Today, and for the next two
weeks, we are going to be studying what,
for many engineers and a few scientists is the most popular
method of solving any differential equation of the
kind that they happen to be, and that is to use the popular
machine called the Laplace transform.
Now, you will get proficient in using it by the end of the two
weeks. But, there is always a certain
amount of mystery that hangs around it.
People scratch their heads and can't figure out where it comes
from. And, that bothers them a lot.
In the past, I've usually promised to tell
you, the students at the end of the two weeks,
but I almost never have time. So, I'm going to break that
glorious tradition and tell you up front at the beginning,
where it comes from, and then talk very fast for the
rest of the period. Okay, a good way of thinking of
where the Laplace transform comes from, and a way which I
think dispels some of its mystery is by thinking of power
series. I think virtually all of you
have studied power series except possibly a few students who just
had 18.01 here last semester, and probably shouldn't be
taking 18.03 anyway, now.
But anyway, a power series looks like this:
summation (a)n x to the n. And, you sum that from, let's say, zero to infinity.
And, the typical thing you want to do with it is add it up to
find out what its sum is. Now, the only way I will depart
from tradition, instead of calling the sum some
generic name like f of x, in order to identify
the sum with the coefficients, a, I'll call it a of x. Now, I want to make just one
slight change in that. I want to use computer
notation, which doesn't use the subscript (a)n.
Instead, this, it thinks of as a function of
the discreet variable, n.
In other words, it's a function which assigns
to n equals zero, one, two, three real numbers.
That's what this sequence of coefficients really is.
So, the computer notation will look almost the same.
It's just that I will write this in functional notation as a
of n instead of (a)n. But, it still means the real
number associated with the positive integer,
n, and everything else is the same.
See, what I'm thinking of this as doing is taking this discreet
function, which gives me the sequence of coefficients of the
power series, and associating that with the
sum of the power series. Let me give you some very
simple examples, two very simple examples,
which I think you know. Suppose this is a function one.
Now, what do I mean by that? I mean it's the constant
function, one. To every positive integer,
it assigns the number one. Okay, what's a of x?
What I'm saying is, in other words,
in this fancy, mystifying form,
is all of these guys are one, what's a of x?
One plus x plus x squared plus x cubed.
Look, you are supposed to be born knowing what that adds up
to. It adds up to one over one
minus x, except that's the wrong answer.
What's wrong about it? It's not true for every value
of x. That's only true when x is such
that that series converges, and that is only true when x
lies between negative one and one.
So, it's not this function. It's this function with its
domain restricted to be less than one in absolute value.
What does that converge to? If x is bigger than one,
the answer is it doesn't converge.
There's nothing else you can put here.
Okay, let's take another function.
Suppose this is, let's see, one over n
you probably won't know. Let's take one you will know,
one over n factorial. Suppose a of n is the function one over n factorial, what's a of x?
So, what I'm asking is, what does this add up to when
the coefficient here is one over n factorial? What's summation x to the n
over n factorial? It is e to the x. And, this doesn't have to be
qualified because this is true for all values of x.
So, in other words, from this peculiar point of
view, I think of a power as summing the operation,
of summing a power series as taking a discreet function
defined for positive integers, or nonnegative integers,
and doing this funny process. And, out of it comes a
continuous function of some sort.
And, notice what goes in is the variable, n.
But, what comes out is the variable, x.
Well, that's perfectly natural. That's the way a power series
is set up. So, the question I ask is,
this is a discreet situation, a discreet summation.
Suppose I made the summation continuous instead of discreet.
So, I want the continuous analog of what I did over there.
Okay, what would a continuous analog be?
Well, instead of, I'll replace n zero,
one, two, that will be replaced by a continued,
that's a discreet variable. I'll replace it by a continuous
variable, t, which runs from zero to infinity,
and is allowed to take every real value in between instead of
being only allowed to take the values of the positive
nonnegative integers. Okay, well, if I want to use t
instead of n, I clearly cannot sum in the
usual way over all real numbers. But, the way the procedure
which replaces summation over all real numbers is integration.
So, what I'm going to do is replace that sum by the integral
from zero to infinity. That's like the sum from zero
to infinity of what? Well, of some function,
but now n is being replaced by the continuous variable,
t. So, this is going to be a
function of t. And, how about the rest of it?
The rest I will just copy, x to the n'th.
Well, instead of n I have to write t and dt.
And, what's the sum? Well, I'll call the sum,
what's the sum a function of? I integrate out the t.
So, that doesn't appear in the answer.
All that appears is this number, x, this parameter,
x. For each value of x,
like one, two, or 26.3, this integral has a
certain value, and I can calculate it.
So, this is going to end up as a function of x,
just as it did before. Now, I could leave it in that
form, but no mathematician would like to do that,
and very few engineers either. The reason is,
in general, when you do integration and differentiation,
you do not want to have as the base of an exponential something
like x. The only convenient thing to
have is e, and the reason is because it's only e that people
really like to differentiate, e to the something.
The only thing is that people really like to differentiate or
integrate. So, I'm going to make this look
a little better by converting x to the t to the base e.
I remember how to do that. You write x equals e to the log
x and so x to the t will be e to the
log x times t, if you want.
Now, the only problem is I want to make one more little change.
After all, I want to be able to calculate this integral.
And, it's clear that if t is going to infinity,
if I have a number here, for example,
like x equals two, that integral is really quite
unlikely to converge. For example,
if a of t were just the constant function,
one, the integral certainly wouldn't converge.
It would be horrible. That integral only has a chance
of converging if x is a number less than one,
so that when I take bigger and bigger powers of it,
I get smaller and smaller numbers.
Don't forget, this is an improper integral
going all the way up to infinity.
Those need treatment, delicate handling.
All right, so I really want x to be less than one.
Otherwise, that integral is very unlikely to converge.
I'd better have it positive, because if I allow it to be
negative I'm going to get into trouble with negative powers,
see what's minus one, for example,
to the one half when t is one half.
That's already imaginary. I don't want that.
If you've got an exponential, the base has got to be a
positive number. So, I want x to be a positive
number. All right, if x in my actual
practices going to lie between zero and one in order to make
the integral converge, how about log x?
Well, log x, if x is less than one,
so log x is going to be less than zero,
and it's going to go all the way down to negative infinity.
So, this means log x is negative.
In this interesting range of x, the log x is always going to be
negative. And now, I don't like that.
The first place I'd like to call this by a new variable
since no one uses log x as a variable.
And, it would make sense to make it a negative,
to make it negative, that is, to write log x is
equal to negative s. Let's put it on the other side,
in order that since log x is always going to be less than
zero, then s will always be positive.
And it's always more convenient to work with positive numbers
instead of negative numbers. So, if I make those changes,
what happens to the integral? Well, I stress,
all these changes are just cosmetic to make things a little
easier to work with in terms of symbols.
First of all, the a I'm going to change.
I don't want to call it a of t because most people
don't call functions a of t. They call them f of t.
So, I'll call it f of t. x is e to the log x,
which is e to the minus s. So, x has its name changed to e to the minus s.
In other words, I'm using as the new variable
not x any longer but s in order that the base be e.
t, I now raise this to the t'th power, but by the laws of
exponents, that means I simply multiply the exponent by t,
and dt. And now, since I'm calling the
function f of t, the output ought to be called
capital F. But it's now a function,
since I've changed the variable, of s.
It's no longer a function of x. If you like,
you may think of this as a of, what's x?
x is e to the negative s, I guess.
I mean, no one would leave a function in that form.
It's simply a function of s. And, what is that?
So, what have we got, finally?
What we have, dear hearts,
is this thing, which I stress is nothing more
than the continuous analog of the summation of a power series.
This is the discrete version. This is by these perfectly
natural transformations the continuous version of the same
thing. It starts with a function
defined for positive values of t, and turns it into a function
of s. And, this is called the Laplace
transform. Now, if I've done my work
correctly, you should all be saying, oh, is that all?
But, I know you aren't. So, it's okay.
You'll get used to it. The first thing you have to get
used to is one thing some people never get used to,
which is you put in a function of t, and you get out a function
of s. How could that be?
You know, for an operator, you put in 3x,
and you get out three if it's a differentiation operator.
In other words, when you have an operator,
the things we've been talking about the last two or three
weeks in one form or another, at least the variable doesn't
get changed. Well, but for a transform it
does, and that's why it's called a transform.
So, the difference between a transform and an operator is
that for a transform a function of t comes in,
but a function of s comes out. The variable gets changed,
whereas for an operator, f of t goes in and
what comes out is g of t, a function using the
same variable like differentiation is a typical
example of an operator, or the linear differential
operators we've been talking about.
Well, but this doesn't behave that way.
The variable does get changed. That's, in fact,
extremely important in the applications.
In the applications, t usually means the time,
and s very often, not always, but very often is a
variable measuring frequency, for instance.
But, so that's a peculiar thing that's hard to get used to.
But, a good thing is the fact that it's a linear transform.
In other words, it obeys the laws we'd love and
like that the Laplace transform-- oh,
I never gave you any notation for the laplace transform.
Hey, I'd better do that. Okay, so, some notation:
there are two notations that are used.
Your book mostly uses the notation that the laplace
transform of f of t is capital F of s,
uses the same letter but with the same capital.
Now, as you will see, there are some places you
absolutely cannot use that notation.
It may seem strange, looks perfectly natural.
There are certain laws you cannot express using that
notation. It's baffling.
But, if you can't do it this way, you can do it using this
notation instead. One or the other will almost
always work. So, I'll use my little squiggly
notation, but that's what I use. I think it's a little more
vivid, and the trouble is that this piles up too many
parentheses. And, that's always hard to
read. So, I like this better.
So, these are two alternate ways of saying the same thing.
The Laplace transform of this function is that one.
Okay, well, let's use, for the linearity law,
it's definitely best. I really cannot express the
linearity law using the second notation, but using the first
notation, it's a breeze. The Laplace transform of the
sum of two functions is the sum of their Laplace transforms of
each of them separately. Or, better yet,
you could write it that way. Let's write it this way.
That way, it looks more like an operator, L of f plus L of g. And, of the same way,
if you take a function and multiply it by a constant and
take the laplace transform, you can pull the constant
outside. And, of course,
why are these true? These are true just because of
the form of the transform. If I add up f and g,
I simply add up the two corresponding integrals.
In other words, I'm using the fact that the
integral, this definite integral, is itself a linear
operator. Well, that's the general
setting. That's where it comes from,
and that's the notation for it. And, now we have to get to
work. The first thing to do to get
familiar with this is, obviously what we want to do is
say, okay, these were the transforms of some simple
discreet functions. Okay, suppose I put in some
familiar functions, f of t.
What do their Laplace transforms look like?
So, let's do that. So, one of the boards I should
keep stored. Why don't I store on this
board? I'll store on this board the
formulas as we get them. So, let's see,
what should we aim at, first?
Let's first find, and I'll do the calculations on
the sideboard, and we'll see how it works out.
I'm not very sure. In other words,
what's the Laplace transform of the function,
one? Well, there's an even easier
one. What's the Laplace transform of
the function zero? Answer: zero.
Very exciting. What's the Laplace transform of
one? Well, it doesn't turn out the
constant anymore than it turned out to be a constant up there.
Let's calculate it. Now, you can do these
calculations carefully, dotting all the i's,
or pretty carefully, or not carefully at all,
i.e. sloppily.
I'll let you be sloppy after, generally speaking,
you could be sloppy unless the directions tell you to be less
sloppy or to be careful, okay?
So, I'll do one carefully. Let's calculate the Laplace
transform of one carefully. Okay, in the beginning,
you've got nothing to use with the definition.
So, I have to calculate the integral from zero to infinity
of one, that's the f of t times e to the negative s t,
so I don't have to put in the one,
dt. All right, now,
let me remind you, this is an improper integral.
This is just about the first time in the course we've had an
improper integral. But, there are going to be a
lot of them over the next couple of weeks, nothing but.
All right, it's an improper integral.
That means we have to go back to the definition.
If you want to be careful, you have to go back to the
definition of improper integral. So, it's the limit,
as R goes to infinity, of what you get by integrating
only up as far as R. That's a definite integral.
That's a nice Riemann integral. So, this is what I have to
calculate. And, I have to take the limit
as R goes to infinity. Now, how do I calculate that?
Well, this integral is equal to, that's easy.
It's just integrating. Remember that you're
integrating with respect to t. So, s is a parameter.
It's like a constant, in other words.
So, it's e to the minus s t, and when I differentiated, the derivative of this would
have negative s. So, to get rid of that negative
s, so the derivative is e to the minus s t.
You have to put minus s in the denominator.
And now, I'll want to evaluate that between zero and R.
And, what do I get? Well it is at the upper limit.
So, it's e to the minus s times R minus, at the lower limit,
it's t is equal to zero, so whatever s is,
it's one. And that's divided by this
constant up front, negative s. So, the answer is,
it is equal to the limit of, as R goes to infinity,
of e to the negative s R minus one divided by minus s. Now, what's that?
Well, as R goes to infinity, e to the minus 2R,
or minus 5R goes to zero, and the answer is minus one
over minus s. So, that's one over s.
And so, that's our answer. Let's put it up here.
It's one over s, except it isn't.
I made a mistake. Well, not mistake,
a little oversight. What's the oversight?
This is okay. This is okay.
This is okay. This is not okay.
This is okay. But that's not okay.
What's wrong? I did slight a verbal hand.
Maybe some of you have picked it up and were too embarrassed
to correct me, but I said like e to the minus
2R obviously goes to zero, and e to the minus 5R
goes to zero. How about e to the minus minus
3 R? Does that go to zero?
No, that's e to the 3R, which goes to infinity. The only time this goes to zero
is if s is a positive number. Minus s looks like a negative
number, but it's not, if s is equal to minus two.
So, this is only true if s is positive because only if s is
positive is this exponent really negative and large,
and therefore going to infinity, going to zero as R
goes to infinity. So, the answer is not one over
s. It is one over s,
s must positive. Now, once again,
here, people don't worry about this sort of thing with power
series because it seems very obvious, you know,
one over x, absolute value of x is less
than one, when it gets to be the Laplace
transform, just because the Laplace transform is mysterious,
the question is, okay, the Laplace transform is
one over s of one, well, Laplace transform of one
I understand is one over s if s is positive.
What is it if s is negative? Okay, right down in your little
books, this, but that down, what is it if s is negative,
and write underneath that, this question is meaningless.
It doesn't mean anything. I'll draw you a picture.
This is a picture of the Laplace transform of one.
It is that. It's one branch of this curve.
It does not include the branch on the left.
It doesn't because I showed you it doesn't.
That's all there is to it. Okay, so I did that carefully.
Now I'm going to get a little less careful.
What's the Laplace transform of e to the a t?
First of all, in general, the kind of
functions for which people like to calculate the Laplace
transform, and basically the only ones there will be in the
tables are exactly the sort of functions that you used in
solving linear equations with constant coefficients.
What kinds of functions entered in there?
Exponentials, sines and cosines,
but they were really complex exponentials,
right? e to the t sine t,
but that was really a complex exponential,
too, just a little more complicated one,
polynomials, and that's about it.
t times e to the t, that was okay,
too. These are the functions for
which people calculate the Laplace transform,
and all the other functions they don't calculate the Laplace
transforms. So, I don't mean to disappoint
you here. You're going to say,
oh, what, that same old stuff? For two more weeks,
we've got that same, well, the Laplace transform
does a lot of things much better than the methods we've been
using. And, I won't.
I'll sell it when I get a chance to, for now,
let's just get familiar with it.
All right, so while I'm not going to calculate e to the a t
for you, because I'd like instead to
just prove a simple formula which will just give that,
and will also give us e to the a t sine t.
It will give us a lot more, instead.
I'm going to calculate a formula for the Laplace
transform of this guy if you already know the Laplace
transform of it. Now, see, this falls in that
category because this is really e to the a t times one. But, I already know the Laplace
transform of one. So that's, if I can get a
general formula for this, I'll be able to get the formula
for e to the a t as a consequence.
So, let's look for this Laplace transform.
Now, it's really easy. Let's see, where am I doing
calculations? Over here.
Okay, so we've got e. So, I want to calculate the
Laplace transform e to the a t f of t.
So I'm going to say that's the integral from zero to infinity
of e to the a t times f of t. And now, the rest I copy. That's the function part of it
that goes to the input, and then there's the other
part. This part is called the kernel,
by the way, but don't worry about that.
However, if you drop it in conversation,
people will look at you and say, gee, they know something I
don't. And you will.
You know that it's the kernel. Okay, well, now,
what kind of formula can I be looking for?
Clearly, I can only be looking for a formula which expresses it
in terms of the Laplace transform of f of t.
Let's calculate and see what we get.
Now, what would you do to that thing to make?
Well, obviously, the thing to do is to combine
the two exponentials. So, that's going to be the
integral from zero to infinity of f of t.
e, now, I'd like to put it, to combine the exponentials in
such a way that it has, still, that same form,
so, I'm going to begin with that negative sign,
and then see what the rest of it has to be.
What is it going to be? Well, minus s t and
plus a t, but I can make that minus a
here, and it will come out right.
So, it's minus s t plus a t, and there are the two parts, those two factors,
dt. So, what's that?
That's the Laplace transform. If the a weren't there,
this would be the Laplace transform of f of t.
What is it with the a there? It's the Laplace transform of f
of t, except that instead of the
variable, s has been replaced by the variable s minus a. I'll give you a second to
digest that. Well, you digest it while I'm
writing it because that's the answer.
And, the way this is most often used, I have to qualify it for
the value. So, if F of s is good
for s positive, the way it would be,
for example, if I used the function one
here, then to finish that off, then, F of s minus a will be, this will be good when s is
bigger than a. Why is that?
Well, because this is true. This is true.
If s minus a is positive, that's the condition.
That's what this Laplace transform is good.
But that simply says that s should be bigger than a. And, since this doesn't look
pretty, let me try to make it look a little bit prettier.
So, let's write it. So, this is assuming F of s is
for s greater than zero. Now, this is called something. This is called,
well, what would you call it? On the left side,
you multiply by an exponential. On the right,
you translate. You shift the argument over by
a. So, this is called,
gulp, the exponential shift. What?
Well, I'll call it the formula. The thing before,
when we talked about operators, we called it the exponential
shift rule or the exponential shift law.
But, in fact, this is, in a way,
a disguised form of the same law.
And, engineers who typically do all their work using the Laplace
transform and don't use operators, this is the form of
the exponential shift law that they would know.
What you can do with one, you can do with the other.
You can now use both. So, what's the answer to e to
the a t? Well, the answer is,
I'm supposed to, e to the a t times one,
the Laplace transform of one is one over s.
And, therefore, what I do is to multiply by e
to the a t, I change s to s minus a .
And so, that's the answer. Let's see, what else don't we
know? Well, how about sines and
cosines? Well, the way to do sines and
cosines is by making the observation that this formula
also works when a is a complex number.
So, can use also for a a complex number, for e to
the a plus b i times t. The Laplace transform of e to the a plus b i times t is one
over s minus a plus b i. And again, it will be for s
bigger than a. So, let's calculate the Laplace
transform of, let's say, well,
I've got to cover up something. Okay, so, that's the Laplace
transform. I've got to remember that.
So, let's calculate the Laplace transform of,
let's say, sine of a t and cosine a t. What do you get for that?
Well, just for a little variety, we could do it by using
that formula, and taking its real and
imaginary parts. Since some of you had so much
difficulty with the backwards Euler formula,
he is a good case where you could use it.
Suppose you want to calculate the Laplace transform of cosine
a t. Well, I'm going to write that
using, I want to calculate using complex exponentials.
The way I will do it is by using the backwards Euler
formula. So, this is e to the i a t plus
e to the minus i a t divided by two. Remember, the foreword Euler
formula would say e to the i a t equals cosine a t plus i sine a
t. That expresses the complex exponential in terms of sines
and cosines. This is the backward formula,
which just read it backwards, expressing cosines and sines in
terms of complex exponentials instead.
Both formulas are useful, almost equally useful,
in fact. And anyway, just remind you of
it, let's use this one. Okay, what's the Laplace
transform, then, of cosine a t?
Well, by linearity, it's equal to one half the
Laplace transform of this guy plus the Laplace transform of
that guy. And, what are those?
Well, the Laplace transform of e to the i a t is one over s
minus i a, and the Laplace
transform of the other guy is one divided by s plus i a. Now, of course,
this has become out to be a real function.
This is real. Every integral is real.
This must come out to be real. This looks kind of complex,
but it isn't. I know automatically that this
is going to be a real function. How I know that?
Well, mentally, you can combine the terms and
calculate. But, I know even before that.
Remember, there are two ways to see that something is real.
You can calculate it and see that its imaginary part is zero,
hack, or without any calculation, if you change i to
minus i, and you get the same thing,
it must be real. Now, if I change i to minus i
in this expression, what happens?
If I change i to minus i, this term turns into that one,
and this one turns into that one.
Conclusion: the sum of the two is unchanged.
And therefore, this is real.
Well, of course, in the time I took to make that
argument, I could have actually calculated it.
So, what the heck, let's calculate it?
So, you do the high school thing, and it's this guy plus
that guy on top, which makes 2s.
I on the bottom is the product of those, which by now you
should know the product of two complex numbers.
A product of a number and its complex conjugate is the sum of
the squares. So, what's the answer?
The twos cancel, and the answer is that the
Laplace transform of cosine a t is s over s squared plus a
squared. And, that will be true as, in general, it's true up there
for positive values of s only. And, the sine a t,
you can calculate that in recitation tomorrow.
The answer to that is a divided by s squared plus a squared. You would get the same answers
if you took the real and imaginary parts of that
expression. It's another way of getting at
the recitations tomorrow; we'll get practice in
calculating other functions related to these by using these
formulas, and also from scratch directly from the definition of
the Laplace transform. Well, there are two things
which we still should do. The first is I want to get you
started with calculating inverse Laplace transforms.
And, the reason for doing that is, in other words,
I've started with f of t, and we've been focusing
on what is capital F of s? But, you will find that when you go to solve differential
equations, by far, the hardest part of the
procedure is you get F of s. The Laplace transform of the
answer, and you have to convert that back into the answer in
terms of t that you were looking for.
In other words, the main step in the procedure
that you are going to be using for solving differential
equations is, and the hardest part of the
step will be to calculate inverse laplace transforms.
Now, you think that could be done by tables,
but, in fact, it can't unless the tables are
too long to be useful. You have to do a certain amount
of work yourself. And, the certain amount of work
that you have to do yourself involves partial fractions
decompositions. And, in case you were wondering
which you are not, the reason you learned partial
fractions in 18.01 was not to learn those silly integrals,
but he learned it so that when you got to 18.03 you would be
able to calculate, solve differential equations by
using Laplace transforms. Sorry.
That's life. Now, so a certain amount of the
recitation time tomorrow will be devoted to reminding you how to
do partial fractions since you haven't done it in a while,
and I assume, yeah, we had that,
I think. Okay, now, they also remind you
of the most efficient method, which about half of you have
had, and the rest think you might have had,
but really aren't sure. So, here's the answer.
We want to find out what it's inverse Laplace transform is.
What you have to do, it normally won't be in the
tables like this. You have to put it in a form in
which it will be in the tables. As you do that,
you have to make partial fractions decompositions,
which, to do it quickly, so if you don't know what I'm
doing now, or you think you once knew but don't quite remember,
go to recitation tomorrow. To get the coefficient here,
I cover up s, and I put s equals zero
because that's the law. To get this coefficient,
I cover up s plus three and I put s equals a negative three
because that's what you're supposed to do.
Put s equal negative three, you get minus one third.
This is equal to that. In this form,
I don't know what the inverse Laplace form is,
but in this form, I certainly do know with the
inverse Laplace transform because the inverse Laplace
transform is linear, and because each of these guys
especially occurs in those tables.
Well, what's this? Well, it's whatever the Laplace
transform of, inverse Laplace transform of
one over s is multiplied by one third.
Well, the inverse Laplace transform of one over s is one.
So, it's one third times one. How about the other guy?
Minus one third, the inverse Laplace transform
of one over s plus three, that's this formula.
a is negative three, and that makes e to the minus
3t. So, if this was the Laplace
transform of the solution to the differential equation,
then the solution in terms of t was this function.
Now, you'll get lots of practice in that.
All I'm doing now is signaling that that's the most important
and difficult step of the procedure, and that,
please, start getting practice. Get up to snuff doing that
procedure. Okay, in the time remaining,
I want to add one formula to this list, and that is going to
be the Laplace transform of, we still haven't done
polynomials. And now, to polynomials,
because the Laplace transform is linear, all I have to do is
know what the Laplace transform of, the individual term of a
polynomial. In other words,
what the Laplace transform of t to the n,
where n is some positive integer?
Well, let's bravely start trying to calculate it.
Integral from zero to infinity t to the n e to the negative st
dt. Now, I think you can see that the method you should use is
integration by part because this is a product of two things,
one of which you would like to differentiate a lot of times,
in fact, and the other won't hurt to integrate it because
it's very easy to integrate. So, this factor is going to be the one that's to be
differentiated, and this is the factor that
will be pleased to integrate it. Let's get started and see what
we can get out of it. Well, this time I'm going to
be, well, I'd better be a little careful because there's a point
here that's tricky. Okay, the first step of
integration by parts is you only do the integration.
You don't do the differentiation.
Remember, the variable is t. The s is just a parameter.
It's just a constant. It's hanging around,
not knowing what to do. Okay, so the first step is you
don't do the differentiation. You only do the integration.
Evaluate it between limits, and then you put a minus sign
before you forget to do it. And then, integral zero to
infinity. Now you do both operations.
So, it's n t to the n minus one,
and you also do the integration.
Okay, let's consider each of these pieces in turn.
Now, this piece, well, there's no problem with
the lower limit, zero, because when t is equal
to zero, this factor is zero, and the thing disappears as
long as n is one or higher. So, it's minus zero here at the
lower limit. The question is,
what is at the upper limit? So, what I have to do is find
out, what is the limit? The limit, as t goes to
infinity, that's what's happening up there,
of t to the n times e to the negative s t divided
by minus s. Well, as t goes to infinity,
this goes to infinity, of course.
This had better go to zero unless I want an answer,
infinity, which won't do me any good.
If this goes to zero, s had better be positive.
So, I'd better be restricting myself to that case.
Okay, so let's assume that s is positive so that this minus s
really is a negative number. Okay, then I have a chance.
So, this is going to be the limit.
Let's write it in a more familiar form with that down
below. So, it's t to the n.
That's going to infinity. But, the bottom is e to the
minus s t. But now, it's plus s t.
And, that's going to infinity, too, because s is positive.
So, the two guys are racing, and the question is,
oh, I lost a minus s here. So, oh... equals minus
one over s. How's that?
So, the question is only, which guy wins?
In the race to infinity, which one wins,
and how do you decide? And, the answer,
of course, is that's the bottom that wins.
The exponential always wins, and it's because of L'Hopital's
rule. You differentiate top and
bottom. Nothing much happens to the
bottom. It gets another factor of s,
but the top goes down to t to the n minus one.
L'Hopital it again, and again, and again,
and again, and again until finally you've reduced the top
to t to the zero where it's defenseless and just sitting
there, and nothing's happened to the bottom.
It's still got e to the s t. and that goes to infinity. So, the answer is,
this is zero by n applications of L'Hopital's rule.
Or, if you're very clever, you can do it in one,
but I won't tell you how. So, the answer is that this is
zero. At the upper limit,
it's also zero at least if s is positive, which is the case
we're considering. That leaves the rest of this.
All right, let's pull the constants out front.
That's plus. Two negatives make a plus.
n over s, now, what's left?
The integral from zero to infinity of t to the n minus
one, e to the minus s t dt. But, what on Earth is that?
That is n over s times the Laplace transform of t to
the n minus one. We got a reduction for it.
We don't get the answer in one step.
But, we get a reduction formula.
And, it says that the Laplace transform, let me write it this
way for once. The first way is now better,
is equal to n over s times the Laplace transform of n minus t
to the n minus one. Okay, the next step, this would be n over s times n
minus one over s times the Laplace transform of t to the n
minus two. If I can continue, I finally get in the top n
times n minus one times all the way down to one divided by the
same number of s's, n of them, times the Laplace
transform of t to the zero, finally.
See, one, zero, n minus one.
And so, what's the final answer?
It is n factorial over s to the what power?
Well, the Laplace transform of this is one over s.
So, the answer is it's s to the n plus one,
n of them here plus an extra one coming from the one over s
here. And, that's the answer.
The Laplace transform of t to the n, oddly enough, is more complicated,
and looks a little different from these.
It's n factorial over s to the n plus one. And, with that,
you can now calculate the Laplace transform of anything in
sight, and tomorrow you will.
