We know that the Fourier transform of the Dirac Delta function is defined as $$\int_{-\infty}^{\infty} \delta(t) e^{-i\omega t} dt = 1,$$

and if I were to reconstruct the function back in time domain, the inverse Fourier transform is defined as

$$\delta(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega t} d\omega.$$

How do I compute this integral analytically?

  • $\begingroup$ Hint: look at the LHS of the second equation ;) $\endgroup$
    – Bosoneando
    Jun 22 '15 at 19:14
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    $\begingroup$ The delta "function" isn't a function, but a distribution. I think what you think of as "computing analytically" is not possible due to that. $\endgroup$ Jun 22 '15 at 19:31
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    $\begingroup$ It cannot be computed, as $\delta$ is not a function, but a distribution, or a generalized function. It is a functional, such that given some function $f$, $\delta[f]=f(0)$. What you have written is indeed its Fourier integral representation, but this integral does not converge in any sense. Its true meaning is only derived by what it does to functions - likewise, this integral representation has its use only when it's "nested" inside yet another integral! $\endgroup$
    – krvolok
    Jun 22 '15 at 19:42

You can view this as a limiting process. Start with the truncated integral: $$ \frac{1}{2\pi}\int_{R}^{R}e^{j\omega t}dt = \frac{1}{\pi}\frac{\sin(R\omega)}{\omega} $$ If you integrate this against a function and take the limit as $R\rightarrow\infty$, then $$ \lim_{R\rightarrow\infty}\int_{-\infty}^{\infty}f(\omega)\frac{1}{2\pi}\int_{-R}^{R}e^{j\omega t}dt d\omega = \lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-R}^{R}\int_{-\infty}^{\infty}f(\omega)e^{j\omega t}d\omega dt. $$ The above is $(f^{\wedge})^{\vee}(0)=f(0)$ if $f$ has some smoothness at $0$.

  • $\begingroup$ I'd say that $\displaystyle \lim_{\epsilon \to 0^+}\frac{1}{2 \pi} \int_{-\infty}^\infty F(\omega) e^{i \omega t} e^{-\epsilon t^2} dt $ is easier for proving the Fourier inversion theorem (when $f \in L^2$ and is continuous) $\endgroup$
    – reuns
    Feb 18 '16 at 2:10
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    $\begingroup$ Why did you switch the integration variable in your first equation from $\omega$ to $t$? $\endgroup$
    – GnomeSort
    Feb 13 '17 at 1:42

You can't really compute that integral analytically, since the delta function isn't really a "function," and the integral is not really well-defined as you've written it down. What you CAN do is prove that when integrated against a test-function, the integral behaves like a delta function.

So let's test it. We want $\int\delta(t)f(t)dt = f(0)$ for any function $f$. We know, from your equation two, that we want

$$\int\delta(t)f(t)dt =\int\int f(t)e^{j\omega t} dt d\omega$$ Now we note that doing the t-integral merely gives us the fourier transform of f: $$=\int \tilde{f}(-\omega) d\omega$$ Finally, we note that this final integral is just the reverse fourier transform of $f$, centered at $t=0$: $$=\int \tilde{f}(-\omega)e^{j(-\omega)0} d\omega=f(0)$$ Thus, the integral you wrote down above behaves just like we want a $\delta$-fucntion to behave. Hope that helps!


Without any limiting procedures, you can use the definition of the distributional Fourier transform, which is defined by

$$\langle \mathcal{F}(d),\mathcal{F}(f) \rangle = \langle d,f \rangle$$

for $d \in \mathcal{S}',f \in \mathcal{S}$. Here $\mathcal{S}$ is the space of Schwartz functions (smooth functions all of whose derivatives decay faster than any rational function at infinity) and $\mathcal{S}'$ is its dual space, the space of "tempered distributions". The equality above says that the Fourier transform is unitary with respect to the dual pairing between tempered distributions and Schwartz functions. This is a fairly natural property, seeing as the Fourier transform is naturally unitary with respect to the $L^2$ inner product, and the $L^2$ inner product is a restriction of this dual pairing to $L^2 \times L^2 \subset \mathcal{S}' \times \mathcal{S}$.

Replacing $f$ with $\mathcal{F}^{-1}(f)$ you get

$$\langle \mathcal{F}(d),f \rangle = \langle d,\mathcal{F}^{-1}(f) \rangle.$$

You can do the same thing for the inverse Fourier transform of a distribution, in which case you get

$$\langle \mathcal{F}^{-1}(d),f \rangle = \langle d,\mathcal{F}(f) \rangle.$$

Since $\langle \delta,f \rangle = f(0)$ (this is the definition of $\delta$), the unitary inverse Fourier transform of the Dirac delta is a distribution which, given a function $f$, evaluates the Fourier transform of $f$ at zero. In other words, $\langle \mathcal{F}^{-1}(\delta),f \rangle = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(x) dx$. Somewhat roughly speaking, this means that the unitary inverse Fourier transform of the Dirac delta is the constant function $\frac{1}{\sqrt{2 \pi}}$.

Note that this is all under the unitary normalization of the Fourier transform. Under other normalizations, the $2 \pi$ factors wind up in different places. Also note that this can be extended somewhat by a density/continuity argument (so that the domain of $\mathcal{F}^{-1}(\delta)$ is larger than $\mathcal{S}$).

Edit: I notice that the actual calculation I did is not what was asked; you asked for a calculation of the inverse Fourier transform of $1$. But the setup is the same, so I'll leave that task to you.


I would try considering the integral

$$\delta_{\epsilon}(t) = \frac{1}{2\pi}\int_{0}^{\infty} e^{(jt-\epsilon)\omega} d\omega+\frac{1}{2\pi}\int_{-\infty}^{0} e^{(jt+\epsilon)\omega} d\omega,$$

do the integral and take the limit $\epsilon\to 0$. You should end up with something like

$$\lim_{\epsilon\to 0}\frac{1}{\pi}\frac{\epsilon}{t^2+\epsilon^2}$$

which is a possible representation of the delta distribution.


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