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

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.
