Let $f:[a,\infty)\to\mathbb R$ be a continuous function such that $\int_a^\infty f(x)dx$ exists finitely , then is it true that $\lim_{x \to \infty}f(x)$ exists and is equal to $0$ ? If not , then what if we restrict the range of $f$ to be non-negative only ?
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2$\int_1^{\infty}\cos (x^2),dx$ exists and is finite (integrate by parts and apply Abels test). But $\lim_{x\to \infty}\cos(x^2)$ is not zero and in fact, does not exist. – Mark Viola Nov 13 '15 at 17:28
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You can build a strictly positive $C^\infty$ and integrable function on $\Bbb R$ such that $\lim_{x\to\infty}f(x)$ does not exist. There were at lesat two different explicit examples of such functions on this site. – TZakrevskiy Nov 13 '15 at 17:29
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No, it's not true even if you assume that $f$ is nonnegative. To see this, pick your favorite convergent series with nonnegative terms, e.g. $\sum a_n = \sum 1/n^2$. Define $f$ by placing a triangle centered at each positive integer $n$, with the triangle's area equal to $a_n$. You can do this in such a way that the heights of the triangle remain constant or even grow, by adjusting the widths appropriately (while keeping the widths less than $1$ so the triangles do not overlap).
You can replace the triangles with smooth functions with compact support to get a smooth counterexample.
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Solid answer! +1 ... and one can use "Bump" functions to accomplish the goal of your last sentence. – Mark Viola Nov 13 '15 at 17:31