I'm not exactly sure how to ask my question, or if this is the right place:
But I was talking to my Fourier Analysis professor after class today, and in the conversation he had mentioned that a function being in the set of absolutely integrable functions, that is, $\int_{-\infty}^{\infty} f(x)\,dx$ exists and is finite, does NOT imply that the $f$ vanishes at infinity.
He very briefly described a function which has rectangles with widths of $1, \frac{1}{2^2}, \frac{1}{3^2},..$, all with height $1$, and being able to "smooth it out" to show the integral converges by comparison to the series of $\frac{1}{n^2}$, but the limit at infinity of the function we made would be $1$. I mean, of course, I see that the widths are tending to $0$, and the height is remaining at $1$, but how can we prove such a thing?
Does this make sense? Could anyone perhaps explain further? I'm sorry if the phrasing is inadequate.
Here is a similar idea to the one your professor was describing. At each positive integer $n>3$, put a triangle of height $2n$ and width $1/n^3$. Then the function $f$ is continuous and its integral is $$ \int f(x)\,dx = \sum_{n=4}^\infty\frac{1}{n^2} < \infty. $$ However, $\limsup_{x\to\infty}f(x) = \infty$, which is to say that $f$ does not vanish at infinity. The problem with $f(x)$ is that it varies a lot, so you might imagine that if you impose some more regularity on it, then it might vanish at infinity, and this is true. If $f$ is uniformly continuous and $\int |f(x)|\,dx<\infty$, then $|f(x)|\to0$ as $|x|\to\infty$.