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Current time:0:00Total duration:10:16

Video transcript

in the last video I showed that the Laplace transform of T or could view that as T to the first power is equal to 1 over s squared if we assume that s is greater than 0 in this video we're going to see what 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 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 s T 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 U V prime is equal to UV minus the integral of iview this is the kind of the swap so u prime 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 now I did it in the last video because I hadn't used it for a while so I had to read arrive it so let's apply it here so what do we want to make our V prime it's always good to use the exponent exponential function because that's easy to take the antiderivative of so this is our V prime which in which case our V is just the antiderivative of that so it's e to the minus st over minus s all right if we take the derivative of this minus s divided by minus s cancels out you just get that and then if we make if we make our you let me pick a good color here if we make this equal to ru 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 R this is T to the N so u is T to the N that's R u times V which is e let me write this down so it's e - there's a minus sign there so let me put the - let me do it in that color - just rewriting this e to the minus s T over s and okay so that's the UV term right there let me make that clear let me pick a good color here so this term right here is this term right here and of course this isn't 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 gonna 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 u prime is n times T to the N minus 1 that's our u prime times V times minus e to the months let me put this minus out here so minus e to the minus s T over s and then all of that of course we have our DT and you have a minus - these things become pluses and 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 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 in value ate it in infinity when you in value ate this and infinity you're going to get 0 and then you're going to subtract this evaluated at zero this evaluated at zero when it evaluated at zero is just minus zero to the N times e to the minus s times zero over s well this becomes zero as well so this whole term evaluated from zero to infinity is all zero which is 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're constant respect with respect to T so you have plus n over s times the integral from 0 to infinity of T to the N minus 1 times e to the minus s T DT now this should look reasonably familiar to you this should look reasonably 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 stdt well we have an e to the minus stdt 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 have a we've simplified things we get we get the Laplace transform of T to the N is equal to this all 0 it's equal to n over 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 than 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 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 over s squared where s is greater than where s is greater than 0 now what happens if we take the Laplace transform Laplace transform of T squared well we can just use this formula appear the Laplace transform of T squared is equal to 2 over s times the Laplace transform of T of just 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 over s times this times 1 over s squared which is equal to 2 over 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 over s in this case n is 3 so it's 3 over s times the Laplace transform of n to the T to the N minus 1 so T squared T squared we know what the Laplace transform of this one was this is just this right there so it's equal to 3 over s times times this thing and I'm gonna actually write it this way because I think it's interesting so I'll write the numerator times 2 times 1 over s over s squared which is we can write it as 3 factorial over what is this s to the fourth power s to the fourth power let's do another one I think you you already are getting the idea of what's going on Laplace transform of T to the fourth power is what it's equal to 4 over s times the Laplace transform of T to the third power and that's just 4 over s times this is 4 over s times 3 factorial over s to the fourth so now for time three factorial that's just four factorial over s to the fifth and so you can just get a general I general principle we can prove this by induction it's almost trivial based on what we've already done that the Laplace transform the Laplace transform of T to the N is equal to n factorial n factorial over s to the n plus 1 we tried it out for it we proved it directly for this base case right here right this is one factorial over s to the 1 plus 1 and then for if we know if we know it's true for this we know it's going to be true for the next increment so an induction proof is almost obvious but you can even see it based on this 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 of itself yeah you don't you know outside of the fact that will 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 have to make on early on when we took our limits as as T approaches infinity anyway hopefully you found that useful