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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 $\delta(t-u)$?

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Many thanks.

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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)

$$

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.

$$

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