Integration of $f(x)^n$, given $f(x)$ is continuous and $0 \le f(x) \le 1$

Question

Given $f(x)$ is continuous and $0 \le f(x) \le 1$, and $\int f(x) dx = g(x)$, is there indefinite or definite integration formula for $$\int f(x)^n dx$$ or an approximation or expansion?

Answer 1

I don't think there is a general formula using only elementary functions.

Just consider $n=2$.

Partial integration leads to $$\int f(y)^2\mathrm{d}y=f(x)g(x)-\int f'(y)g(y)\mathrm{d}y,$$ and only if $f$ is differentiable and $f'$ integrable. But from there, you won't get to far.

On the other hand, let's consider $f$ to be infintely differentiable and let $$h(x)=\intop_{x_0}^x f(y)^2\mathrm{d}y.$$

For Taylor series we need derivatives of $h$. It is clear $h'=f^2$. Iterated product rule yields $$h^{(m+1)}=(f^2)^{(m)}=\sum_{k=0}^m \binom{m}{k}f^{(k)}f^{(m-k)}. $$

So our Taylor series of $h$ at $x_0$ (for example) is $$T(x)=\sum_{m=0}^\infty \frac{1}{m!}h^{(m)}(x_0)(x-x_0)^m=\sum_{m=1}^\infty \sum_{k=0}^{m-1} \frac{f^{(k)}(x_0)f^{(m-1-k)}(x_0)}{k!(m-1-k)!m}(x-x_0)^{m},$$because $h(x_0)=0$. Further index shift might ease it up a bit but this formular is far from handy. And for $n\geq 3$ it is not going to be any better.

Answer 2

Consider as a counterexample, the function $$f(x)=e^{-x^2}$$ defined in $[0,+\infty)$ which is a constraint of a Gaussian Function, and is integrable but its antiderivative cannot be written in terms of elementary functions.

For another example consider $$\sqrt{1-x^4}$$ defined in $[0,1)$ for which the integral can be approached this way.