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This is a direct quote from page 472 of this book:

From Fourier's Inversion theorem $$f(t)= \int_{-\infty}^\infty f(u) \, \mathrm{d}{u} \left( \frac{1}{2\pi}\int_{-\infty}^\infty e^{-i\omega(t-{u})} \,\mathrm{d}\omega \right) \tag{1}$$ comparison of $(1)$ with the Dirac-Delta property: $$f(a)= \int f(x) \, \mathrm{d}x \, \delta(x-a)$$ shows we may write the $\delta$ function as $$\delta(t-u)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{i\omega(t-{u})} \, \mathrm{d}\omega$$

My question is what is the part in the large parentheses of $(1)$ got to do with $\delta(t-u)$?

Many thanks.

BLAZE
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    My answer here might be helpful, http://math.stackexchange.com/questions/991263/when-is-it-insufficient-to-treat-the-dirac-delta-as-an-evaluation-map/991299#991299 – Spencer Oct 10 '15 at 06:22
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    I suggest that you switch book. The book you are using is apparently confusing you over and over again, and those (similar) questions will never stop. See this question and this question for some suggestions. – mickep Oct 10 '15 at 06:28
  • @mickep Maybe, but it has served me well in the past, it's only causing confusion regarding the delta function. There is a simple answer to this question do you know it mickep? – BLAZE Oct 10 '15 at 06:31
  • Yes, there is a simple answer. Learn how the Fourier transform of a distribution is defined. From that it will follow that $\mathcal F 1=\delta$ (in the sense of distributions), modulo the $2\pi$ which can be put in different places. – mickep Oct 10 '15 at 06:39
  • Nope, I did that in another question. Now, I really think it is time for you to study one of the books in the links I posted to actually learn distribution theory. Once you have done so, I suggest you go back to these kind of questions, and you will see how and why one uses this abuse of notation. – mickep Oct 10 '15 at 07:15

2 Answers2

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You can interpret this classically instead. Assuming Fourier conditions on $f$, the inverse Fourier transform applied to the Fourier transform gives you back $f$: $$ f(t)=\lim_{R\rightarrow\infty}\frac{1}{2\pi} \int_{-R}^R e^{i\omega t} \int_{-\infty}^\infty f(u)e^{-i\omega u} \, du \, d\omega. $$ Interchanging orders of integration, $$ f(t) = \lim_{R\rightarrow\infty}\int_{-\infty}^\infty f(u)\left(\frac{1}{2\pi} \int_{-R}^R e^{i\omega(t-u)} \, d\omega\right) \, du. $$

Michael Hardy
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Disintegrating By Parts
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    Is there a non classical interpretation as well? Just because it said that it can be classically be interpreted as, made me wonder weather there are superseding theories. – jimjim Oct 11 '15 at 01:17
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    @Arjang : If you are looking for a way to interpret $\int_{-\infty}^{\infty}e^{i\omega(t-u)}d\omega$ as a scalar integral, you can stop looking. As mentioned in the comments, one can think of such an integral as representing the Fourier transform of an exponential, but that interpretation doesn't treat this object as a scalar integral, and it requires the Theory of Distributions. Many theories have been devised to make sense of what you're asking about and other related $\delta$-function like objects. Trying to make sense of $\delta$ functions has created a lot of Mathematics. – Disintegrating By Parts Oct 11 '15 at 11:34
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If you have $$ f(t)= \int_{-\infty}^\infty f(u) \Big(\cdots\cdots\text{blah blah} \cdots\cdots\Big) \, \mathrm{d}{u} $$ then you can conclude that $$ \Big(\cdots\cdots\text{blah blah} \cdots\cdots\Big) = \delta(u-t) $$

Michael Hardy
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  • Is it really that simple? – BLAZE Oct 11 '15 at 01:18
  • @BLAZE: If "blah blah" means a distribution and this holds for all test functions $f$, then yes; a distribution is defined by the integrals you get with test functions, and if two distributions give the same integrals, they are the same distribution. –  Oct 11 '15 at 12:20
  • @Michael Okay I guess it is that simple, thanks Michael – BLAZE Oct 12 '15 at 00:25
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    I suppose I should say "for ALL test functions $f$". Or Schartz functions. ${}\qquad{}$ – Michael Hardy Oct 12 '15 at 02:32