Convergence of an infinite Riemann sum to an integral Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be smooth, bounded, uniformly continuous, and $|f(x)| \leq 1/|x|^{N}$ for any $N$. Then is it true that $$\frac{1}{n}\sum_{k = -\infty}^{\infty}f(k/n) \rightarrow \int_{-\infty}^{\infty}f(x)\, dx$$ as $n \rightarrow \infty$?
 A: I would say yes. The condition $|f(x)| \leq \frac{1}{|x|^N}$ for any $N\in \mathbb{N}$ implies that the function $f\equiv 0$ outside of the closed interval $[-1,1]$. Since $f$ is continuous and bounded  on $[-1,1]$;  it is Riemann integrable, which means
$$\lim_{n\rightarrow \infty} \sum_{k=-n}^{n-1} f\left(\frac{k}{n}\right)\frac{1}{n} = \int_{[0,1]} f(x) dx = \int_{\mathbb{R}} f(x) dx $$
A: As Xiao mentions, if $|f(x)|\le|x|^{-N}$ for any $N\in\mathbb{N}$, then $f$ is supported in $[-1,1]$, and then this becomes a standard Riemann integral. However, if you simply meant that $f(x)=O\left(|x|^{-N}\right)$, then the conclusion is not necessarily true.

Note that for a smooth function,
$$
\hspace{-1cm}\lim_{n\to\infty}\frac{n}{b-a}\left(\int_a^bf(x)\,\mathrm{d}x-\sum_{k=1}^nf\left(a+(b-a)\frac kn\right)\frac{b-a}n\right)=-\frac12\int_a^bf'(x)\,\mathrm{d}x\tag{1}
$$
Thus, even if the increment $\Delta x=\frac{b-a}{n}$ gets smaller, the difference
$$
\int_a^bf(x)\,\mathrm{d}x-\sum_{k=1}^nf\left(a+(b-a)\frac kn\right)\frac{b-a}n\tag{2}
$$
might not converge unless $f'\in L^1$.
Consider the function $f(x)=e^{-x}\sin\left(e^{2x}\right)$ on $[0,\infty)$. $f(x)$ decays faster than any power of $x$, yet
$$
f'(x)=2e^x\cos\left(e^{2x}\right)-e^{-x}\sin\left(e^{2x}\right)
$$
is not in $L^1$.
