Asymptotic behavior of an integrable function

Question

Assume that $f(x)\geq 0$ for $0\le x<\infty$ and $\int_{0}^{\infty} f(x)\, dx<\infty$. Could we conclude that $\lim_{x\to\infty} f(x)=0$? My idea is that if we can find a counter example, then $f(x)$ must be an oscillating function in $x$. Any suggestion, idea, or comment is welcome, thanks!

Answer 1

Let $m\ge0$ and $n\gt2m+2$. Then $$ \begin{align} \int_0^\infty x^m\left(\frac{1+\cos(2\pi x)}2\right)^{x^n}\,\mathrm{d}x &=\sum_{k=0}^\infty\int_0^1(x+k)^m\left(\frac{1+\cos(2\pi(x+k))}2\right)^{(x+k)^n}\,\mathrm{d}x\\ &\le\sum_{k=0}^\infty(k+1)^m\int_0^1\left(\frac{1+\cos(2\pi x)}2\right)^{k^n}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty(k+1)^m\int_0^1\cos^{2k^n}(\pi x)\,\mathrm{d}x\\ &=1+\sum_{k=1}^\infty(k+1)^m\frac1{4^{k^n}}\binom{2k^n}{k^n}\\ &\le1+\sum_{k=1}^\infty\frac{2^mk^m}{\sqrt{\pi k^n}}\\ &=1+\frac{2^m}{\sqrt\pi}\zeta\left(\frac n2-m\right) \end{align} $$ where we've used inequality $(10)$ from this answer.

Thus, for $m\ge0$ and $n\gt2m+2$, $$ \int_0^\infty x^m\left(\frac{1+\cos(2\pi x)}2\right)^{x^n}\,\mathrm{d}x\lt\infty $$ However, for $x\in\mathbb{Z}$, we have $x^m\left(\frac{1+\cos(2\pi x)}2\right)^{x^n}=x^m$, so the function does not tend to $0$.

Here is the plot of this function for $m=1$ and $n=5$:

Answer 2

No, you are right there is a counterexample.

Join the following dots: $$\left\{\left(n-\frac{1}{2^n},0\right),(n,1),\left(n+\frac{1}{2^n},0\right)\;:\;n\in\mathbb{N}^+\right\}.$$ and you will obtain the graph of a piecewise linear function with the desired property.

P.S. Here is an unbounded version: $$\left\{\left(n-\frac{1}{4^n},0\right),(n,2^n),\left(n+\frac{1}{4^n},0\right)\;:\;n\in\mathbb{N}^+\right\}. $$