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I have a problem in which I have to find a continuous function $f(x)$ that:

  1. Is absolutely integrable on $\mathbb{R}$.
  2. Does not converge to $0$ as $x \to \infty$.

From what I understand, it is an oscillating function that I am after, and one function that sufficiently fulfills these conditions is $f(x) = \dfrac{\sin(x^2)}{(1+x^2)}$.

However, I have a problem understanding why this is a sufficient function and not e.g. $g(x) = \dfrac{\sin(x)}{(1+x^2)}$. As I understand it, the $\sin(x^2)$ term in $f$ will have a quadratically decreasing period of time that will ultimately hinder $f$ from converging to $0$. This is not a problem in $g$, which will make it converge to $0$.

How can I prove that $f$ does not converge to $0$ formally? I have only seen a lot of handwaving when trying to find an answer to this.

M. A.
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Warnstad
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1 Answers1

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The functions $\frac{\sin(x)}{1 + x^2}$ and $\frac{\sin(x^2)}{1 + x^2}$ both tend to $0$ as $x \to \infty$. Both have absolutely bounded numerators and quickly growing denominators.

It's not necessary for the function you're looking for to oscillate at all. One way to make an integrable function that doesn't tend to zero would be to have occasional "bumps" appear. For example, for each $n \geq 1$, put a triangle raising from $0$ to $1$ over the interval $[2^n - 1/n^2, 2^n]$, and then falling from $1$ to $0$ on $[2^n, 2^n + 1/n^2]$. Everywhere else you can have your function be $0$. This is continuous (though not differentiable), the total area is finite, it's integrable, and the function doesn't tend to $0$.

davidlowryduda
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