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Let $f$ be a continuous, real-valued function on $[1,\infty)$ such that $\int_{1}^{\infty} f(x) \, dx$ is finite.

Does this necessarily imply that there exists $\kappa > 0$ such that $f(x) = O\!\left( \frac{1}{x^{1+\kappa}} \right)$ ?

That is (expanding the big-O notation), there exists $\kappa > 0$, $M > 0$, $x_0 > a$ such that $|f(x)| \leq \frac{M}{x^{1+\kappa}}$ for all $x \geq x_0$?

Remark: When $\kappa > 0$ we have $\int_{1}^{\infty} \frac{1}{x^{1+\kappa}} \, dx = \left[ \frac{-1}{\kappa x^{\kappa}}\right]_{1}^{\infty} = \frac{1}{\kappa}$ which is finite.

Why I am asking: I have a situation where I have a convergent integral, and want to use feature of the integrand to establish whether another integral is convergent. 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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1 Answers1

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You can't.

Example: Let $f$ be mostly zero, except centered at every half-integer $n+\frac12$, $n\geq 1$, we have a triangle function of height $1$ and base $1/n^2$. In other words, $$ f(x)=\sum_{n\geq 1}\max(0,1-2n^2\lvert n+\tfrac12-x\rvert) $$ Then $\sup_{x\geq N}f(x)=1$ for all $N$ so it doesn't decay. But the total area of triangles, $\frac12\zeta(2)$, is finite.

In fact, you can tweak this example so that $\sup_{x\geq N}f(x)=\infty$, and if you know about bump functions, get $f$ infinitely differentiable too.

user10354138
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