For $a<b,\,\{a,\,b,\,k\}\in\mathbb{Z}$, Are there examples of integrals $\int_a^b f(x)dx = e^k$ with a positive $f(x)>0$ unrelated with the exponential function? (so no scalings of the form $\frac{\text{result}}{e^k}\cdot e^k$ are allowed)
To explain what I mean, the only similar examples I have found so far (that don't fulfill neither the possitivity of the function neither the integer integration limits conditions, and are the same integral with a just change of variable) are:
- $$\frac{1}{\pi}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\left(\tan(x)\right)\ dx = \frac{1}{\pi}\int\limits_{0}^{1} \cos\left(\tan\left(\frac{\pi x}{2}-\frac{\pi}{2}\right)\right)\ dx =\frac{1}{e}$$
- $$\frac{1}{\pi}\int\limits_{-\infty}^{\infty} \frac{\cos(x)}{1+x^2}\ dx = \frac{1}{e}$$
And I am wondering if its possible to have a some function $F(x) = \int f(x)\ dx$ such as $F(b) - F(a) \propto e^k$ with integer numbers as its limits of integration.
Obviously you can always made a function $f^*(x) = \frac{f(x)}{e^k}$ to make it fit so this is not allowed, and also trigonometric functions are related with the exponential function due $\cos(x)=\frac{\exp(ix)+\exp(-ix)}{2}$, so try to understand what I am asking with unrelated as thinking as classical functions taught at school - don't know how to formally state it.
The objective is recreational: if I know this function $f(x)$, I could draw it in a square section of some unknown height $M>0$ and width $b-a$ and estimate $e$ as some operation using the landing of uniformly distributed random point in the rectangle and operating on the proportion below the area of $f(x)$
PS: I am aware of the video example $\int_1^e \frac{1}{x}\ dx=1$, is from where this doubt have rise.
Last findings
Through the properties $\lim_{x \to 0} (1+x)^{\frac{1}{x}} = e$ and $\lim_{x \to 1} (1+x)^{\frac{1}{x}}= 2$, I get to: $$\int\limits_0^1 \left(2-(1+x)^{\frac{1}{x}}\left(\frac{1}{x(1+x)}-\frac{\log(1+x)}{x^2}\right)\right) \,dx = e$$
where I believe the only problematic value is the logarithm since is not clear if I need to know beforehand the value of $e$ in order to define them and calculate its value. Since $\log(x) = \log(2)\log_2(x)$, and it is possible to find $\log(2)=\int_1^2 \frac{dx}{x}$, and the Binary Logarithm $\log_2(x)$ was tabulated previous the discovery of the constant $e$ as is stated in Wikipedia, I believe the formula could be a valid way to find $e$ if it wasn't been knowing yet (at least it is not in Wikipedia), or also finding $\log(x) = \int_1^x \frac{du}{u}$ I think don't require to know $e$ beforehand, but maybe I am wrong, if its that the case, please explain why. Maybe this makes someone to have an alternative idea of how to describe it with other elementary functions.
Also I tried to approximate $\log(1+x)\approx \log(2)x+(x-x^2)\frac{\log(2)(1-\log(2))(2\log(2)-1)}{3-8\log(2)+6 (\log(2))^2}$ on $x=[0,\,1]$ trying to find a logarithms-free version but it didn't work. Neither using small exponents for approximating $\log(1+x) = \lim\limits_{a \to 0} \displaystyle{\frac{\log(2)}{1-\frac{1}{2^a}}}\left( 1-\frac{1}{(1+x)^a}\right) $
By the way, the founded example is just $$\int_0^1 \frac{\partial}{\partial x}\left[2x-(1+x)^{\frac{1}{x}} \right]\,dx = e$$
It is possible to use also terms of the form $q(x)=\left(1+\tan\left(\frac{\pi x}{2}\right)\right)^{\cot\left(\frac{\pi x}{2}\right)}$ to find that $$\int_0^1 1-q'(x)\ dx=e$$ but since is almost the same function but even more complicated, and also has the same limitations due the logarithm appearing on the derivative, it doesn't deserves more analysis (at least without founding a way to avoid the logarithm).
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Later I have realized that when even the constant $e$ is not explicetly involved, it is still hidden since it still requires to find the limit $\lim_{x \to 0} (1+x)^{\frac{1}{x}} = e$ so I don't know if its valid (this problematic point when under the integral is just a point of zero meassure, so maybe the problem is avoided, but I don't really know).
Another related ones: $$\int_0^1 \frac{\partial}{\partial x}\left[2x+(2-x)^{\frac{1}{1-x}} \right]\,dx = e$$
$$\int_0^1 \frac{\partial}{\partial x}\left[2x+\frac{1}{2}\left((2-x)^{\frac{1}{1-x}}-(1+x)^{\frac{1}{x}}\right) \right]\,dx = e$$
In principle, it should be infinitely many strictly possitive functions which area below the curve in the interval $[0;\ 1]$ have a value of $e$, so its kind of frustrating I was not able to find any were the constant $e$ was not involved beforehand... Hope you could understand now what I am trying to find, a way to find the value of $e$ as it were unkown by integrating some function $f(x)$ made by classic functions you could draw with Straightedge and compass.
Do you figure out any? Or they are impossible to find without knowing the value of $e$ beforehand?
