Nonnegative $C^{\omega}$ function $f$ with $\int_\mathbb{R} f(t) \ dt < + \infty$ but $\lim f(x) \neq 0$

Question

Is there a nonnegative $C^{\omega}$ function $f$ with $\int_\mathbb{R} f(t) \ dt < + \infty$ (in the Riemann sense) but $\lim f(x) \neq 0$ as $x \to +\infty$ or $x \to -\infty$?

Additionally, what can we say about stronger conditions? For example, if $f$ is positive (as opposed to merely nonnegative)? Or instead of $\lim f(x) \neq 0$, $\limsup f(x) = +\infty$?

Unless I'm mistaken, here is an example for the $C^{\infty}$ case, which is, of course, weaker.

Consider the bump function

$$\sigma(x) = \begin{cases} \exp\left( -\frac{1}{1 - x^2}\right) & \mbox |x| < 1\\ 0 & \mbox{$x \in [-2, -1] \cup [1, 2)$} \end{cases}$$

extended $4$ periodically to $\mathbb{R}$. Since this is even, we will only consider $x \geq 0$. We alter this as follows. The "bump" portion of each period (ie. where $f(x) \neq 0$) is scaled horizontally so it's width is very small. A consequence of this is that the "$0$" portion of each period gets relatively larger within each period. We must make the $nth$ bump so thin that the area underneath the bump is less than $2^{-n}$. This can be done since the height is bounded. Note that, since $x \geq 0$, the first bump is sort of a half-bump, but this makes no difference.

Our original function $\sigma$ was $C^{\infty}$, so our new function, which has only been scaled, is as well.

Answer 1

The function

$$f(x) = (x^2+1)\left ( \frac{2+\cos x}{3}\right )^{x^8}$$

is real analytic on $\mathbb R,$ is strictly positive everwhere, and satisfies $\int_{\mathbb R} f(x) \, dx < \infty.$ Note that $f(2n\pi) = (2n\pi)^2 + 1 \to \infty.$

How do we know $\int_{\mathbb R} f(x) \, dx < \infty?$ Laplace's method shows

$$\int_0^{2\pi}\left ( \frac{2+\cos x}{3}\right )^{n}\,dx = O\left ( \frac{1}{\sqrt n}\right )$$

as $n\to \infty.$ Once we know that, we can obtain

$$\int_{2n\pi}^{2(n+1)\pi} f(x)\, dx = O(1/n^2),$$

which gives the result.

A lot more is true actually: A beautiful result of Carleman shows the following: Suppose $g:\mathbb R \to \mathbb R$ is continuous, and $\epsilon: \mathbb R \to (0,\infty)$ is any positive continuous function. Then there exists an entire function $f$ such that $|f(x)-g(x)| < \epsilon(x)$ for all $x\in \mathbb R.$ Since it is easy to find a continuous function that does what you asked for, taking $\epsilon (x) = e^{-x^2}$ and applying Carleman will give you an entire function that does the same thing.

Answer 2

I believe the function $$ f(x) = 1 + \tanh\left[\left(\sin^2x-1\right)e^{x^2}\right]\coth\left(e^{x^2}\right) $$ satisfies your condition. Analytic functions are closed under addition, multiplication, and composition, so the above is analytic itself. The limit as $x\rightarrow \pm\infty$ does not exist, because $f(x)$ returns to 1 every odd multiple of $\pi/2$, but the area under each period vanishes quickly enough to converge.

In fact, the periods vanish quickly enough that I could probably add a growing multiplicative factor in front and make the lim sup infinite.