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Let $\varphi$ be a real smooth function with sufficient decay and vanishing mean. Then there exist a functions $\psi$ such that: $$\psi(x)'=\varphi(x)$$ then $$\hat{\psi}(\xi) = \frac{\hat{\varphi}(\xi)}{\xi} + \hat{\varphi}(0)\delta(\xi) = \frac{\hat{\varphi}(\xi)}{\xi}$$

Here i am using the following $$\hat\psi(\xi)=\int_{\mathbb{R}}\psi(x)e^{-2\pi i x\xi}\:dx$$

Can somehow help me understanding this result? Can someone give me a hint of how to prove this formula? Where does the delta function comes from?

riemannfanboy
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  • Can you give more context? For example, which definition of the Fourier transform are you using? Is this in \Bbb{R}$? – Danilo Gregorin Afonso Apr 06 '20 at 21:12
  • If you are referring to my answer on your post several days ago, this is a well known Fourier transform property. But there needs to be an $i$ in the denominator. Not to mention as stated what you have written is not true because you've ignored the rest of the context of the answer. – Ninad Munshi Apr 06 '20 at 21:20
  • @NinadMunshi i'm sorry, i've new in this fields (distribution theory/Fourier Analysis) and i do not know yet what is necessary and what is not. And yes, it is. I have been searching throw out my books and i've not found a answer, and the answers i read here are somehow poor and i find them not satisfying. If you could somehow, help me or recomend me some book i would appreciate it! Thanks anyway for your answer. – riemannfanboy Apr 07 '20 at 07:20
  • @DaniloGregorin Sure i will edit the question. – riemannfanboy Apr 07 '20 at 07:20
  • Have a look at this and accept my answer if the only confusion you have is a property which you can look up in a table (the standard method of dealing with Fourier transform properties) – Ninad Munshi Apr 07 '20 at 07:29
  • If the relevance of the linked post confuses you, note that $\int f(x) :dx = f\star H$ where $H$ is the Heaviside function (aka integration is equivalent to convolution with a Heaviside). This means that the Fourier transform of an antiderivative is equal to the product of Fourier transforms of the original function and the Heaviside. – Ninad Munshi Apr 07 '20 at 07:33
  • @NinadMunshi omg thank you soo much, this is what i needed. Not forcing the answer with the delta fucntion but where actually if comes from, it is as simple as the convulution with the heaviside, thank you so much! – riemannfanboy Apr 07 '20 at 07:44

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