence such that $ f(x_n)$ will converge to zero. Yet I'm not 100% sure about this or how to find these points. So any help will be great.

Thanks.

Answer 1

The answer is NO without further assumptions on $f$ other than it is integrable. For a counter-example, consider the function

$$f(x) = (-1)^{\left\lfloor x^2 \right\rfloor}$$

For any $k \in \mathbb{Z}_{+}$, define $$F_k = \int_{\sqrt{2k-2}}^{\sqrt{2k}} f(x) dx = -\sqrt{2k-2} + 2\sqrt{2k-1} -\sqrt{2k}$$ Notice for large $k$,${}^{\color{blue}{[1]}}$

$$\begin{align}F_k &= (\sqrt{2k-1} - \sqrt{2k-2})-(\sqrt{2k}-\sqrt{2k-1})\\ &= \frac{1}{\sqrt{2k-1}+\sqrt{2k-2}} - \frac{1}{\sqrt{2k}+\sqrt{2k-1}}\\ &= \frac{2}{(\sqrt{2k-1}+\sqrt{2k-2})(\sqrt{2k}+\sqrt{2k-1})(\sqrt{2k}+\sqrt{2k-2})}\\ &\sim \frac{1}{8\sqrt{2k^3}} \end{align} $$ We find as a sequence in $n$, $\displaystyle\;\int_0^{\sqrt{2n}} f(x) dx = \sum_{k=1}^n F_k$ converges to some limit $\Delta$ as $n \to \infty$.${}^{\color{blue}{[2]}}$

For any $\epsilon > 0$, choose a $N \in \mathbb{Z}_{+}$ so large such that

$$\sqrt{2N+1}-\sqrt{2N} < \frac{\epsilon}{2} \quad\text{ and }\quad \left|\sum_{k=1}^m F_k - \Delta \right| < \frac{\epsilon}{2},\forall m \ge N$$

For any $y > \sqrt{2N}$, let $n \ge N$ be the integer such that $y \in \left[\sqrt{2n},\sqrt{2n+2}\right]$, we have

$$\left|\int_0^y f(x) dx - \int_0^{\sqrt{2n}} f(x) dx\right| = \left|\int_{\sqrt{2n}}^y f(x) dx\right| \le \sqrt{2n+1}-\sqrt{2n}\\ \le \sqrt{2N+1}-\sqrt{2N} < \frac{\epsilon}{2}$$

This leads to $$\left| \int_0^y f(x) dx - \Delta \right| \le \left| \int_{\sqrt{2n}}^{y} f(x) dx \right| + \left| \sum_{k=1}^n F_k - \Delta \right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

Since $\epsilon$ is arbitrary, $\displaystyle\;\lim\limits_{y\to\infty}\int_0^y f(x)dx\;$ exists and equal to $\displaystyle\;\Delta = \sum_{k=1}^\infty F_k\;$.

By definition, $|f(x)| = 1$ for all $x$. It is impossible to build a sequence $x_n$ such that $f(x_n)$ converges to $0$.

Random Notes

• $\color{blue}{[1]}$ - Another way to see the $F_k \sim O\left(\frac{1}{\sqrt{k^3}}\right)$ dependence goes like this. The expression $$F_k = -\sqrt{2k-2} + 2\sqrt{2k-1} - \sqrt{2k}$$ has the form of $2^{nd}$ order finite difference for function $-\sqrt{x}$ at $x = 2k-1$. By a generalization of mean value theorem to higher order finite differences, there is a $\xi \in (2k-2,2k)$ such that $$F_k = \left.\frac{d^2}{dx^2}(-\sqrt{x})\right|_{x=\xi} = \frac{1}{4\sqrt{\xi^3}} \sim \frac{1}{8\sqrt{2k^3}}$$

• $\color{blue}{[2]}$ - It can be shown $\Delta = 2(1-\sqrt{8})\zeta\left(-\frac12\right) \approx 0.76020962521937$.

Answer 2

If $\int_0^\infty f(t) dt$ converges, then the sequence $(\int_0^n f(t) dt)_{n \in \mathbb{N}}$ is convergent and in particular a cauchy sequence. Therefore, for any $n \in \mathbb{N}$ you find a $m \in \mathbb{N}$, with

$$ \int_m^{m+1} f(t) dt < \frac{1}{n}$$

this implies a $x_n \in [m,m+1]$ with $f(x_n) \leq \frac{1}{n}$.

Therefore $\lim f(x_n) \leq 0$. If $f$ is non-negative you get $\lim f(x_n) = 0$. Also note that by construction you can choose $m$ arbitrary large, therefore $x_n \to \infty$.

Edit: This solution only works for non-negative f. However, you can find a sequence $(x_n),(y_n)$ with $ \lim_{n \to \infty} f(x_n) \leq 0 $ and $\lim_{n\to\infty} f(y_n) \geq 0$. Maybe one can use a similar argument to find a solution.