Laplace transform of t: L{t}

let's try to fill in our Laplace transform table a little bit more and a good place to start is just to write our definition of the Laplace transform the Laplace transform of some function f of T is equal to the integral from zero to infinity of e to the minus st times our function f of T DT that's our definition and the very first one we solved for can even do it on the side right here was the Laplace transform of one just you know we can almost view that as T to the 0 and that was equal to the integral from 0 to infinity f of T was just 1 so it's e to the minus s T DT which is equal to the antiderivative of e to the minus s T is minus 1 over s minus 1 over s e to the minus s T and then you have to evaluate that from 0 to infinity when you take the limit is this term approaches infinity this e to the minus this becomes e to the minus infinity if we assume s is greater than 0 so if we assume s is greater than 0 this whole term goes to 0 so you can end up with 0 minus this thing evaluated at 0 so when you evaluate T is equal to 0 this term right here becomes 1 right e to the 0 becomes 1 so it's minus minus 1 over s which is the same thing as plus 1 over s so the plus transform of one of just the you know the constant function 1 is 1 over s we already solved that now let's increment it a little bit let's see if we can figure out the Laplace transform of T so we could view this as T to the 0 now this is T to the 1 this is going to be equal to the integral from 0 to infinity of e to the minus s T times T DT now I can tell you right now that you know this isn't I don't have the antiderivative of this memorize I don't know what it is but there's a sense that the integration by parts could be useful because integration by parts kind of decompose this into a simpler problem and I always forget integrate by parts so I'll read arrive it here in in this purple color so if we have u times V if we take the derivative with respect to say T of that that's equal to the derivative of the first times the second function plus the first function times the derivative of the second function just the product rule now if we take the integral of both sides of this equation we get u v is equal to the antiderivative of U prime V plus the antiderivative of U V Prime now since we want to apply this to an integral maybe let's make this what we want to solve for so you can get the integral of u V prime is equal to we can just subtract this from that side of the equation so it's equal to em to switch swapping the side so I'm just solving for this and to solve for this I just subtract this from that so it's equal to u V minus the integral of U prime V so there you go even though I have trouble memorizing this formula it's not too hard to read arrive as long as you remember as long as you remember the product rule right there so if we're going to do integration by part it's good to define our our V prime to be something it's easy to take the antiderivative of because we have to figure out V later on and it's good to take u to be something that's easy to take the derivative of so let's make T is equal to R U and let's make e to the minus st as being our v prime if that's the case then what is what is V well V is just the antiderivative of that in fact we've done it before it's minus 1 over s e to the minus st that's V and then if we want to figure out u prime because we're going to figure out that later anyway u prime is just the derivative of T that's just equal to 1 so let's apply this let's see so the Laplace transform of T the Laplace transform of T is equal to UV u is T viii is this right here it's it's x minus one over s e to the minus s T e to the minus s T that's the UV term right there and this isn't this is a definite integral right so we're going to evaluate this term right here from zero to infinity and then it's minus minus the integral minus the integral from 0 to infinity of U prime which is just 1 times V V we just figured out here is minus let me write it in the V in these color times minus 1 over s this is my V right here minus 1 over s e to the minus s T DT of that is DT DT and this let me see if I can so let's let's see if we can simplify this so this is equal to let me minus T over s e to the minus s T evaluated from 0 to infinity and then see we can take thus well this is just 1 we could you know this 1 times anything is 1 we can just not write that and then bring the minus one of our s out so if we bring the minus 1 over s out this becomes plus 1 over s times the integral from 0 to infinity of e to the minus s T DT and this should look familiar to you this is exactly what we solved for right here it was a little applause transform of 1 so let me let's keep that in mind so this right here is little plus transform of 1 and I want to write it that way because we're going to see a pattern of this in the next video I'm going to write that as a plus transfer 1 but what is this equal to so we're going to evaluate this as it add to infinity and then subtract from that evaluated at 0 so I'm just going to make you can kind of view it as the substitution so this is equal to and well let me write it this way it's limit as a approaches infinity of minus a over s e to the minus s a so that's this evaluated at infinity and then from that we're going to subtract this evaluate at 0 so minus all of this but we already have a minus sign here so we could write a plus plus we could write zero over s times e to the minus s times 0 and then of course we have this term right here so let me write that term I'll do it in yellow let me do it in blue plus 1 over s that's this right there times the Laplace transform of 1 and what do we get what do we get so what's the limit is this as a approaches infinity you might say well you know this as a approaches infinity right here this becomes a really big number there's a minus sign in there so it'd be a really big negative number but this is going to this is an exponent a is an exponent right here so e to the minus infinity is going to go to 0 much faster than this is going to go to infinity this term right here is a much stronger function I guess you can the way you can see you could try it out on your calculator if you don't believe me this term is going to overpower this term and so this whole thing is going to go to 0 this whole thing is going to go to 0 likewise e to the minus this e to the 0 this is 1 but you're multiplying it times a 0 so this is also going to go to 0 which is convenient because all of this stuff just disappears and we're left with the Laplace transform of t is equal to 1 over s times the Laplace transform of 1 and we know what the Laplace transform of 1 is the Laplace transform of 1 we just did at the beginning of the video which is equal to 1 over s if we assume that s is greater than 0 and in fact we have to assume that s was greater than 0 here in order to sassoon that this goes to 0 only if s is greater than 0 then when you get a minus infinity here does this approach to 0 so fair enough so Laplace transform of T is equal to 1 over s times 1 over s which is equal to 1 s squared for s is greater than zero so we have one more entry in our table and then we can use this we're going to do in the next video is build up to the Laplace transform Laplace transform of T to any arbitrary exponent and we'll do this in the next video