L{sin(at)} - transform of sin(at)

Let's keep building our table of Laplace transforms. And now we'll do a fairly hairy problem, so I'm going to have to focus so that I don't make a careless mistake. But let's say we want to take the Laplace transform-- and this is a useful one. Actually, all of them we've done so far are useful. I'll tell you when we start doing not-so-useful ones. Let's say we want to take the Laplace transform of the sine of some constant times t. Well, our definition of the Laplace transform, that says that it's the improper integral. And remember, the Laplace transform is just a definition. It's just a tool that has turned out to be extremely useful. And we'll do more on that intuition later on. But anyway, it's the integral from 0 to infinity of e to the minus st, times-- whatever we're taking the Laplace transform of-- times sine of at, dt. And now, we have to go back and find our integration by parts neuron. And mine always disappears, so we have to reprove integration by parts. I don't recommend you do this all the time. If you have to do this on an exam, you might want to memorize it before the exam. But always remember, integration by parts is just the product rule in reverse. So I'll just do that in this corner. So the product rule tells us if we have two functions, u times v. And if I were take the derivative of u times v. Let's say that they're functions of t. These are both functions of t. I could have written u of x times v of x. Then that equals the derivative of the first times the second function, plus the first function times the derivative of the second. Now, if I were to integrate both sides, I get uv-- this should be review-- is equal to the integral of u prime v, with respect to dt-- but I'm just doing a little bit of shorthand now-- plus the integral of uv prime. I'm just trying to help myself remember this thing. And let's take this and subtract it from both sides. So we have this integral of u prime v is going to be equal to this, uv minus the integral of uv prime. And, of course, this is a function of t. There's a dt here and all of that. But I just have to do this in the corner of my page a lot, because I always forget this, and with the primes and the integrals and all that, I always forget it. One way, if you did want to memorize it, you said, OK, the integration by parts says if I take the integral of the derivative of one thing and then just a regular function of another, it equals the two functions times each other, minus the integral of the reverse. Right? Here, when you take the subtraction, you're taking the one that had a derivative, now it doesn't. And the one that didn't have a derivative, now it does. But anyway, let's apply that to our problem at hand, to this one. Well, we could go either way about it. Let's make u prime is equal to-- we'll do our definition-- u prime is equal to e to the minus st, in which case you would be the antiderivative of that, which is equal to minus 1 over s e to the minus st, right? And actually, this is going to be an integration by parts twice problem, so I'm just actually going to define the Laplace transform as y. That'll come in useful later on. And I think I actually did a very similar example to this when we did integration by parts. But anyway, back to the integration by parts. So that's u. And let me do v in a different color. So when v-- if this is u prime, right? This is u prime, then this is v. So v is equal to sine of at. And then what is v prime? Well, that's just a cosine of at, right? The chain rule. And now, we're ready to do our integration. So the Laplace transform, and I'll just say that's y, y is equal to-- y is what we're trying to solve for, the Laplace transform of sine of at-- that is equal to u prime v. I defined u prime in v, right? That's equal to that. The integral of u prime times v. That equals uv. So that's minus 1 over s e to the minus st, times v, sine of at, minus the integral. And when you do the integration by parts, this could be an indefinite integral, an improper integral, a definite integral, whatever. But the boundary stays. And we can still say, from 0 to infinity of uv prime. So u is minus 1 over s e to the minus st, times v prime, times a cosine of at-- fair enough-- dt. Well, now we have another hairy integral we need to solve. So this might involve another integration by parts, and it does. Let's see if we can simplify it at [? all. ?] Let's take the constants out first. Let me just rewrite this. So we get y is equal to minus e to the minus st over s, sine of at. So you have a minus minus plus a over s-- a divided by s, and then these two negative signs cancel out-- times the integral from 0 to infinity, e to the minus st, cosine of at, dt. Let's do another integration by parts. And I'll do this in a purple color, just so you know this is our second integration by parts. Over here. Let's define once again, u prime is equal to e the minus st. So this is u prime. Then u is equal to minus 1 over s e to the minus st. We'll make v equal to cosine of at. The hardest part about this is not making careless mistakes. And then v prime-- I just want it to be on the same row-- is equal to minus a sine of at, right? The chain rule, derivative of cosine is minus sine. So let's substitute that back in, and we get-- this is going to get hairy; actually, it already is hairy-- y is equal to minus e to the minus st over s, sine of at, plus a over s, times-- OK. Integration by parts. uv. So that's minus 1 over s e to the minus st, times v, times cosine at, minus the integral from 0 to infinity. This problem is making me hungry. It's taking so much glucose from my bloodstream. I'm focusing so much not to make careless mistakes. Anyway, integral from 0 to infinity. And now, we have uv prime, so u is minus 1 over s e to the minus st. That's u. And then v prime times minus a. So let's make that minus cancel out with this one. So that becomes a plus. a sine of at, dt. I'm starting to see the light at the end of the tunnel. So then, let's simplify this thing. And, of course, we're going to have to evaluate this whole thing, right? Actually, we're going to have to evaluate everything. Let's just focus on the indefinite integral for now. We're going to have to take this whole thing and evaluate-- let's just say that y is the antiderivative and then evaluate it from infinity to 0. From 0 to infinity. So y is equal to minus e to the minus st over s, sine of at. Now let's distribute this. Minus a over s squared, e to the minus st, cosine of at. Right? OK, now I want to make sure I don't make a careless mistake. OK. Now, let's multiply this times this and take all of the constants out. So we have an a and an s. a over s. There's a minus sign. We have a plus a to the s. So we'll have a minus a squared over s squared, times the integral from 0-- well, I said I'm just worrying about the indefinite integral right now, and we'll evaluate the boundaries later. e to the minus st, sine of at, dt. Now, this is the part, and we've done this before, it's a little bit of a trick with integration by parts. But this expression, notice, is the same thing as our original y. Right? This is our original y. And we're assuming we're doing the indefinite integral, and we'll evaluate the boundaries later. Although we could have kept the boundaries the whole time, but it would have made it even hairier. So we can rewrite this integral as y. That was our definition. And actually, I just realized I'm running out of time, so I'll continue this hairy problem in the next video. See you soon.