Laplace transform of the dirac delta function

In the last video, I introduced you to what is probably the most bizarro function that you've encountered so far. And that was the Dirac delta function. And I defined it to be-- and I'll do the shifted version of it. You're already hopefully reasonably familiar with it. So Dirac delta of t minus c. We can say that it equals 0, when t does not equal c, so it equals 0 everywhere, but it essentially pops up to infinity. And we have to be careful with this infinity. I'm going to write it in quotes. It pops up to infinity. And we even saw in the previous video, it's kind of different degrees of infinity, because you can still multiply this by other numbers to get larger Dirac delta functions when t is equal to c. But more important than this, and this is kind of a pseudodefinition here, is the idea that when we take the integral, when we take the area under the curve over the entire x- or the entire t-axis, I guess we could say, when we take the area under this curve, and obviously, it equals zero everywhere except at t is equal to c, when we take this area, this is the important point, that the area is equal to 1. And so this is what I meant by pseudoinfinity, because if I have 2 times the Dirac delta function, and if I'm taking the area under the curve of that, of 2 times the Dirac delta function t minus c dt, this should be equal to 2 times-- the area of just under the Dirac delta function 2 times from minus infinity to infinity of the delta function shifted by c dt, which is just 2 times-- we already showed you, I just said, by definition, this is 1, so this will be equal to 2. So if I put a 2 out here, this infinity will have to be twice as high, so that the area is now 2. That's why I put that infinity in parentheses. But it's an interesting function. I talked about it at the end of the last video that it can help model things that kind of jar things all of a sudden, but they impart a fixed amount of impulse on something and a fixed amount of change in momentum. And we'll understand that a little bit more in the future. But let's kind of get the mathematical tools completely understood. And let's try to figure out what the Dirac delta function does when we multiply it, what it does to the Laplace transform when we multiply it times some function. So let's say I have my Dirac delta function and I'm going to shift it. That's a little bit more interesting. And if you want to unshift it, you just say, OK, well, c equals 0. What happens when c equals 0? And I'm going to shift it and multiply it times some arbitrary function f of t. If I wanted to figure out the Laplace transform of just the delta function by itself, I could say f of t is equal to 1. So let's take our Laplace transform of this. And we can just use the definition of the Laplace transform, so this is equal to the area from 0 to infinity, or we could call it the integral from 0 to infinity of e to the minus -- that's just part of the Laplace transform definition-- times this thing-- and I'll just write it in this order-- times f of t times our Dirac delta function. Delta t minus c and times dt. Now, here I'm going to make a little bit of an intuitive argument. A lot of the math we do is kind of-- especially if you want to be very rigorous and formal, the Dirac delta function starts to break down a lot of tools that you might have not realized it would break down, but I think intuitively, we can still work with it. So I'm going to solve this integral for you intuitively, and I think it'll make some sense. So let's draw this. Let me draw this, what we're trying to do. So let me draw what we're trying to take the integral of. And we only care from zero to infinity, so I'll only do it from zero to infinity. And I'll assume that c is greater than zero, that the delta function pops up someplace in the positive t-axis. So what is this first part going to look like? What is that going to look like? e to the minus st times f of t? I don't know. It's going to be some function. e to the minus st starts at 1 and drops down, but we're multiplying it times some arbitrary function, so I'll just draw it like this. Maybe it looks something like this. This right here is e to the minus st times f of t. And the f of t is what kind of gives it its arbitrary shape. Fair enough. Now, let's graph our Dirac delta function. With zero everywhere except right at c, right at c right there, it pops up infinitely high, but we only draw an arrow that is of height 1 to show that its area is 1. I mean, normally when you graph things you don't draw arrows, but this arrow shows that the area under this infinitely high thing is 1. So we do a 1 there. So if we multiply this, we care about the area under this whole thing. When we multiply these two functions, when we multiply this times this times the delta function, this is-- let me write this. This is the delta function shifted to c. If I multiply that times that, what do I get? This is kind of the key intuition here. Let me redraw my axes. Let me see if I can do it a little bit straighter. Don't judge me by the straightness of my axes. So that's t. So what happens when I multiply these two? Everywhere, when t equals anything other than c, the Dirac delta function is zero. So it's zero times anything. I don't care what this function is going to do, it's going to be zero. So it's going to be zero everywhere, except something interesting happens at t is equal to c. At t equals c, what's the value of the function? Well, it's going to be the value of the Dirac delta function. It's going to be the Dirac delta function times whatever height this is. This is going to be this point right here or this right there, that point. This is going to be this function evaluated at c. I'll mark it right here on the y-axis, or on the f of t, whatever you want to call it. This is going to be e to the minus sc times f of c. All I'm doing is I'm just evaluating this function at c, so that's the point right there. So if you take this point, which is just some number, it could be 5, 5 times this, you're just getting 5 times the Dirac delta function. Or in this case, it's not 5. It's this little more abstract thing. I could just draw it like this. When I multiply this thing times my little delta function there, I get this. The height, it's a delta function, but it's scaled now. It's scaled, so now my new thing is going to look like this. If I just multiply that times that, I essentially get e to the minus sc times f of c. This might look like some fancy function, but it's just a number when we consider it in terms of t. s, it becomes something when we go into the Laplace world, but from t's point of view, it's just a constant. All of these are just constants, so this might as well just be 5. So it's this constant times my Dirac delta function, times delta of t minus c. When I multiply that thing times that thing, all I'm left with is this thing. And this height is still going to be infinitely high, but it's infinitely high scaled in such a way that its area is going to be not 1. And I'll show it to you. So what's the integral of this thing? Taking the integral of this thing from minus infinity to infinity, since this thing is this thing, it should be the same thing as taking the integral of this thing from minus infinity to infinity. So let's do that. Actually, we don't have to do it from minus infinity. I said from zero to infinity. So if we take from zero to infinity, what I'm saying is taking this integral is equivalent to taking this integral. So e to the minus sc f of c times my delta function t minus c dt. Now, this thing right here, let me make this very clear, I'm claiming that this is equivalent to this. Because everywhere else, the delta function zeroes out this function, so we only care about this function, or e to the minus st f of t when t is equal to c. And so that's why we were able to turn it into a constant. But since this is a constant, we can bring it out of the integral, and so this is equal to-- I'm going to go backwards here just to kind of save space and still give you these things to look at. If we take out the constants from inside of the integral, we get e to the minus sc times f of c times the integral from 0 to infinity of f of t minus c dt. Oh sorry, not f of t minus c. This is not an f. I have to be very careful. This is a delta. Let me do that in a different color. I took out the constant terms there, and it's going to be a delta of t minus c dt. Let me get the right color. Now, what is this thing by definition? This thing is 1. I mean, we could put it from minus infinity to infinity, it doesn't matter. The only time where it has any area is right under c. So this thing is equal to 1. So this whole integral right there has been reduced to this right there, because this is just equal to 1. So the Laplace transform of our shifted delta function times some other function is equal to e to the minus sc times f of c. Let me write that again down here. Let me write it all at once. So the Laplace transform of our shifted delta function t minus c times some function f of t, it equals e to the minus c. Essentially, we're just evaluating e to the minus st evaluated at c. So e to the minus cs times f of c. We're essentially just evaluating these things at c. This is what it equals. So from this we can get a lot of interesting things. What is the Laplace transform-- actually, what is the Laplace transform of just the plain vanilla delta function? Well, in this case, we have c is equal to 0, and f of t is equal to 1. It's just a constant term. So if we do that, then the Laplace transform of this thing is just going to be e to the minus 0 times s times 1, which is just equal to 1. So the Laplace transform of our delta function is 1, which is a nice clean thing to find out. And then if we wanted to just figure out the Laplace transform of our shifted function, the Laplace transform of our shifted delta function, this is just a special case where f of t is equal to 1. We could write it times 1, where f of t is equal to 1. So this is going to be equal to e to the minus cs times f of c, but f is just a constant, f is just 1 here. So it's times 1, or it's just e to the mine cs. So just like that, using a kind of visual evaluation of the integral, we were able to figure out the Laplace transforms for a bunch of different situations involving the Dirac delta function. Anyway, hopefully, you found that reasonably useful.