TITLE: If $f(x)$ is integrable can we say that $f(x)^n$ is integrable?
QUESTION [8 upvotes]: Suppose $f(x)$ is a positive and continuously differentiable functions. In addition, it is well-known that      $\int_{0}^{\infty} f(x)dx$ is bounded. My point of view is that $\int_{0}^{\infty} f(x)^mdx$ (where $m \in \mathbb N$) is bounded. I will be gratefull if you would propose me a conterexample. If this claim is true, how can I prove it?

REPLY [5 votes]: Define the following function:
$$f(x)=\begin{cases}
2/\sqrt{x}-1,&0<x\leq1 \\
e^{1-x},&x>1.
\end{cases}$$
Then $f$ is continuously differentiable, and we have:
$$\int_0^\infty f(x)\,dx=\int_0^1\frac{2}{\sqrt{x}}-1\,dx+\int_1^\infty e^{1-x}\,dx=4,$$
but:
$$\int_0^\infty f^2(x)\,dx=\int_0^1 1-\frac{4}{\sqrt x}+\frac{4}{x}\,dx+\int_1^\infty e^{2-2x}\,dx,$$
and $\displaystyle\int_0^11-\frac{4}{\sqrt x}+\frac{4}{x}\,dx$ diverges.