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# Laplace transform 2

Video transcript

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.