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# Laplace transform of t^n: L{t^n}

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In the last video, I showed the Laplace transform of t, or we could view that as t the first power, is equal to 1/s squared, if we assume that s is greater than 0. In this video, we're going to see if we can generalize this by trying to figure out the Laplace transform of t to the n, where n is any integer power greater than 0, so n is any positive integer greater than 0. So let's try it out. So we know from our definition of the Laplace transform that the Laplace transform of t to the n is equal to the integral from 0 to infinity of our function-- well, let me write t to the n-- times, and this is just the definition of the transform, e to the minus st, dt. And similar to when we figured out this Laplace transform, your intuition might be that, hey, we should use integration by parts, and I showed it in the last video. I always forget it, but I just recorded it, so I do happen to remember it. So the integration by parts just tells us that the integral of uv prime is equal to uv minus the integral of-- I view this as kind of the swap-- so u prime v. So this is just our integration by parts formula. If you ever forget it, you can derive it in about 30 seconds from the product rule. And I did it in the last video because I hadn't used it for awhile, so I had to rederive it. So let's apply it here. So what do we want to make our v prime? It's always good to use the exponential function, because that's easy to take the antiderivative of. So this is our v prime, in which case our v is just the antiderivative of that. So it's e to the minus st over minus s. If we take the derivative of this, minus s divided by minus s cancels out, and you just get that. And then if we make our u-- let me pick a good color here. If we make this equal to our u, what's our u prime? u prime is just going to be n times t to the n minus 1. Fair enough. So let's apply the integration by parts. So this is going to be equal to uv. u, I'll use this t to the n, so u is t to the n, that's our u, times v, which is e-- let me write this down-- so it's minus-- there's a minus sign there, so we put the minus. Let me do it in that color. I'm just rewriting this. e to the minus st/s. So that's the uv term right there. Let me make that clear. And let me pick a good color here. So this term right here is this term right here. And, of course, I'm going to have to evaluate this from 0 to infinity, so let me write that: 0 to infinity. I could put a little bracket there or something, but you know we're going to have to evaluate that. And then from that, we're going to have to subtract the integral. And let me not forget our boundaries. 0 to infinity of u prime is n times t to the n minus 1-- that's our u prime-- times v times minus-- so let me put this minus out here-- so minus e to the minus st/s. And then all of that. Of course, we have our dt, and you have a minus minus. These things become pluses. Let's see if we can simplify this a little bit. So we get our Laplace transform of t to the n is equal to this evaluated at infinity and evaluated at 0. So when you've evaluate-- what's the limit of this as t approaches infinity? As t approaches infinity, this term, you might say, oh, this becomes really big. And I went over this in the last video. But this term overpowers it, because you're going to have e to the minus infinity, if we assume that s is greater than 0. So if s is greater than 0, this term is going to win out and go to 0 much faster than this term is going to go to infinity. So when you evaluate it at infinity, when you evaluate this at infinity, you're going to get 0. And then you're going to subtract this evaluated at 0. This evaluated at 0, when it's evaluated at 0 is just minus 0 to the n times e to the minus s times 0/s Well, this becomes 0 as well. So this whole term evaluated from 0 to infinity is all 0, which is a nice, convenient thing for us. And then we're going to have this next term right there. So let's take out the constant terms. This n and this s are constant. They are constant with respect to t. So you have plus n/s times the integral from 0 to infinity of t to the n minus 1 times e to the minus st, dt. Now, this should look reasonably familiar to you. What's the definition of the Laplace transform? The Laplace transform of any function is equal to the integral from 0 to infinity of that function times e to the minus st, dt. Well, when we have an e to the minus st, dt, we're taking the integral from 0 to infinity, so this whole integral is equal to the Laplace transform of this, of t to the n minus 1. So just that easily, because this term went to 0, we've simplified things. We get the Laplace transform of t to the n is equal to-- this is all 0-- it's equal to n/s-- that's right there-- times this integral right here, which we just figured out was the Laplace transform of t to the n minus 1. Well, this is a nice, neat simplification. We can now figure out the Laplace transform of a higher power in terms of the one power lower that, but it still doesn't give me a generalized formula. So let's see if we can use this with this information to get a generalized formula. So the Laplace transform of just t-- so let me write that down; I wrote that at the beginning of the problem. We get the Laplace transform, I could write this as t to the 1, which is just t, is equal to 1/s squared, where s is greater than 0. Now, what happens if we take the Laplace transform of t squared? Well, we can just use this formula up here. The Laplace transform of t squared is equal to 2/s times the Laplace transform of t, of just t to the 1, right? 2 minus 1. So times the Laplace transform of t to the 1. Well, t, we know what that is. This is equal to 2/s times this, times 1/s squared, which is equal to 2/s to the third. Interesting. Let's see if we can do another one. What is-- I'll do it in the dark blue-- the Laplace transform of t to the third? Well, we just use this formula up here. It's n/s. In this case, n is 3. So it's 3/s times the Laplace transform of t to the n minus 1, so t squared. We know what the Laplace transform of this one was. This is just this right there. So it's equal to 3/s times this thing. And I'm going to actually write it this way, because I think it's interesting. So I'll write the numerator. Times 2 times 1/s over s squared, which is-- we could write it as 3 factorial over-- what is this? s to the fourth power. Let's do another one. And I think you already are getting the idea of what's going on. The Laplace transform of t to the fourth power is what? It's equal to 4/s times the Laplace transform of t the third power. And that's just 4/s times this. So it's 4/s times 3 factorial over s to the fourth. So now 4 times 3 factorial, that's just 4 factorial over s to the fifth. And so you can just get the general principle. And we can prove this by induction. It's almost trivial based on what we've already done. That the Laplace transform of t to the n is equal to n factorial over s to the n plus 1. We proved it directly for this base case right here. This is 1 factorial over s to the 1 plus 1. And then if we know it's true for this, we know it's going to be true for the next increment. So induction proof is almost obvious, but you can even see it based on this. If you have to figure out the Laplace transform of t to the tenth, you could just keep doing this over and over again, but I think you see the pattern pretty clearly. So anyway, I thought that was a neat problem in and of itself, outside of the fact, it'll be useful when we figure out inverse and Laplace transforms. But this is a pretty neat result. The Laplace transform of t to the n, where n is some integer greater than 0 is equal to n factorial over s to the n plus 1, where s is also greater than 0. That was an assumption we had to make early on when we took our limits as t approaches infinity. Anyway, hopefully you found that useful.