Integral Identity of Real Functions

Question

I recently encountered the following beautiful integral and sum :

$$\int_{0}^{\infty} \frac{\sin x}{x} dx = \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} dx = \frac{\pi}{2}$$

and

$$\sum_{n=1}^{\infty} \frac{\sin n}{n} = \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}} = \frac{\pi-1}{2}.$$

You can find the solutions to each integral or sum here and here.

Inspired by the first integral, I wanted to find functions $f$ such that

$$\int_{a}^{b} f(x) dx = \int_{a}^{b} \left[ f(x) \right]^{2} dx$$

for real numbers (or $\pm \infty$) $a$ and $b$. I asked this question on AoPS and here's how one of the replies solved it :

For any continuous function $g$, let constant $k$ be

$$k := \frac{\int_{a}^{b} g(x) dx}{\int_{a}^{b} \left[ g(x) \right]^{2} dx}$$

and define $f$ as $f := kg$ then we get

$$\int_{a}^{b} f(x) dx = \frac{\left[ \int_{a}^{b} g(x) dx \right]^{2}}{\int_{a}^{b} \left[ g(x) \right]^{2} dx} = \int_{a}^{b} \left[ f(x) \right]^{2} dx.$$

Using this solution, we can similarly find functions $f$ such that

$$\int_{a}^{b} f(x) dx = \int_{a}^{b} \left[ f(x) \right]^{n} dx$$

for any natural number $n$. However, for the function $\frac{\sin x}{x}$, I have no idea whether it can be derived from this general solution. In other words, would this general solution give all possible functions $f$? For any natural number $n$ or maybe $n = 2$ for now.

My second question is about the sum. How could one find function $f$ that satisfies

$$\int_{a}^{b} f(x) dx = \int_{a}^{b} \left[ f(x) \right]^{2} dx$$

'And'

$$\sum_{n=1}^{\infty} f(n) = \sum_{n=1}^{\infty} \left[ f(n) \right]^{2}?$$

This can be generalized to $\int_{a}^{b} f(x) dx = \int_{a}^{b} \left[ f(x) \right]^{m} dx$ and $\sum_{n=1}^{\infty} f(n) = \sum_{n=1}^{\infty} \left[ f(n) \right]^{m}$ for natural number $m$, but it's too much at this point.

Answer 1

However, for the function $ \sin(x) / x $, I have no idea whether it can be derived from this general solution. In other words, would this general solution give all possible functions f? For any natural number n or maybe n=2 for now.

Nope. It dosen't work for all natural number n: We can prove this easily: $$ \begin{align*} \lim_{{x} \to {\infty}} \int_{0}^{x} \frac{\sin(x)}{x} ~\mathrm{d}x = \lim_{{x} \to {\infty}} \operatorname{Si}(x) &= \frac{1}{2}\pi\\ \lim_{{x} \to {\infty}} \int_{0}^{x} \frac{\sin^{2}(x)}{x^{2}} ~\mathrm{d}x &= \frac{1}{2}\pi\\ \lim_{{x} \to {\infty}} \int_{0}^{x} \frac{\sin^{3}(x)}{x^{3}} ~\mathrm{d}x &= \frac{3}{8}\pi\\ \lim_{{x} \to {\infty}} \int_{0}^{x} \frac{\sin^{4}(x)}{x^{4}} ~\mathrm{d}x &= \frac{1}{3}\pi\\ \lim_{{x} \to {\infty}} \int_{0}^{x} \frac{\sin^{6}(x)}{x^{6}} ~\mathrm{d}x &= \frac{11}{40}\pi\\ \Rightarrow \lim_{{x} \to {\infty}} \int_{0}^{x} \frac{\sin^{n}(x)}{x^{n}} ~\mathrm{d}x &= \begin{cases} \begin{matrix} \frac{1}{2}\pi, ~\text{ if n = 1 or n = 2}\\ y \ne \frac{1}{2}\pi, ~\text{ if n > 2} \end{matrix}\end{cases} \\ \end{align*} $$

Aka it's only true if n = 1 or n = 2!

My second question is about the sum. How could one find function f that satisfies $ \int_{a}^{b} f(x) ~\mathrm{d}x = \int_{a}^{b} \left[ f(x) \right]^{2} ~\mathrm{d}x $ 'And' $ \sum_{n=1}^{\infty} f(n) = \sum_{n=1}^{\infty} \left[ f(n) \right]^{2} $?

That's a bit more complex:

If we got $ \int_{a}^{b} f(x) ~\mathrm{d}x = \int_{a}^{b} \left[ f(x) \right]^{m} ~\mathrm{d}x $ and $ \sum_{n=1}^{\infty} f(n) = \sum_{n=1}^{\infty} \left[ f(n) \right]^{m} $, we can say: $$ \begin{align*} \sum_{n=1}^{\infty} f(n) &= \sum_{n=1}^{\infty} \left[ f(n) \right]^{m}\\ \sum_{n=1}^{\infty} f(n) &= \sum_{n=1}^{\infty} f(n)^{m} \quad\mid\quad \div (\sum_{n=1}^{\infty} f(n)^{m})\\ \frac{\sum_{n=1}^{\infty} f(n)}{\sum_{n=1}^{\infty} f(n)^{m}} &= 1\\ \sum_{n=1}^{\infty} \frac{f(n)}{f(n)^{m}} &= 1\\ \Rightarrow f(n) &= f(n)^{m} \ne 0\\ \end{align*} $$

So they must be a function that gives itself from a to b aka they must be equal in the interval $ I = [a, ~b]$!