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Is there a real valued function

$f: (0, \infty) \rightarrow \mathbb{R}$

s.th.

$\int_{0}^\infty \cos(kx) f(x) dx$

does not exist as a reasonable function/distribution, but

$\int_{0}^\infty \sin(kx) f(x) dx$

exists (or vice versa)?

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    Try $f(x)=x^{-(1+\alpha)}$ for $\alpha\in(0,1)$? – Feng Oct 28 '22 at 11:28
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    @Feng The Fourier transform as a distribution exists for all $|x|^s$. See THIS. Therefore, both the sine and cosine transforms exist. – Mark Viola Oct 28 '22 at 13:20
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    @MarkViola Of course. I thought OP was concerning something like the convergence of improper integral. Thanks for your reply! Btw, nice answer :) – Feng Oct 28 '22 at 13:57
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    @Feng Indeed. The OP might have been considering functions under the PV distribution in the sense of improper Riemann integrals. And thank you for the nice comment. – Mark Viola Oct 28 '22 at 17:38

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