Find a function $f$ with given conditions

Question

I need a function $f(x)$ such that

1. $f(x) $ is non-negative continuous

2. $\int\limits_0^{\infty}f(x)\ dx$ exists

3. $\lim\limits_{x \rightarrow \infty }f(x) $ does not exists

I think such a function doesn't exist because condition $(2)$ gives that corresponding summation exist which imply as $\lim\limits_{n \rightarrow \infty }f(n)=0 $

Answer 1

What about $$ g_n(x)\stackrel{\text{def}}{=}n e^{-n^6(x-n)^2},\qquad f(x) \stackrel{\text{def}}{=} \sum_{n\geq 1} g_n(x) $$ ? $g_n(x)$ is a positive continuous function, concentrated around $x=n$. It is simple to check that $f(x)$ is positive and continuous. For any $n\in\mathbb{N}^+$ we have $f(n)\geq g_n(n) = n$, hence $\lim_{x\to +\infty}f(x)$ does not exist. However $$ \int_{0}^{+\infty}f(x)\,dx \leq \sum_{n\geq 1}\int_{-\infty}^{+\infty}g_n(x)\,dx = \sum_{n\geq 1}\frac{\sqrt{\pi}}{n^2} = \frac{\pi^{5/2}}{6}$$ so $f(x)$ fulfills the wanted constraints.

Answer 2

Hint: Build a function that in every interval of type $[n, n+1], n\in N$, the function defines a triangle that becomes smaller and smaller as $n$ grows, and maybe you can write that area of the triangles as $\sum a_n$ with $a_n$ defining the triangle's area in $[n, n+1]$.

Maybe the series is a geometric series?