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There is the following proof for the Fourier transform of a derivative: $$ f(t) = \int_{-\infty}^\infty S(\nu) e^{2\pi i \nu t}d\nu $$ then $$ f'(t) = \frac{d}{dt} \int_{-\infty}^\infty S(\nu)e^{2\pi i \nu t} d\nu = 2\pi i\int_{-\infty}^\infty \nu S(\nu)e^{2\pi i \nu t}d\nu \longleftrightarrow 2\pi i\nu S(\nu) $$

Why in the proof are we able to move a derivative inside the integral? It seems that we apply Leibniz's rule, but the integral is improper. Thank you for your help!

saz
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Konstantin
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    In general, you cannot just pull the derivative inside. You need some assumption like $x \mapsto x \cdot S (x) \in L^1$. – PhoemueX Mar 05 '17 at 11:43
  • @PhoemueX Thank you for the reply! So the proof is incorrect or we have some necessary conditions given that the original Fourier transform exists? – Konstantin Mar 05 '17 at 11:45
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    Your proof is incorrect since you cannot simply pull $\nu$ outside the integral. it should read $$f'(t) = 2\pi i \int_{-\infty}^{\infty} \nu S(\nu) e^{2\pi i \nu t} , d\nu.$$ In order to interchange differentiability and integration you have to assume that $\nu \mapsto \nu S(\nu)$ is integrable, as PhoemueX pointed out. – saz Mar 05 '17 at 12:21
  • @saz Thank you for the corrections, I updated the question! – Konstantin Mar 05 '17 at 12:24
  • @PhoemueX Why the assumption $xS(x)\in L_1$ is enough to move the derivative inside? – Konstantin Mar 05 '17 at 20:55
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    @Konstantin: My answer here (http://math.stackexchange.com/questions/246380/difference-of-differentiation-under-integral-sign-between-lebesgue-and-riemann) explains this in some detail. Essentially, we want to apply the dominated convergence theorem and this can be justified if $x S(x) \in L^1$. – PhoemueX Mar 06 '17 at 09:46

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