# Show that Riemann sum equals integral

Suppose we have some continuous function $$\textbf{F}\colon \mathbb{R}^{n} \to \mathbb{R}^n$$ (i.e. a force field) and a curve $$\gamma\colon [a, b]\subseteq \mathbb{R} \to \mathbb{R}^{n}$$ in $$C^1$$. Considering the partition $$P$$ of the interval $$[a, b]$$ \begin{align*} P= \left\{t_{0}=a, t_{1}=a + \frac{b-a}{n}, a + 2 \frac{b-a}{n} , \ldots,t_{n}= b\right\} \implies \Delta t = \frac{b-a}{n} .\end{align*} I want to show that \begin{align*} \lim_{n \to\infty} \sum_{k=1}^{n}\left\langle \textbf{F}\left(\gamma(t_{k}^{*})\right), \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)\right\rangle \Delta t &= \int_{a}^{b}\left\langle \textbf{F}\left(\gamma(t)\right),\gamma'(t)\right\rangle \mathrm{d}t \\ &= \lim_{n \to\infty} \sum_{k=1}^{n}\left\langle \textbf{F}\left(\gamma(t_{k}^{*})\right),\gamma'(t_{k}^{*}) \right\rangle\Delta t .\end{align*} So let $$\epsilon>0$$ be arbitrary. We can apply the mean value theorem componentwise, namely for all $$1\le i\le m$$ we have \begin{align*} \forall x,y \in [a, b] \colon \gamma_{i}(y)-\gamma_{i}(x)=\gamma_{i}'(\xi_i) (y-x), \quad \xi_i \in [x, y] .\end{align*}

By uniform continuity of $$\gamma'$$ there exists some $$\delta$$ s.t. for all $$x$$ and for all $$y \in (x-\delta, x+\delta)$$ \begin{align*} \left\|\gamma'(y)-\gamma'(x) \right\|_{2} <\epsilon .\end{align*} Now, we choose $$N>0$$ s.t. \begin{align*} \frac{b-a}{N}< \delta .\end{align*} Then, for all $$n\ge N$$ we find \begin{align*} &\left| \sum_{k=1}^{n}\left(\left\langle \textbf{F}\left(\gamma(t_{k}^{*})\right), \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)\right\rangle \Delta t -\left\langle \textbf{F}\left(\gamma(t_{k}^{*})\right),\gamma'(t_{k}^{*}) \right\rangle\Delta t \right) \right | \\ = &\left| \sum_{k=1}^{n}\Delta t\left\langle \textbf{F} \left(\gamma(t_{k}^{*})\right), \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right) -\gamma'(t_{k-1})+\gamma'(t_{k-1}) -\gamma'(t_{k}^{*})\right\rangle \right | \\ \le & \Delta t\sum_{k=1}^{n}\left\| \textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2}\left\| \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)-\gamma'(t_{k-1})+\gamma'(t_{k-1}) -\gamma'(t_{k}^{*})\right\|_{2} \\ \le &\Delta t\sum_{k=1}^{n} \left\| \textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2}\left(\left\| \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)-\gamma'(t_{k-1})\right\|_2+\left\|\gamma'(t_{k-1}) -\gamma'(t_{k}^{*})\right\|_{2}\right) \\ < &\Delta t \sum_{k=1}^{n} \left\|\textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2} \cdot \epsilon(\sqrt{m} +1) .\end{align*} The last inequality follows from the fact that \begin{align*} \frac{\gamma_{i}(t_{k})-\gamma_{i}(t_{k-1})}{t_{k}-t_{k-1}} = \gamma_{i}'(\xi_{i}), \quad \xi_{i} \in (t_{k-1}, t_{k}) \end{align*} and since $$\xi_i-t_{k-1}<\delta$$ we have \begin{align*} |\gamma_{i}'(\xi_{i})-\gamma_{i}'(t_{k-1})| < \epsilon \end{align*} for all $$1\le i\le m$$.

Since $$\left\|\textbf{F}\left(\gamma(t)\right)\right\|_{2}$$ is a continuous function on a compact set it takes a maximal value, let this be $$M$$. It follows that \begin{align*} \Delta t \sum_{k=1}^{n} \left\|\textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2} \cdot \epsilon(\sqrt{m} +1) \le (b-a)^{2} m\cdot M\cdot \epsilon(\sqrt{m} +1) .\end{align*} Choosing $$\epsilon$$ in the beginning accordingly concludes the proof.

The overall structure of your proof is good but there is a problem with applying the MVT since $$\gamma$$ is a vector valued function.

So I just write what you need to fix avoiding some detail also.

Then $$\left |\sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma(t_k)-\gamma(t_{k-1})\right> - \sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma' (t_k^*) \right> \Delta t\right|$$$$=\left |\sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma(t_k)-\gamma(t_{k-1})-\gamma' (t_k^*) \Delta t\right> \right|$$$$\le\sum_{k=1}^n \|\mathbf F(\gamma(t_k^*))\|_2\cdot \|\gamma(t_k)-\gamma(t_{k-1})-\gamma' (t_k^*) \Delta t\|_2$$$$=\sum_{k=1}^n \|\mathbf F(\gamma(t_k^*))\|_2\cdot \|\theta_k(t_k)-\theta_k(t_{k-1})\|_2$$where $$\,\theta_k (t)=\gamma (t)-\gamma'(t_k^*)\;t$$ .

Apply now to $$\,\theta_k\,$$ a weakened version of the MVT valid for a vector valued function.
See Baby Rudin, third ed., p. 113 .

Then for every $$k$$ there exists $$\xi_k \in [t_{k-1},t_k]$$ such that$$\|\theta_k(t_k)-\theta_k(t_{k-1})\|_2\le \|\theta_k'(\xi_k)\|_2\,\Delta t$$

Since $$\theta_k'(\xi_k)=\gamma' (\xi_k)-\gamma' (t_k^*)$$you get finally$$\left |\sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma(t_k)-\gamma(t_{k-1})\right> - \sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma' (t_k^*) \right>\Delta t \right|$$$$\le \sum_{k=1}^n \|\mathbf F(\gamma(t_k^*))\|_2\cdot \|\gamma' (\xi_k)-\gamma' (t_k^*)\|_2\,\Delta t$$Anyway know that you can succeed even using just the classic monodimensional MVT: this means you must work componentwise of course

Edit

By the uniform continuity of $$\gamma'$$, for every $$\varepsilon>0$$ there exists $$\delta_i>0$$ such that $$|\gamma_i'(x)-\gamma_i'(y)|<\varepsilon\quad\text{if}\quad |x-y|<\delta_i$$Let $$\,\delta=\min\, {\delta_i}$$ .

Then, by the monodimensional MVT, there exists $$\,\xi_{ki}\in[t_{k-1},t_k]\,$$ such that $$\|\gamma(t_k)-\gamma(t_{k-1})-\gamma' (t_k^*) \Delta t\|_2$$$$= \sqrt {\sum_{i=1}^m (\gamma_i' (\xi_{ki})-\gamma_i' (t_k^*))^2}\,\Delta t < \sqrt m\, \varepsilon\, \Delta t\quad \text{if}\quad \Delta t < \delta$$

• I edited my solution, do you think the application of the mean value theorem is correct? Dec 1 '21 at 0:00
• @Hilberto1 You don't need to subtract and add $\gamma'(t_{k-1})$: see my edit. Dec 1 '21 at 13:40
• I see, but it doesn't invalidate the result. Thanks a lot though! Dec 1 '21 at 15:49

The Edit is wrong. If $$\delta_1$$ depends on $$n$$ then so does $$N$$ and you have a problem. Formula (2) is also not correct. You are subtracting a scalar and a vector. You should have $$\|\frac{\gamma(x)-\gamma(y)}{x-y}-\gamma’(x)\| \le\epsilon$$ It’s not clear how you get (2) using uniform continuity of the derivative. I think you need to use either the fundamental theorem of calculus combined with the uniform continuity of the derivative or the mean value theorem applied to each component combined with the uniform continuity of the derivative… Something like that.

• You are right, I will look over this later. Is the rest ok? Nov 27 '21 at 13:02
• Yes, everything else is correct Nov 27 '21 at 14:10
• I edited my answer, is this what you meant? Nov 27 '21 at 14:31
• Almost. You have to apply the mvt to each component $\gamma_k$ of $\gamma$. You will get a different $\xi_k$ for each component. Nov 27 '21 at 21:16