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- Let's keep doing some Laplace transforms. For one, it's good
- to see where a lot of those Laplace transform tables
- you'll see later on actually come from, and it just makes
- you comfortable with the mathematics.
- Which is really just kind of your second semester calculus
- mathematics, but it makes you comfortable with the whole
- notion of what we're doing.
- So first of all, let me just rewrite the definition of the
- Laplace transform.
- So it's the L from Laverne & Shirley.
- So the Laplace transform of some function of t is equal to
- the improper integral from 0 to infinity of e to the minus
- st times our function.
- Times our function of t, and that's with respect to dt.
- So let's do another Laplace transform.
- Let's say that we want to take the Laplace transform-- and
- now our function f of t, let's say it is e to the at.
- Laplace transform of e to the at.
- Well we just substituted it into this definition of the
- Laplace transform.
- And this is all going to be really good integration
- practice for us.
- Especially integration by parts.
- Almost every Laplace transform problem turns into an
- integration by parts problem.
- Which, as we learned long ago, integration by parts is just
- the reverse product rule.
- So anyway.
- This is equal to the integral from 0 to infinity.
- e to the minus st times e to the at, right?
- That's our f of t.
- dt.
- Well this is equal to just adding the exponents because
- we have the same base.
- The integral from 0 to infinity of e
- to the a minus stdt.
- And what's the antiderivative of this?
- Well that's equal to what?
- With respect to C.
- So it's equal to-- a minus s, that's just going to be a
- constant, right?
- So we can just leave it out on the outside.
- 1/a minus s times e to the a minus st. And we're going to
- evaluate that from t is equal to infinity or the limit as t
- approaches infinity to t is equal to 0.
- And I could have put this inside the brackets, but it's
- just a constant term, right?
- None of them have t's in them, so I can just pull them out.
- And so this is equal to 1/a minus s times-- now we
- essentially have to evaluate t at infinity.
- So what is the limit at infinity?
- Well we have two cases here, right?
- If this exponent-- if this a minus s is a positive number,
- if a minus s is greater than 0, what's going to happen?
- Well as we approach infinity, e to the infinity just gets
- bigger and bigger and bigger, right?
- Because it's e to an infinitely positive exponent.
- So we don't get an answer.
- And when you do improper integrals, when you take the
- limit to infinity and it doesn't come to a finite
- number, the limit doesn't approach anything, that means
- that k the improper integral diverges.
- And so there is no limit.
- And to some degree, we can say that the Laplace transform is
- not defined with a minus s is greater than 0 or when a is
- greater than s.
- Now what happens if a minus s is less than 0?
- Well then this is going to be some negative
- number here, right?
- And then if we take e to an infinitely negative number,
- well then that does approach something.
- That approaches 0.
- And we saw that in the previous video.
- And I hope you understand what I'm saying, right?
- e to an infinity negative number approaches 0, while e
- to an infinitely positive number is just infinity.
- So that doesn't really converge on anything.
- So anyway.
- If I assumed that a minus s is less than 0, or a is less than
- s, and this is the assumption I will make, just so that this
- improper integral actually converges to something.
- So if a minus s is less than 0, and this is a negative
- number, e to the a minus s times-- well t, where t
- approaches infinity will be 0.
- Minus this integral evaluated at 0.
- So when you value this at 0, what happens?
- T equals 0.
- This whole thing becomes e to the 0 is 1.
- And we are left with what?
- Minus 1/a minus s.
- And that's just the same thing as 1/s s minus a.
- So we have our next entry in our Laplace transform table.
- And that is the Laplace transform.
- The Laplace transform of e to the at is equal to 1/s s minus
- a, as long as we make the assumption that s is
- greater than a.
- This is true when s is greater than a, or a is less than s.
- You could view it either way.
- So that's our second entry in our Laplace transform table.
- Fascinating.
- And actually, let's relate this to our previous entry in
- our Laplace transform table, right?
- What was our first entry in our Laplace transform table?
- It was Laplace transform of 1 is equal to 1/s, right?
- Well isn't 1 just the same thing as e to the 0?
- So we could have said that this is the Laplace-- I know
- I'm running out of space, but I'll do it here in purple.
- We could have said Laplace transform of 1 is the same
- thing as e to the 0 times t, right?
- And that equals 1/s.
- And luckily it's good to see that that is consistent.
- And actually, remember, we even made the condition when s
- is greater than 0, right?
- We assumed that s is greater than 0 this example.
- Here again, you say s is greater than 0.
- This is completely consistent with this one, right?
- Because if a is equal to 0, then the Laplace transform of
- e to the 0 is just 1/s minus 0.
- That's just 1/s.
- And we have to assume that s is greater than zero.
- So really these are kind of the same entry in our Laplace
- transform table.
- But it's always nice in mathematics when we see that
- two results we got in trying to do slightly different
- problems actually are, in some ways,
- connected or the same result.
- Anyway I'll see you in the next video and we'll keep
- trying to build our table of Laplace transforms. And maybe
- three or four videos from now I'll actually show you how
- these transforms are extremely useful in solving all sorts of
- differential equations.
- See you soon.