From an infinite summation to an integral It is a well known formula that, given a suitably well behaved function, we have
$$
\lim_{n\to\infty} \dfrac{1}{n}\sum_{k=1}^{n} f(k/n) = \int_0^1 f(x)dx.
$$
I am interested in an infinite version of this. Is it true that
$$
\lim_{n\to\infty} \dfrac{1}{n}\sum_{k=-\infty}^{\infty} f(k/n) = \int_{-\infty}^\infty f(x)dx?
$$
Edit: We can assume that the function has sufficient smoothness and decays sufficiently fast for the summations and integrations to make sense.
 A: Let $\varphi_n(t)=f\left(\frac{\lfloor nt\rfloor}{n}\right)$, then
$$ \int_{-\infty}^{+\infty}\varphi_n(t)dt=\sum_{k=-\infty}^{+\infty}\int_{\frac{k}{n}}^{\frac{k+1}{n}}\varphi_n(t)dt=\frac{1}{n}\sum_{k=-\infty}^{+\infty}f\left(\frac{k}{n}\right) $$
However for all $t\in\mathbb{R}$, $\lim\limits_{n\rightarrow +\infty}\varphi_n(t)=f(t)$ (I suppose that $f$ is continuous). If we get an upper bound $|\varphi_n|\leqslant g$ where $g\in L^1(\mathbb{R})$ then we can use the dominated convergence theorem, therefore
$$ \lim\limits_{n\rightarrow+\infty}\int_{-\infty}^{+\infty}\varphi_n(t)dt=\int_{-\infty}^{+\infty}f(t)dt $$
and you get the desired equality. In order to have such an upper bound, we can suppose that $f$ is monotone, thus we have $|\varphi_n(t)|\leqslant |f(t)|$ if $f$ is non-decreasing and $|\varphi_n(t)|\leqslant |f(t-1)|$ if $f$ is decreasing. We can also suppose that there exists $\delta>0$ such that $f(t)=\underset{t\rightarrow\pm\infty}{\mathcal{O}}\left(\frac{1}{t^{1+\delta}}\right)$ because there would exist $M>0$ such that $|f(t)|\leqslant\frac{M}{(1+|t|)^{1+\delta}}$ for all $t\in\mathbb{R}$ and therefore $|\varphi_n(t)|\leqslant\frac{C}{(1+|t|)^{1+\delta}}$ where $C>0$.
A: In my opinion it would not work.
You should do something like this:
$$\lim_{n\to\infty} \sum_{h=-\infty}^{+\infty}\frac{1}{n}\sum_{k=1}^{n}f\bigg(\frac{k}{n}+h\bigg)$$
