Riemann sum from $0$ to $\infty$

Question

I wonder which limit correctly defines the integral $I= \displaystyle\int_0^\infty u(x)dx $. Is it:

$$ I = \lim_{h \rightarrow 0} \sum_{r=1}^\infty u(rh)h = \color{red}{\lim_{h \rightarrow 0}} \color{blue}{\lim_{n \rightarrow \infty}}u(h)h + u(2h)h + \ldots +u(nh)h$$

Or:

$$ I = \lim_{n \rightarrow \infty} \lim_{h \rightarrow 0} \sum_{r=1}^n u(rh)h = \color{blue}{\lim_{n \rightarrow \infty}}\color{red}{\lim_{h \rightarrow 0}} u(h)h + u(2h)h + \ldots + u(nh)h$$

Answer 1

Neither definition is correct. By definition, we have

\begin{equation} I = \int_0^\infty u(x)\,dx = \lim_{R\to\infty} \int_0^R u(x)\,dx. \end{equation}

Writing the integral as a limit of left Riemann sums, we get

\begin{equation} I = \lim_{R\to\infty} \int_0^R u(x)\,dx =\lim_{R\to\infty}\lim_{n\to\infty} \sum_{r=0}^{n-1} u(rR/n)\frac{R}{n}. \end{equation}

So as in your second definition the outer limit is the one controlling the domain of integration, but you also have to make sure that your Riemann sums reflect the expanding domain.