Does $\int_{0}^{\infty} f(x)^n dx$ always shrink if $\int_{0}^{\infty} f(x) dx=1$ and $n$ is as large as you want. This is similar to can $\int_{0}^{\infty} (f(z)^n+f(z)^{n+1})$ grow forever? but I think this follows the guide lines. SuperWarm's motivation says that he/she wants to understand how taking a function that converges when you do the indefinite integral to the power of a large number $n$ changes the result of the indefinite integral. I found this interesting and I wonder if her/his thought that it would always shrink was true, But I couldn't find a proof of that it anywhere.
So are there functions that no matter how high $n$ is you can take the indefinite integral of $f(x)^n$ and it will grow, where $n>1$ and $\int_{0}^{\infty} f(x)dx=1$ and $f(z)$ is differentiable everywhere.
$$\int_{0}^{\infty} f(x)^n dx>1$$
Not necessarily. Take $f$ to satisfy $f(x)$ to be a bounded function satisfying $\int_{x=0}^{1/3} f(x) = 2/3$, and then $f$ to be a continuous function on $(1/3,\infty)$ satisfying $\int_{1/3}^{\infty} f(x) = 1/3$, such that $f$ is continuous on the entire positive real line. Then here $\int_0^{\infty}f^{n}(x)$ will grow with $n$ eventually.