Where does the $2\pi$ in Fourier Transform Dirac delta identity come from?

Question

$$ \chi( \omega - \omega ')= \int_{-\infty} ^ {\infty} dt e^{j( \omega - \omega ')t} = 2 \pi \delta ( \omega - \omega ') $$

That is the identity to proof. I have seen different ways to proof this, and also here in stack. But my most important question is the $2\pi$ , I believe I know where it comes from, I just believe it could also work without that $2\pi$.

Can someone help me and make a simple derivation.

I DO understand that $\int_{-\infty} ^ {\infty} dt e^{j( \omega - \omega ')t} = \delta ( \omega - \omega ')$

Just to make sure, $ \chi( \omega - \omega ')$ is the fourier transform of $x(t)$

Answer 1

To begin with, the identity in the OP is not true for all functions, so there is no hope of proving it generally. If you want to prove the statement to see where the $2\pi$ comes from, you'll need to say which set of functions you're working with.

So, instead of giving a general argument, let's look at one particular function for which the Fourier inversion theorem holds and see if we can identify where the $2\pi$ enters in. Consider the integral \begin{align} \int_{-\infty}^{\infty} \left(\int_{-\infty}^{\infty} e^{i(k - k')x} dx \right) e^{-k^2} dk \end{align} We compute \begin{align} \int_{-\infty}^{\infty} \left(\int_{-\infty}^{\infty} e^{i(k - k')x} dx \right) e^{-k^2} dk &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{i(k - k')x} e^{-k^2} dx dk \\&= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{i(k - k')x - k^2}dx dk \\&= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(k^2 - ixk + ixk')}dx dk \\&= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(\left(k - \frac{ix}{2}\right)^2 + \frac{x^2}{4} + ixk'\right)}dxdk \\&= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(\left(k - \frac{ix}{2}\right)^2\right)} e^{-\left(\frac{x^2}{4} + ixk'\right)}dxdk \\&= \int_{-\infty}^{\infty} e^{-\left(\frac{x^2}{4} + ixk'\right)} dx \int_{-\infty}^{\infty} e^{-\left(\left(k - \frac{ix}{2}\right)^2\right)} dk \\&= \int_{-\infty}^{\infty} e^{-\frac{1}{4}\left(x^2 + 4ixk'\right)} dx \left( \sqrt{\pi} \right) \\&= \sqrt{\pi}\int_{-\infty}^{\infty} e^{-\frac{1}{4}\left(\left(x + 2ik'\right)^2 + 4k'^2 \right)} dx \\&= \sqrt{\pi} e^{-k'^2} \int_{-\infty}^{\infty} e^{-\frac{1}{4}\left(x + 2ik'\right)^2} dx \\&= \sqrt{\pi} e^{-k'^2} \sqrt{4\pi} = 2\pi e^{-k'^2} \end{align} In other words $$ \int_{-\infty}^{\infty} \left(\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i(k - k')x} dx \right) e^{-k^2} dk = e^{-k'^2} $$ So, if it is indeed the case that $$ \int_{-\infty}^{\infty} e^{i(k - k')x} dx = C \delta(k - k') $$ for some constant $C$, then $$ e^{-k'^2} = \int_{-\infty}^{\infty} \left(\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i(k - k')x} dx \right) e^{-k^2} dk = \int_{-\infty}^{\infty} \left(\frac{1}{2\pi}\left(C\delta(k - k')\right) \right) e^{-k^2} dk = \frac{C}{2\pi} e^{-k'^2} $$ which implies $$ C = 2\pi $$

We have shown that the prefactor in the OP's identity must be $2\pi$. But it is perhaps still not clear why. I think the best I can say at this time is that all functions for which the identity is true behave like $e^{-x^2}$ for large $x$. See, for example, https://en.wikipedia.org/wiki/Schwartz_space.

Answer 2

In some courses the there's a 2pi factor in the fourier transformation, in others there isn't, it depends on how you define the fourier transform. See https://mathoverflow.net/questions/265299/the-2-pi-in-the-definition-of-the-fourier-transform/265366

Answer 3

Note that$$\int_{-1/\epsilon}^{1/\epsilon}\exp\left[j(\omega-\omega^\prime)t\right]dt=\frac{2}{\epsilon}\operatorname{sinc}\frac{\omega-\omega^\prime}{\epsilon}.$$This is a "nascent delta function", an expression of the form $\frac{1}{\epsilon}f\left(\frac{\omega-\omega^\prime}{\epsilon}\right)$ with $\epsilon$-independent image under $\int_{\Bbb R}d\omega$ for $\epsilon>0$. As $\epsilon\to0^+$, such expressions $\to\infty$ for $\omega\to\omega^\prime$ and $\to0$ otherwise (if $f$ satisfies certain mild conditions that certainly hold here), i.e. we get a convergence in measures to a multiple of $\delta(\omega-\omega^\prime)$. The integral over $\omega$ gives you the multiplicative constant. We want to prove$$\int_{\Bbb R}\frac{2}{\epsilon}\operatorname{sinc}\frac{\omega-\omega^\prime}{\epsilon}d\omega=2\pi,$$or equivalently$$\int_0^\infty\frac{\sin x}{x}dx=\frac{\pi}{2}.$$This has been asked a few times before on math.se. My favourite approach is to rewrite the $\frac{1}{x}$ factor as $\int_0^\infty\exp(-xy)dy$, giving$$\Im\int_0^\infty\frac{dy}{y-j}=\int_0^\infty\frac{dy}{1+y^2}=\frac{\pi}{2}.$$