Integrating a function with $\mathrm dx$ as an exponent

Question

I am a high school student learning calculus. I encountered this what seems to be a challange problem:$$\int(x^{\mathrm dx}-1)$$We have learned some integrating techniques, but we still didn’t learn integration by parts. All the previous problems I encountered didn’t have $\mathrm dx$ as an exponent, so I really don’t know where to start.

Update: Here's how my teacher approached the question

-Multiply the numerator and the denominator by $dx$ then find the integral

$$ \int \frac{\left(x^{d x}-1\right) d x}{d x} $$

Answer 1

This is more a recreational quiz than a real question, as the notation is unorthodoxical and in principle meaningless.

Considering the Taylor development of the exponential,

$$e^t=1+t+\frac{t^2}2+\frac{t^3}{3!}+\cdots$$

we can expand symbolically

$$\int (x^{dx}-1)=\int (e^{\ln x\,dx}-1)=\int(\ln x\,dx+\frac12\ln^2 x\,dx^2+\frac1{3!}\ln^2 x\,dx^3+\cdots).$$

Then as the $dx^k$ represent higher degree infinitesimals, they could be neglected and

$$\int (x^{dx}-1)=\int\ln x\,dx.$$

Anyway, this statement is not rigourous and it is not true that these terms can just be neglected. An expression like $\displaystyle\int f(x)\,dx^2$ is still undefined.

Answer 2

When $dx$ is a true infinitesimal (e.g. interpreted as an infinitesimal in hyperreals) , it is possible to make sense out of the "integral". However, I have strong doubt that this is the intended interpretation.

Since the Riemann like sum $$S(\Delta x) \stackrel{def}{=} \sum x^{\Delta x} - 1$$ is defined for finite $\Delta x > 0$. By transfer principle of hyperreals, the function $S(dx)$ is defined for all infinitesimal $dx > 0$. Like the ordinary construction of integral in the framework of hyperreals, if the standard part of $S(dx)$ is a real number independent of choice of $dx$, we can use it as a definition of the "integral" $\int x^{dx} - 1$.

Translate this back to standard analysis, we can interpret the "integral" as a limit of Riemann like sum $$\int_a^b x^{dx} - 1 = \lim_{\delta(P) \to 0 } \sum_{i=1}^n (x_i^*)^{\Delta x_i} - 1$$

where $P$ stands for any tagged partition of $[a,b]$:

$$a = x_0 < x_1 < \cdots < x_n = b\quad\text{ and }\quad x_k^* \in [ x_{k-1}, x_k ]\quad\text{ for } 1 \le k \le n$$ and $\delta(P) = \max_k \{ x_k - x_{k-1} : 1 \le k \le n \}$ is the mesh of the partition.

If I'm not mistaken, the indefinte "integral" evaluates to $$\int x^{dx} - 1 \stackrel{?}{=} \int \log x dx = x\log x - x + \text{constant}$$

For more details about this sort of approach to calculus through infinitesimals, Jerome Keisler's classic Elementary Calculus - an infinitesimal approarch will be an excellent reference.

Answer 3

(Note: This is NOT a rigorous treatment)

We have $$lim_{x\to 0}\frac{a^x - 1}{x} = \ln(a)$$ $$\implies (a^x - 1) \approx x \ln(a)$$ for very small $x$.

Note: what we want to basically show is, for very small $\Delta x$: $$x^{\Delta x} - 1 \approx ln(x) {\Delta x}$$ For example: $$(10000^{0.001} - 1) - (\ln(10000)*0.001) = 0.0000425$$ $$(0.5^{0.001} - 1) - (\ln(0.5)*0.001) = 0.00000024$$

Hence $$\int(x^{dx} - 1) \approx \int \ln(x) dx = \boxed{ x\ln(x) - x + C}$$

where $C$ is a constant.