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.