# If $\int_0^\alpha x^p f(x)dx$ converges and $f$ is monotone, then $\lim_{x \to 0} x^{p+1} f(x) = 0$

$f(x)$ is monotone on $(0, \alpha)$ and improper integral $\int_0^\alpha x^p f(x)dx$ converges. Prove that $$\lim_{x \to 0} x^{p+1} f(x) = 0$$

My attempt:

Assume that $\lim_{x \to 0} x^{p+1} f(x) \neq 0$, i.e. there exists $\{x_n\}_{n=1}^\infty$ strictly decreasing, goes to $0$ and $|x_n^{p+1}f(x_n)| \geq \epsilon_0$

WLOG, assume $f(x) \geq 0$ for $x$ small. Then $f(x)$ cannot be increasing since $x^p f(x) \geq \dfrac{\epsilon_0}{x_n}$ is unbounded for $x$ small.

Now $f(x)$ can only be decreasing. I tried $$\begin{split} \int_{x_{n+k}}^{x_n} x^p f(x)dx & = \sum_{i=1}^{n+k-1}\int_{x_{i+1}}^{x_i} x^p f(x)dx \\ & \geq \sum_{n=1}^{n+k-1} f(x_{i+1})\int_{x_{i+1}}^{x_i} x^pdx \\ & \geq \frac{1}{p+1}\sum_{n=1}^{n+k-1} f(x_{i+1})(x_{i}^{p+1} - x_{i+1}^{p+1}) \\ & \geq \frac{1}{p+1}\sum_{n=1}^{n+k-1} (f(x_{i})x_{i}^{p+1} - f(x_{i+1})x_{i+1}^{p+1}) \\ & \geq \frac{1}{p+1} (f(x_{n})x_{n}^{p+1} - f(x_{n+k})x_{n+k}^{p+1}) \end{split}$$

But I couldn't find a contradiction and got stuck here.

I'll assume $f$ is positive and decreasing and at first $p\ge 0.$ Suppose $x_k^{p+1}f(x_k) > \epsilon$ for some sequence $x_k \to 0.$ Then
$$\int_{x_k/2}^{x_k} x^pf(x)\,dx \ge (x_k/2)^pf(x_k)\cdot(x_k/2) \ge \epsilon/2^{p+1}$$
for all $k.$ This is a contradiction, because the integral on the left $\to 0$ as $k\to \infty.$
The case $p<0$ is easier because $x^pf(x)$ is decreasing. Here we have $x^{p+1}f(x) = x^pf(x)\cdot x \le \int_0^x f(t)t^p\,dt \to 0.$