If $f(x)$ is integrable can we say that $f(x)^n$ is integrable? 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?
 A: 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.
A: Even if $f(x)$ is continuous at $x=0$, it is not necessarily true that the convergence of $\int_0^\infty f(x)\,dx$ implies the convergence of $\int_0^\infty f(x)^m\,dx$. The following is a counterexample for every $m\ge2$:
Let $I_1,I_2,I_3,\dots$ be a sequence of disjoint intervals of lengths $\frac1{1^4},\frac1{2^4},\frac1{3^4},\dots$. Define
$$
f(x) = \begin{cases}
j^2, &\text{if $x\in I_j$ for some $j\ge1$}, \\ 0, &\text{otherwise}.
\end{cases}
$$
Then $\int_0^\infty f(x)\,dx = \sum_{j=1}^\infty j^2 \frac1{j^4} = \frac{\pi^2}6 < \infty$, but $\int_0^\infty f(x)^2\,dx = \sum_{j=1}^\infty j^4 \frac1{j^4} = \infty$.
Admittedly this $f(x)$ is not continuous, but it is easy to approximate it arbitrarily closely (in the $L^1$ and $L^2$ means) by arbitrarily smooth functions, which will still serve as counterexamples.
If one adds the additional assumption that the positive function $f(x)$ tends to $0$ as $x\to\infty$, then the convergence of $\int_0^\infty f(x)\,dx$ does imply the convergence of $\int_0^\infty f(x)^m\,dx$ (even without an assumption of continuity or differentiability).
