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But anyway, it's the integral from 0 to infinity of e to the minus st, times whatever we're taking the Laplace

Colombo

Actually, let's factor out a negative e to the minus st.

That is equal to what?

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So what is this first part going to look like?

From 0 to infinity.

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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.

Castries

Let me just rewrite this.

So this is u prime.

So that makes this 1 over s, sine of at.

This s is just a constant, so we can bring it out.

That would all cancel out, and you'd just be left with e to the minus st.

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Anyway, integral from 0 to infinity.

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 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 we can still say, from 0 to infinity of uv prime.

I don't know.

And then that is equal to the integral from 0 to infinity.

This is going to be equal to the integral from 0 to infinity, of e to the minus st times t dt.

Maybe it looks something like this.

So let's factor out an e to the minus st.

It's just going to be 1 this entire time, so our integral simplifies to the definite integral from t is equal to c to t is equal to infinity of e to the minus st times f of t minus is c dt.

If we take the derivative of this, minus s divided by minus s cancels out, and you just get that.

Integration by parts. uv.

Now let's distribute this.

You know, we could 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 st, dt, which is equal to the antiderivative of e to the minus st, which is minus 1 over s, e to the minus st.

This is f of t. e to the minus st times f of t dt.

Let me just say what this is, first of all.

So that's the improper integral from 0 to infinity of e to the minus st times 1 here.

So we get y is equal to minus e to the minus st over s, sine of at.

So this is our v prime, in which case our v is just the antiderivative of that.

That is equal to c1e to the minus st, f of t, plus c2e to the minus st, g of t, and all of that times dt.

So it's s squared plus a squared, over s squared, y is equal to minus e to the minus st, times this whole thing, sine of at, minus 1 over s squared, cosine of at.

So this is going to be equal to c1 times the integral from 0 to infinity of e to the minus st, times f of t, d of t, plus c2 times the integral from 0 to infinity of e to the minus st, g of t, dt.

Let me do it in that color.

So if we bring the minus 1 s out, this becomes plus 1 s times the integral from 0 to infinity of e to the minus st, dt.

And let's see if we can simplify this.

So let's make t is equal to our u and let's make e to the minus st as being our v prime.

I know I haven't actually done improper integrals just yet, but I'll explain them in a few seconds.