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Plancherel's Theorem states that for $f \in L^{2}(\mathbb{R})$ we have

$$f(x) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}F(k)e^{ikx}dk \Longleftrightarrow F(k) = \frac{1}{\sqrt{2 \pi}}\int^{\infty}_{\infty}f(x)e^{-ikx}dx.$$

If we consider $f(x) := \delta(x)$ (the delta function) then using this theorem it follows simply that $$\delta(x) = \frac{1}{2 \pi}\int_{-\infty}^{\infty}e^{ikx}dk.$$

Clearly then for $x = 0$ and $x \neq 0$ the integral is divergent. Also, apparently $\delta(x)$ is not square integrable, hence I'm not sure that we can even use Plancherel's Theorem. But having said that I understand that this result does hold. Is it incorrect to show that this result is true using Plancherel's Theorem?

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    How is this a physics question? (If you're asking if this way to show that $\delta$ is the Fourier transform of $1$ that physicists often use is rigorously wrong, then yes, it is. That's physicists doing math for you ;) ) – ACuriousMind Mar 03 '16 at 19:48
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    I would vote to migrate this to Math, but I'm pretty sure it's been asked before. – DanielSank Mar 03 '16 at 19:48
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    @DanielSank This is a question from Griffith's book "Introduction to Quantum Mechanics" second edition. Well related to a question. He asks to prove it using Plancherel's Theorem, which is easy, but I don't understand how you can use the Theorem if $\delta$ is not square integrable. –  Mar 03 '16 at 19:50
  • Just showing up in a physics textbook doesn't mean the question belongs on this site. The "sorting rules" here are what they are regardless of where the problem shows up in a book. There is zero physics in this question. – DanielSank Mar 03 '16 at 19:53
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    @DanielSank Yeah but as AcuriosMind states the fact that these proofs seem to be dealt with differently (possibly in terms of rigour, I'm not sure) in maths and physics, seems to be related to physics...and maths. –  Mar 03 '16 at 19:55
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    Having said that I will post on maths site, but I think it's missing the point being overly pedantic about maths and physics like this. –  Mar 03 '16 at 20:01
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    @JohnDoe, $\delta$-function is not even a function, so it really does not make sense to try to be rigorous about this question, unless you are willing to go into what $\delta$-function really is. If you are, then in the theory of distributions you can make this argument precise in several ways. – Peter Kravchuk Mar 03 '16 at 21:01

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