 Let's see if we can do something a bit more difficult. Let's set the f of t equals t So let's get the Laplace transform. You never know how to do these curly places properly of t Now we do have our Short-hand of doing that that is going to be the improper integral going from 0 to infinity of e to the power negative s t times t dt So here I have the product of two functions of t and I'm taking the integral with respect to that variable In other words, I better use the power rule of integration here I'm going to select these various methods of doing this. I'll do it long-hand this way. I'm going to select This t as being u Which means v prime has got to be the e to the power negative s t That makes u prime equal to one and that means v should have been one Negative one over s e to the power negative s t if I take the first derivative of that with respect to t I'm going to end up with that This leaves me with u v minus u prime v So this is going to leave me With u v that's negative of one over s it's going to leave me with a t And it's going to leave me with an e to the power negative s t and that it goes from 0 to infinity Remember it goes from 0 to b and I'm taking the limit of that Then I'm going to have a negative Improper integral and now it becomes u prime v That's why I chose t because I could Simplify it so the new integral must be a bit simpler than the old one was One over s negative one of s is a constant so I can bring that out that makes this one over s And I'm going to have e to the power negative s t dt Now what is this? This is the Laplace transform of one This is nothing other we've just done it previous video This is the Laplace transform of one and we know what the Laplace transform of ones. Well, this is one over s So yeah, I'm already gonna have one over s times another one over s you can do this whole thing again We just did in previous video So it's going to be one over s squared for these two Remember this is times one so that's the Laplace transform of one which was we showed one of s Let's do this side though. It's going to be Negative one over s Times this is going to be t over e to the power s t going from 0 to infinity So let's do that that is negative one over s. So what are we gonna have? We're gonna have infinity over e to the power s times infinity Minus is zero over e to the power s times zero And I have that Plus one over s squared Now this is in determinant form because I have infinity divided by infinity I have I will have infinity divided by infinity, but remember this wasn't actually that was this actually a b That was actually the limit As b goes to infinity of it was b over e to the power s b And if I put in the limit there it would be Infinity divided by infinity in determinant form. So I take Independently the first derivative of both that would have left me with one over s times e to the power s b and if I now slip in Infinity in a it's one over infinity. That's the zero. So that's zero minus zero This is one That's the zero zero times. This is the zero. So it's just one over s squared There we go the Laplace transform of t to the power one is one over s squared That is the Laplace transform of t Remember we showed before the Laplace transform of we said one, but that's the same as t to the power zero is one over s T to the power one is one over s squared. I see we I think you can see where this is going But we'll get to that and we'll do some more examples next