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Let $h(x)$ be real-valued, monotonically-decreasing function on the non-negative reals such that $\int_{0}^{\infty} x^{k+1} h(x) \, dx$ is convergent for some positive integer $k$.

Does it follow that there exists $\beta > 0$ such that $h(x) = O\!\left( \frac{1}{x^{k+2+\beta}} \right)$ as $x \to \infty$?

Remark: Naively, $h(x)$ needs to decrease to zero sufficiently quickly to override the growth of $x^{k+1}$ ... but I'm worried about whether I've accidentally asserted the existence of a "slowest function with a converging integral" (which doesn't exist - this question and this question).

Why I am asking: I have a function $f(x)$ and am seeking to establish whether $\int_{0}^{\infty} x^{k} f(x) \, dx$ is convergent. I applied integration by parts to the finite integral $$\begin{align*} \int_{0}^{t} x^{k} f(x) \, dx &= \left[ \frac{x^{k+1} f(x)}{k+1} - \int \frac{x^{k+1}}{k+1} f'(x) \, dx \right]_{0}^{t} \\&= \frac{t^{k+1} f(t)}{k+1} - \int_{0}^{t} \frac{x^{k+1}}{k+1} f'(x) \, dx \end{align*}$$ From other information, I know that $f'(x)$ is monotonically decreasing, $\int_{0}^{\infty} x^{k+1} f'(x) \, dx$ is convergent and $f(t) \to 0$ as $t \to \infty$. (In fact in my particular application, $f'(x)$ turns out to be the tail probability of a cumulative distribution function on the non-negative reals.) But I appear to need to know how quickly $f(t)$ tends to zero if I'm to conclude that the first term on the right-hand side has a finite limit as $t \to \infty$. I'm hoping that this information can be deduced from the convergence of $\int_{0}^{\infty} x^{k+1} f'(x) \, dx$.

(This question is a refinement of an earlier one. I am asking a new question as per guidance. The condition that $h(x)$ is monotonically decreasing was added after I realized I forgot it, following comments by @user10354138. Many thanks in advance.)

Edit: See this question & answer for a revised question that got to an answer that I applied.

PtH
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    Basically the same example defeats your attempt to bound $h$. Just change the height to go much faster than your suggested bound but the base compensate by becoming much narrower. – user10354138 Feb 24 '21 at 06:29
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    If you want to get a bound like that, you need to do something to prevent the function from laying really low for a long time and infinitely often create a spike, or that changes sign sufficiently often to avoid a bound by immediately creating nearly opposite spikes (such as $\sin(x^2)$). Something like $h$ is nonincreasing would work, but it may be too restrictive for your application? – user10354138 Feb 24 '21 at 06:37
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    No: consider $h(x) = 1/(x^{k+2}(\log x)^2)$. (Make it equal to $0$ for $0\le x\le1$ to avoid irrelevant problems at the integral's left endpoint.) – Greg Martin Feb 24 '21 at 07:22

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(Writing out the answer as indicated by @[Greg Martin].)

Let $$ h(x) = \begin{cases} \displaystyle\frac{1}{x^{k+2}(\log x)^2} & \text{if $x \geq e$} \\ 0 & \text{otherwise} \end{cases} $$ Then for $x \geq e$, $h(x)$ is monotonically decreasing and $$ \int x^{k+1} h(x) \, dx = \int \frac{1}{x(\log x)^2} \, dx \underset{y \, = \, \log x}{=} \int \frac{1}{y^2} \, dy = -\frac{1}{y} = -\frac{1}{\log x} $$ so $$ \int_{e}^{\infty} x^{k+1} h(x) \, dx = \left[ -\frac{1}{\log x} \right]_{e}^{\infty} = 1 $$ hence $ \int_{0}^{\infty} x^{k+1} h(x) \, dx $ is convergent. But there is no $\beta > 0$ for which $\log(x) = O\!\left(\frac{1}{x^\beta}\right)$ as $x \to \infty$.

PtH
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