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Let's keep doing some Laplace transforms. For one, it's good

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to see where a lot of those Laplace transform tables

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you'll see later on actually come from, and it just makes

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you comfortable with the mathematics.

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Which is really just kind of your second semester calculus

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mathematics, but it makes you comfortable with the whole

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notion of what we're doing.

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So first of all, let me just rewrite the definition of the

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Laplace transform.

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So it's the L from Laverne & Shirley.

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So the Laplace transform of some function of t is equal to

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the improper integral from 0 to infinity of e to the minus

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st times our function.

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Times our function of t, and that's with respect to dt.

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So let's do another Laplace transform.

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Let's say that we want to take the Laplace transform-- and

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now our function f of t, let's say it is e to the at.

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Laplace transform of e to the at.

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Well we just substituted it into this definition of the

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Laplace transform.

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And this is all going to be really good integration

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practice for us.

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Especially integration by parts.

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Almost every Laplace transform problem turns into an

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integration by parts problem.

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Which, as we learned long ago, integration by parts is just

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the reverse product rule.

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So anyway.

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This is equal to the integral from 0 to infinity.

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e to the minus st times e to the at, right?

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That's our f of t.

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dt.

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Well this is equal to just adding the exponents because

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we have the same base.

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The integral from 0 to infinity of e

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to the a minus stdt.

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And what's the antiderivative of this?

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Well that's equal to what?

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With respect to C.

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So it's equal to-- a minus s, that's just going to be a

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constant, right?

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So we can just leave it out on the outside.

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1/a minus s times e to the a minus st. And we're going to

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evaluate that from t is equal to infinity or the limit as t

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approaches infinity to t is equal to 0.

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And I could have put this inside the brackets, but it's

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just a constant term, right?

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None of them have t's in them, so I can just pull them out.

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And so this is equal to 1/a minus s times-- now we

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essentially have to evaluate t at infinity.

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So what is the limit at infinity?

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Well we have two cases here, right?

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If this exponent-- if this a minus s is a positive number,

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if a minus s is greater than 0, what's going to happen?

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Well as we approach infinity, e to the infinity just gets

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bigger and bigger and bigger, right?

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Because it's e to an infinitely positive exponent.

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So we don't get an answer.

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And when you do improper integrals, when you take the

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limit to infinity and it doesn't come to a finite

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number, the limit doesn't approach anything, that means

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that k the improper integral diverges.

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And so there is no limit.

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And to some degree, we can say that the Laplace transform is

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not defined with a minus s is greater than 0 or when a is

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greater than s.

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Now what happens if a minus s is less than 0?

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Well then this is going to be some negative

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number here, right?

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And then if we take e to an infinitely negative number,

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well then that does approach something.

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That approaches 0.

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And we saw that in the previous video.

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And I hope you understand what I'm saying, right?

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e to an infinity negative number approaches 0, while e

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to an infinitely positive number is just infinity.

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So that doesn't really converge on anything.

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So anyway.

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If I assumed that a minus s is less than 0, or a is less than

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s, and this is the assumption I will make, just so that this

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improper integral actually converges to something.

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So if a minus s is less than 0, and this is a negative

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number, e to the a minus s times-- well t, where t

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approaches infinity will be 0.

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Minus this integral evaluated at 0.

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So when you value this at 0, what happens?

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T equals 0.

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This whole thing becomes e to the 0 is 1.

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And we are left with what?

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Minus 1/a minus s.

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And that's just the same thing as 1/s s minus a.

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So we have our next entry in our Laplace transform table.

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And that is the Laplace transform.

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The Laplace transform of e to the at is equal to 1/s s minus

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a, as long as we make the assumption that s is

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greater than a.

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This is true when s is greater than a, or a is less than s.

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You could view it either way.

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So that's our second entry in our Laplace transform table.

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Fascinating.

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And actually, let's relate this to our previous entry in

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our Laplace transform table, right?

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What was our first entry in our Laplace transform table?

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It was Laplace transform of 1 is equal to 1/s, right?

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Well isn't 1 just the same thing as e to the 0?

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So we could have said that this is the Laplace-- I know

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I'm running out of space, but I'll do it here in purple.

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We could have said Laplace transform of 1 is the same

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thing as e to the 0 times t, right?

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And that equals 1/s.

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And luckily it's good to see that that is consistent.

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And actually, remember, we even made the condition when s

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is greater than 0, right?

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We assumed that s is greater than 0 this example.

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Here again, you say s is greater than 0.

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This is completely consistent with this one, right?

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Because if a is equal to 0, then the Laplace transform of

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e to the 0 is just 1/s minus 0.

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That's just 1/s.

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And we have to assume that s is greater than zero.

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So really these are kind of the same entry in our Laplace

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transform table.

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But it's always nice in mathematics when we see that

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two results we got in trying to do slightly different

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problems actually are, in some ways,

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connected or the same result.

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Anyway I'll see you in the next video and we'll keep

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trying to build our table of Laplace transforms. And maybe

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three or four videos from now I'll actually show you how

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these transforms are extremely useful in solving all sorts of

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differential equations.

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See you soon.

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