Existence of function
- Tolaso J Kos
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Existence of function
Examine if there exists a continuous function $f:[1, +\infty) \rightarrow \mathbb{R}$ such that $f(x) >0 , \; \forall x \in [1, +\infty)$ and that the integral $\bigintsss_{1}^{\infty} f(x)\, {\rm d}x$ converges whereas $\bigintsss_{1}^{\infty} f^2(x)\, {\rm d}x$ diverges.
Don't be in a rush to answer.
Don't be in a rush to answer.
Imagination is much more important than knowledge.
Re: Existence of function
Yes, there exists. The basic idea is to construct a train of picks. Picks are constructed to satisfy the following heuristics: Each pick at $x=n$ is of an approximate height $n$ and of an approximate width $n^{-3}$, thus the area is of an approximate order $n^{-2}$, whereas squared its height gets squared thus its area of an approximate order $n^{-1}$.
If we want a specific function , then
$$f(x) = e^{-x} + \sum_{n=2}^{\infty} \max\{0, n - n^4|x - n|\}$$
suffices.
If we want a specific function , then
$$f(x) = e^{-x} + \sum_{n=2}^{\infty} \max\{0, n - n^4|x - n|\}$$
suffices.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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