# Can we change $\triangle$ratio of Riemann integral? $\int_a^b f(x)(a(dx))^{b(dx)}\quad a,b\in \mathbb R^{\neq0}$

I wonder whether we can take integral as following or not? And, do they make any sense?

$$f:\quad\text{is continious in [a,b]}$$$$\displaystyle\int_a^b f(x)(dx)^2\tag1$$

$$\displaystyle\int_a^b f(x)a^{(dx)}\quad\quad (a\in\mathbb R)\tag2$$

$$\int_a^b f(x)(dx)^{(dx)}\tag3$$

We know definition of Riemann integral, https://en.wikipedia.org/wiki/Riemann_integral

Hence,If we write the Riemann form,for example $(1)$;

$$\displaystyle\int_a^b f(x)(dx)^2=\lim\limits_{n\to\infty}\sum_{i=1}^nf\left(a+i\frac{b-a}{n}\right)\left(\frac{b-a}{n}\right)^2=\underbrace{\lim\limits_{n\to\infty}\sum_{i=1}^n}_{\int}\underbrace{f\left(a+i\frac{b-a}{n}\right)\left(\frac{b-a}{n}\right)}_{\text{What is it?}}\underbrace{\left(\frac{b-a}{n}\right)}_{dx}$$

What I gonna ask is how we can evaluate and give a sense them?

• If we let $a=0$, $b=1$ and then we start testing some $f(x)$ we can get an idea. For $f(x)=x$ we clearly get $$\int_0^1 x(dx)^2 = \lim_{n \to \infty} \sum_{k=1}^n \frac{k}{n^3}=\lim_{n \to \infty} \frac{n+1}{2n^2} = 0$$ while for $f(x) = (1/x^2)e^{1/x^2}$ we get $$\int_0^1 (1/x^2)e^{1/x^2}(dx)^2=\lim_{n\to\infty}\sum_{k=1}^n \frac{e^{(n/k)^2}}{k^2}>\lim_{n\to\infty}n^2=\infty$$ – Brevan Ellefsen Apr 1 '17 at 20:02

You can try and do that but it doesn't seem you'll get something that is worth your while. For example, using your definitions we have

$$\int_a^b f(x) \, (dx)^2 = \lim_{n \to \infty} \left( \sum_{i=1}^n f \left( a + i \frac{b - a}{n} \right) \left( \frac{b - a}{n} \right)^2 \right) = \\\lim_{n \to \infty} \frac{b - a}{n} \sum_{i=1}^n f \left( a + i \frac{b - a}{n} \right) \left( \frac{b - a}{n} \right) = \\ \left( \lim_{n \to \infty} \frac{b-a}{n} \right) \left( \lim_{n \to \infty} \sum_{i=1}^n f \left( a + i \frac{b - a}{n} \right) \left( \frac{b - a}{n} \right) \right) = 0 \cdot \int_a^b f(x) \, dx = 0$$

for all continuos $f$. Similarly, if $c \geq 1$ and $f > 0$, then

$$\int_a^b f(x) c^{dx} = \lim_{n \to \infty} \left( \sum_{i=1}^n f \left( a + i \frac{b - a}{n} \right) c^{\frac{b - a}{n}} \right) \geq \lim_{n \to \infty} \sum_{i=1}^n f \left( a + i\frac{b - a}{n} \right) = \infty.$$

• Intuitively, "$dx"$ has the correct order of magnitude to sum against $f(c_i)$ and get a finite result for nice functions. If you replace $dx$ by some function of $dx$, it will screw the order of magnitude and in most cases will yield trivial and meaningless results. – levap Apr 1 '17 at 20:30
• Out of curiosity, in my comment above I attempted to integrate $\int_0^1 f(x) (dx)^2$ where $f(x) = (1/x)\exp(1/x)$ and I got $\infty$... did I make a mistake in my calculations, and if so where? – Brevan Ellefsen Apr 1 '17 at 21:29
• @BrevanEllefsen: In my calculation, I used the fact that $f$ is Riemann integrable on $[a,b]$ (because it is continuous). In your case, the integrand is not Riemann integrable (even in the extended sense) and definitely not continuous which is what the OP asked about. – levap Apr 1 '17 at 21:31
• To me, the really interesting one is $(dx)^{(dx)}$... the sum seems to be an asymptotically linear function in $n$ for most common functions $f(x)$ (logarithms, exponentials, power functions, etc.). Logarithms go to $-\infty$ while any powers of $x$ as well as exponentials seem to go to $+\infty$ – Brevan Ellefsen Apr 1 '17 at 22:02
• @BrevanEllefsen: Since $\lim_{x \to 0^{+}} x^x = 1$, the expression $dx^{dx}$ is close to $1$ as the mesh gets finer. This will imply that for positive continuous functions the "integral" $\int_a^b f(x) \, dx^{dx}$ will be $+\infty$. – levap Apr 1 '17 at 22:07