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

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

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 $$x(t)$$

• Take a look at the definition of the inverse FT. Commented Jun 23, 2019 at 11:41
• Importantly, $\int_{-\infty}^{\infty} dt e^{j(\omega - \omega')t} \neq \delta(\omega - \omega')$. The $2\pi$ must be there for the identity to be correct. Commented Jun 23, 2019 at 11:47
• This is what I have seen and done: Given the inverse fourier $x(t) = { \frac{1}{2\pi} } \int_{ -\infty }^{ \infty } \chi (j \omega _{1}) e^{j \omega _{1} t} dt$ and the Fourier transform as: $\chi (j \omega )= \int_{- \infty }^{ \infty } x(t) e^{-j \omega } dt$ Then substituting $x(t)$ into the FT and a bit of rearranging: $\chi (j \omega )= \int_{- \infty }^{ \infty } \chi ( \omega_{1}) \big( \frac{1}{2 \pi } \int_{- \infty }^{ \infty } e^{-jt( \omega - \omega _{1} )} dt\big) d \omega _{1}$ @RodrigodeAzevedo Commented Jun 23, 2019 at 12:16
• I'll have to less respectfully disagree so that the OP doesn't get the wrong idea. As I showed in my response, $\int_{-\infty}^\infty e^{i(k - k')x} dx = C\delta(k - k')$ is only true when $C = 2\pi$. There is nothing conventional about the $2\pi$. It is a mathematical fact that it must be there. Otherwise, we will get the wrong answer when computing $\int_{-\infty}^\infty \int_{-\infty}^\infty e^{i(k - k')x} f(k)dx dk$ for functions $f$ where the integral is well defined. Commented Jun 23, 2019 at 13:07
• You can normalize the measure any way you want, but, as the link claims, if you want both Haar measures to be the Lebesgue measure, you can't avoid the factor of $2\pi$. Based on this, I would say there definitely is a sense in which the $2\pi$ factor is natural. At the very least, it is the answer you get when you compute the integral in my previous comment for any Schwartz function using the Lebesgue measure on $\mathbb{R}^2$. Commented Jun 23, 2019 at 13:34

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$$\Im\int_0^\infty\frac{dy}{y-j}=\int_0^\infty\frac{dy}{1+y^2}=\frac{\pi}{2}.$$

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.

• This is not a valid computation because the integral $\int_{-\infty}^{\infty} e^{i(k-k')x}dx$ is not well-defined at the outset. This is the same problem the OP has. Tell me how you define that integral, and I will tell you where the $2\pi$ fits in. Commented Jun 23, 2019 at 13:05
• It is a valid computation because, while the integral $\int_{-\infty}^\infty e^{i(k-k')x} dx$ is not well-defined, the integral $\int_{-\infty}^\infty \int_{-\infty}^{\infty} e^{i(k - k')x} e^{-k^2} dxdk$ most certainly is. It has only one answer, namely $e^{-k'^2}$. Thus, if the OP identity is to be true in the sense of distributions, we are forced to choose $C = 2\pi$. Commented Jun 23, 2019 at 13:11
• By true in the sense of distributions, I mean the hacky physics definition (which can be formalized for suitable function spaces): $f = g$ in the sense of distributions if $\int_{-\infty}^\infty f(x)h(x) dx = \int_{-\infty}^{\infty} g(x)h(x)$ for all functions $h$ where both integrals are defined. Commented Jun 23, 2019 at 13:14
• Sure, the computation after the first equality is correct. But still you have to tell us how the first integral is to be understood. Is it a Fourier transform? In which case, how do you define (normalize) the Fourier transform? Commented Jun 23, 2019 at 13:30
• Late edit: $e^{-k'^2}$ in the above comment should read $2\pi^{-k'^2}$. I understood the integral as being defined so that $\int_{-\infty}^\infty \left(\int_{-\infty}^\infty e^{i(k - k')x} dx \right) f(k) dk = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{i(k - k')x} f(k) dx dk$ for all functions $f$ where the latter integral is defined and $dxdk$ is the Lebesgue measure on $\mathbb{R}^2$. While this is merely a convention, I would argue it is a fairly natural one and probably the one the OP was assuming without realizing it. Commented Jun 23, 2019 at 13:39

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

• I don't think this is a matter of convention. The fact is $\int_{-\infty}^\infty e^{ikx} dx = 2\pi \delta(k)$ regardless of the convention you adopt for the Fourier transform. The question remains, where does this $2\pi$ come from? Commented Jun 23, 2019 at 11:46