A Little Pulse…

This past semester I was working as T.A in a Mathematical Physics course. One of the main topics studied was Fourier analysis which I consider a really beautiful subject, but also rather cryptic. This area is fulled with mysterious and magical-looking identities, one of them in particular calls my attention every time I use it and I would like to discuss it here. Also, I think it may be useful to show a more complete (but not formal at all) proof of this fact, because once you “believe” it, the derivation of the Fourier transform identities comes as pretty straight forward. I’m assuming here a certain knowledge or intuition about the Dirac delta function and its sampling property Forgetting about suspense, here is the thing I would like to prove:

\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega x}\mathrm{d}\omega

From the first time you look at this, you know that it has to be true, not because it is obvious (I don’t think it is), but because it looks pretty nice, it is no something that you would usually get by writing random symbols. This of course is not good enough reason to believe this, but instead of proving this right away, we begin with a different integral:

\int_{0}^{\infty}\frac{\sin k x}{x}\mathrm{d}x, k>0

Of course, setting u= kx, \mathrm{d}u=k\mathrm{d}x , we get a nicer integral:

\int_{0}^{\infty}\frac{\sin u}{u}\mathrm{d}u

To proceed, we use the trick of “differentiating under the integral sign”, a.k.a. Feynman’s Trick. we define the following function:

f(\theta)=\int_{0}^{\infty}e^{-u\theta}\frac{\sin u}{u}\mathrm{d}u, 0\leq \theta

Then we differentiate this with respect to \theta . Here  we are assuming that \theta and u are independent variables, we thus obtain the following from Leibniz formula:

\frac{\mathrm{d} f}{\mathrm{d} \theta }= \frac{\mathrm{d} }{\mathrm{d} \theta }\left(\int_{0}^{\infty}e^{-u\theta}\frac{\sin u}{u}\mathrm{d}u \right)= \int_{0}^{\infty}\frac{\partial}{\partial \theta}\left(e^{-u\theta}\frac{\sin u}{u}\right)\mathrm{d}u

\frac{\mathrm{d} f}{\mathrm{d} \theta }= \int_{0}^{\infty}(-u)e^{-u\theta}\frac{\sin u}{u}\mathrm{d}u = - \int_{0}^{\infty}e^{-u\theta}\sin u\mathrm{d}u

Remember that \sin u = \frac{e^{iu}-e^{-iu}}{2 i} , this should simplify the integral:

\frac{\mathrm{d} f}{\mathrm{d} \theta }= \frac{i}{2 } \int_{0}^{\infty}e^{-u\theta}(e^{iu}-e^{-iu})\mathrm{d}u=- Im\{ \int_{0}^{\infty}e^{u(-\theta+i)}\mathrm{d}u \} =- Im\left\{ \left. \frac{e^{u(-\theta+i)}}{-\theta+i} \right|_{0}^{\infty} \right\}

But \lim_{u \rightarrow \infty} |e^{u(-\theta+i)}|=  \lim_{u \rightarrow \infty} |e^{-u\theta}| =0 , leaving us with

\frac{\mathrm{d} f}{\mathrm{d} \theta }=-Im\left\{ \frac{1}{-\theta+i} \right\}=-Im\left\{ \frac{-\theta-i}{\theta^2+1} \right\}=-\frac{1}{\theta^2+1}

If we integrate this again with respect to \theta we get the desired integral:

\frac{\mathrm{d} f}{\mathrm{d} \theta }=-\frac{1}{\theta^2+1} \Rightarrow f(\theta)= C- \arctan \theta

Notice that this function should vanish as its argument tends to infinity (this can be notice from its integral form). From this we can get the value for the integration constant:

f(\theta)= \frac{\pi}{2}- \arctan \theta

In particular, when \theta =0 we get

\int_{0}^{\infty}\frac{\sin u}{u}\mathrm{d}u = \frac{\pi}{2} \

or

\int_{0}^{\infty}\frac{\sin k x}{x}\mathrm{d}x= \frac{\pi}{2} , k>0

But, what happens if k<0 ? Since the sine function is an odd function it should be clear that the integral changes its sign; for the case of it being null, the result is clear, leaving:

\int_{0}^{\infty}\frac{\sin k x}{x}\mathrm{d}x= \left \{ \begin{array}{lr} \frac{\pi}{2} \quad k>0\\o  \quad k=0 \\ -\frac{\pi}{2} \quad k<0 \end{array} \right.

This can be written down in a single line using the Heaviside Step Function, which takes a value of one whenever its argument is non-negative, zero on negatives and one half at zero.

\int_{0}^{\infty}\frac{\sin \omega x}{x}\mathrm{d}x= \pi H(\omega) - \frac{\pi}{2}

It shouldn’t be hard to convince yourself from the parity of this function that

\int_{-\infty}^{\infty}\frac{\sin \omega x}{x}\mathrm{d}x= 2\pi H(\omega) - \pi \qquad (*)

And also, we  can see from its definition that

\int_{-\infty}^{\omega}\delta (x) \mathrm{d}x= H(\omega)

or even better, from the fundamental theorem of calculus

\delta (\omega) =\frac{\mathrm{d}}{\mathrm{d}\omega} H(\omega)

I know I’m stretching the rules here, but please don’t mind that, as I said, I’m not intending to show any formal proof, but rather one that can convince you. This last equation tell us pretty clearly that (*) shall be differentiated:

\frac{\mathrm{d}}{\mathrm{d}\omega}\int_{-\infty}^{\infty}\frac{\sin \omega x}{x}\mathrm{d}x=\int_{-\infty}^{\infty}x\frac{\cos \omega x}{x}\mathrm{d}x= 2\pi \delta(\omega)

\int_{-\infty}^{\infty}\cos \omega x\mathrm{d}x= 2\pi \delta(\omega) \qquad (\Delta)

This one by itself is a pretty interesting identity, but we are not quite there yet. Nevertheless, the rest is quite simple. From parity, we get that \int_{-t}^{t}\sin \omega x\mathrm{d}x= 0. Taking the limit for t \rightarrow \infty and multiplying by the imaginary unit gets

\int_{-\infty}^{\infty}i\sin \omega x\mathrm{d}x= 0

If we add this last equation to (\Delta) and use the Euler’s formula we finally arrive to our result

\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega x}\mathrm{d}\omega

It wasn’t that hard. For the final act we may use a slightly different form of this identity replacing x \rightarrow x-y . This change does not modify anything we did so far, but it’s a lot cleaner to introduce it until now:

\delta(x-y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega (x-y)}\mathrm{d}\omega \qquad(\odot)

Now, assume f:\mathbb{R}\rightarrow \mathbb{C} to be an integrable function. This means that its integral (in the Lesbegue sense) along the real axis is finite. From the sampling property of the Dirac Delta we can write the following identity

f(y) = \int_{-\infty}^{\infty} f(x) \delta (x-y)\mathrm{d}x

And this thing can be rewritten using the equation (\odot)

f(y) = \int_{\infty}^{\infty}f(x) \left( \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega (x-y)}\mathrm{d}\omega\right)  \mathrm{d}x

But of course we can change the order of integration and rearrange some terms

f(y) = \frac{1}{2\pi} \int_{\infty}^{\infty}\int_{-\infty}^{\infty}f(x)e^{i\omega x} e^{-i\omega y} \mathrm{d}x \mathrm{d}\omega = \frac{1}{\sqrt{2\pi}} \int_{\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}f(x)e^{i\omega x}  \mathrm{d}x  \right) e^{-i\omega y} \mathrm{d}\omega

We can rewrite what is inside of the parenthesis to have a much cleaner version defining

\hat{f}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}f(x)e^{i\omega x}  \mathrm{d}x

Leaving us with the following

f(y) = \frac{1}{\sqrt{2\pi}} \int_{\infty}^{\infty}  \hat{f}(\omega) e^{-i\omega y} \mathrm{d}\omega

If we plug the original independent variable on this last equation we get the already known pair:

\hat{f}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}f(x)e^{i\omega x}  \mathrm{d}x

f(x) = \frac{1}{\sqrt{2\pi}} \int_{\infty}^{\infty}  \hat{f}(\omega) e^{-i\omega x} \mathrm{d}\omega

The former equation is the Fourier Transform of the f function and the later is its inverse. In some places you would find that the signs on the exponential function of this definitions are interchanged and usually if you ask about it they would tell you that it is not important as long as you use a consistent convention. The reason for this should be obvious from the symmetry of the Dirac Delta function (\delta(x-y)=\delta(y-x) .

This is it for now. Using this identity you can prove pretty easily some properties like the Parseval’s Formula, which allows you see that this transformation preserves the inner product of vector. There are other things that can be said about this transformation and the studied identity, but I leave it here for today; maybe next time I’ll work out something with a more algebraic approach.

Have a nice day.

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