Can $\int_{0}^{\infty} (f(z)^n+f(z)^{n+1})dz$ go towards infinity as $n$ goes to infinity while $\int_{0}^{\infty} (f(z)+f(z)^2)dz=1$ and any $n$ you pick bigger than or equal to $1$ converges to a finite value?$$\int_{0}^{\infty} (f(z)^n+f(z)^{n+1})dz<\int_{0}^{\infty} (f(z)^{n+1}+f(z)^{n+2})dz$$
My motivation for this problem is to understand what happens to functions that converge when you take an indefinite integral and then you take it to the power of a constant. I'm pretty sure that it has to stay finite for taking it to the power of something bigger than $1$ and I thought it might always shrink for most functions like $f(z)=e^{-x^2}$ then you take the indefinite integral you get $\frac{\sqrt{\pi}}{2}$ when you take it to the power of something like $100$ you get a smaller answer. So I slightly moved the powers and the function to see if it can grow forever.
Yes. A very simple one is $$f(x)=\begin {cases} 2& 0 \le x \lt \frac 16\\0&x \ge \frac 16 \end {cases}$$ I chose the $\frac 16$ to make $\int_0^\infty (f(x)+f(x)^2)dx=1$. For any finite $n$ the integral converges but as $n$ increases the integral gets larger and larger.