The Fourier Transform and the Dirac Delta Function

[@146fallon]

Thank you so much. I hope I watch it earlier. Both the Fourier transform and the Dirac delta are so important. This example can at least convince one to think of the Dirac delta as being a distribution instead of a function.

[@anubhavprakash8150]

sad to see so less views, it deserves a lot more, the explanation is damn good

[@physicsandmathlectures3289]

Thank you!

[@ANJA-mj1to]

The way how is presented with assuming no previous knowledge of the subject is great. In this brief description of the basic idea of The Dirac "delta - function" with following properties which disobeys Dirichlet's conditions (one of contiditions) we can illustrate and draw multidimensional Fourier form implying Cartesian coordinates for i. e. The Dirac wall (drawing 2 graphs with superimpose, let's say, "plane on the plane") 👍

[@brandonstokes5927]

this video saved my life thank you king

[@sanjaythorat]

Nice one. @ 2:55, I think you missed putting 2*pi under square root.

[@physicsandmathlectures3289]

Thank you for your comment. In this case I am combing the 1/sqrt(2 pi) from the definition of the inverse Fourier transform with the constant 1/sqrt(2 pi) that we are taking the inverse Fourier transform of.

[@kimia5390]

thank you so much for these videos! the delta function is finally starting to make sense to me now. I was wondering if you could do a video on " Contour integrals" as well along with some examples? I couldn't find any useful resources that explain it simply..

[@hrperformance]

Thanks so much!

[@mohamedazarudeen6131]

Hi, Thank you for the awesome important playlist you've created for us. I couldn't really get the final part of this problem. What's your conclusion on this undefined integral? Can you please rephrase it in a simple understandable manner?

Nice explanation!

[@physicsandmathlectures3289]

Thank you!

[@61gopalprabhulsm70]

what the integral representation of the derivative of the delta function

[@physicsandmathlectures3289]

In this case we can start from the integral representation of the delta function and take the derivative of both sides. Moving the derivative inside of the integral, we can differentiate the exponential function to pull down a factor of ik. This gives the result you're looking for.

[@beau-payage]

I have a question. Exp (i 2 pi x) = [Exp( i2pi) ]**x = 1**x =1? What is the problem?

[@physicsandmathlectures3289]

The issue here is that 1**x = 1 excludes other possible solutions. Consider the case where x = 1/2. In that case we can have 1**1/2 = 1 or 1**1/2 = -1 since we have to consider both signs when we take a square root. This generalizes for x=1/3 or 1/4 etc. You might find it useful to look up the 'nth roots of unity' since this goes further into how you determine what the possible solutions of 1**x are besides 1.

[@beau-payage]

@@physicsandmathlectures3289 thanks a lot.

[@NeonNotch]

Beautiful lecture and very nice handwriting! What program is it that you use to write?

[@physicsandmathlectures3289]

Thank you! I use SmoothDraw 4 with an older medium sized Wacom Intuos tablet.

[@prasantakumarbiswas3185]

thanks a lot sir.

[@jamaljaffer8412]

perfect well appreciated

[@hershyfishman2929]

Why doesn't it make sense intuitively? If it oscillates forever from -inf to inf between positive and negative it is = 0, except if x = 0, then it doesn't oscillate and the integral yields infinity. That's the delta function.

[@hendrikludwig5303]

Yes, I think one could imagine to take the limit to infinity in a symmetrical fashion (-T to T with T -> inf)

[@hendrikludwig5303]

but wait, there is a Sine still remaining. But on "average", that's 0, too

[@AbdelJalil-pv5kl]

My intuition is as it follows : The integral of the exp(-ikx) from minus infinity to positive infinity when k=0 is the same integral of one in the same interval therfore it gives infinity but the integral when k doesn't equal 0 is undefined although it wouldn't matter because it would be too small compared to infintity which already present so in convention the integral of exp(-ikx) from minus infinity to positive infinity is infinity when k=0 and 0 elsewhere which perfectly lines up with the dirac delta function distribution.

[@individuoenigmatico1990]

The intuition is that, if we suppose x≠0, then the exponential function H(k)=e^ikx has a period of 2pi/x. And every time you complete a period the integral over that period gives you zero (like a sine or cosine). That is to say the integral from k=0 to k=2pi/x of H(k) is equal to zero. And in general the integral from k=n2pi/k to k=(n+1)2pi/x of H(k) is zero. Since for the delta you are doing the integral from k=-infinity to k=+infinity, it's as if you are summing this integral over infinite periods, but each of these periods contributes zero. Hence the full integral (and hence the delta) is zero. Of course this is just intuition, because the integral from k=-infinity to k=+infinity of H(k) is simply not well defined. As a limit it simply doesn't exist. It merely oscillates between -2/x and 2/x.

[@AbuSous2000PR]

no it makes sense many thx all else never explain it

[@yufengyang8527]

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