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PROFESSOR: Now, today
we are continuing
with this last unit.
Unit 5, continued.
The informal title of this unit
is Dealing With Infinity.
That's really the extra little
piece that we're putting in to
our discussions of things like
limits and integrals.
To start out with today, I'd
like to recall for you,
L'Hopital's Rule.
And in keeping with the spirit
here, we're just going to do
the infinity / infinity case.
I stated this a little
differently last time, and I
want to state it again today.
Just to make clear what the
hypotheses are and what the
conclusion is.
We start out with, really,
three hypotheses.
Two of them are kind
of obvious.
The three hypotheses are that
f (x) tends to infinity, g (
x) tends to infinity, that's
what it means to be in this
infinity / infinity case.
And then the last assumption
is that f ' ( x) / g ' (x),
tends to a limit, L. And this
is all as x tends to some a.
Some limit a.
And then the conclusion is that
f( x) / g ( x) also tends
to L, as x goes to a.
Now, so that's the way it is.
So it's three limits.
But presumably these are
obvious, and this one is
exactly what we were going
to check anyway.
Gives us this one limit.
So that's the statement.
And then the other little
interesting point here, which
is consistent with this idea of
dealing with infinity, is
that a equals plus or minus
infinity and L equals plus or
minus infinity are OK.
That is, the numbers capital L,
the limit capital L and the
number a can also be infinite.
Now in recitation yesterday,
you should have discussed
something about rates of growth,
which follow from what
I said in lecture last time and
also maybe from some more
detailed discussions that
you had in recitation.
And I'm going to introduce a
notation to compare functions.
Namely, we say that f ( x) is
a lot less than g ( x).
So this means that the limit,
as it goes to infinity, this
tends to 0.
As x goes to infinity,
this would be.
So this is a notation, a new
notation for us. f is a lot
less than g.
And it's meant to be read
only asymptotically.
It's only in the limit
as x goes to
infinity that this happens.
And implicitly here, I'm always
assuming that these are
positive quantities. f
and g are positive.
What you saw in recitation was
that you can make a systematic
comparison of all the standard
functions that we know about.
For example, the ln function
goes to infinity.
But a lot more slowly
than x to a power.
A lot more slowly then e^ x.
A lot more slowly than,
say, e ^ x ^2.
So this one is slow.
This one is moderate.
And this one is fast.
And this one is very
fast. Going to infinity.
Tends to infinity, and
this is of course
as x goes to infinity.
All of them go to infinity, but
at quite different rates.
And, analogous to this, and
today we're going to be doing
this, needing to do this quite
a bit, is rates of decay,
which are more or less the
opposite of rates of growth.
So rates of decay are rates
at which things tend to 0.
So the rate of decay, and for
that I'm just going to take
reciprocals of these numbers.
So 1 / ln x tends to 0.
But rather slowly.
It's much bigger
than 1 / x ^ p.
Oh, I didn't mention that
this exponent p
is meant to be positive.
That's a convention that I'm
using without saying.
I should've told you that.
So think x ^ 1/2, x ^ 1, x ^2,
they're all in this sort of
moderate intermediate range.
And then that, in turn, goes to
0 but much more slowly then
1 / e ^ x, also known
as e^ - x.
And that, in turn, this guy
here goes to 0 incredibly
fast. e ^ - x ^2 vanishes
really, really fast. So this
is a review of the
L'Hopital's Rule.
What we said last time, and the
application of it, which
is to rates of growth and tells
us what these rates of
growth are.
Today, I want to talk about
improper integrals.
And improper integrals, we've
already really seen one or two
of them on your exercises.
And we mention them a
little bit, briefly.
I'm just going to go through
them more carefully and more
systematically now.
And we want to get just exactly
what's going on with
these rates of decay and
their relationship
with improper integrals.
So I need for you to understand
on the spectrum of
the range of functions like
this, which ones are suitable
for integration as x
goes to infinity.
Well, let's start out
with the definition.
The integral from a to infinity
of f(x) dx is, by
definition the limit as n goes
to infinity of the ordinary
definite integral up to some
fixed, finite level.
That's the definition.
And there's a word that we use
here, which is that we say the
integral, so this is terminology
for it, converges
if the limit exists.
And diverges if not.
Well, these are the key
words for today.
So here's the issue that we're
going to be addressing.
Which is whether the limit
exists or not.
In other words, whether
the integral
converges or diverges.
These notions have a geometric
analog, which you should
always be thinking of at
the same time in the
back of your head.
I'll draw a picture of the
function Here it's
starting out at a.
And maybe it's going
down like this.
And it's interpreting
it geometrically.
This would only work
if f is positive.
Then the convergent
case is the case
where the area is finite.
So the total area is finite
under this curve.
And the other case is the
total area is infinite.
I claim that both of these
things are possible.
Although this thing goes on
forever, if you stop it at one
stage, n, then of course
it's a finite number.
But as you go further and
further and further, there's
more and more and more area.
And there are two
possibilities.
Either as you go all the way
out here to infinity, the
total that you get adds
up to a finite total.
Or else, maybe there's
infinitely much.
For instance, if it's a straight
line going across,
there's clearly infinitely
much area underneath.
So we need to do a bunch
of examples.
And that's really our main job
for the day, and to make sure
that we know exactly what
to expect in all cases.
The first example is the
integral from 0 to infinity of
e^ - kx dx.
Where k is going to be
some positive number.
Some positive constant.
This is the most fundamental,
by far, of
the definite integrals.
Improper integrals.
And in order to handle this, the
thing that I need to do is
to check the integral from
0 up to n. e ^ - kx dx.
And since this is an easy
integral to evaluate, we're
going to do it.
It's - 1 / k e ^ - kx, that's
the antiderivative.
Evaluated at 0 and n.
And that, if I plug in these
values, is - 1 / k, e^ - k N.
Minus, and if I evaluate it at
0, I get a (- 1 / k) e^ 0.
So there's the answer.
And now we have to think
about what happens
as n goes to infinity.
So as n goes to infinity, what's
happening is the second
term here stays unchanged.
But the first term is e to
some negative power.
And the exponent is getting
larger and larger.
That's because k is
positive here.
You've definitely got
to pay attention.
Even though I'm doing this with
general variables here,
you've got to pay attention
to signs of things.
Because otherwise you'll always
get the wrong answer.
So you have to pay very
close attention here.
So this is, if you like, e ^
minus infinity in the limit,
which is 0.
And so in the limit, this
thing tends to 0.
And this thing is just
equal to 1 / k.
And so all told, the answer
is 1 / k And that's it.
Now we're going to abbreviate
this a little bit.
This thought process, you're
going to have to go through
every single time you do this.
But after a while you also get
good enough at it that you can
make it a little bit
less cluttered.
So let me show you a shorthand
for this same calculation.
Namely, I write 0 to infinity
e ^ - kx dx.
And that's equal to - 1 / k
e ^ - kx 0 to infinity.
That was cute.
Not small enough, however.
So, here we are.
We have the same calculation
as we had before.
But now we're thinking, really,
in our minds that this
infinity is some very,
very enormous number.
And we're going to plug it in.
And you can either do this
in your head or not.
You say - 1 / k e^ - infinity.
Here's where I've used the
fact that k is positive.
Because e ^ - k times a large
number is minus infinity.
And then here + 1
/ k - (- 1 / k).
Let me write it the same
way I did before.
And that's just equal to 0 + 1
/ k, which is what we want.
So this is the same calculation,
just slightly
abbreviated.
Yeah.
Question.
STUDENT: [INAUDIBLE]
PROFESSOR: Good question.
The question is, what
about the case when
the limit is infinity?
I'm distinguishing between
something existing and its
limit being infinity here.
Whenever I make a discussion
of limits, I say a finite
limit, or in this case, it works
for infinite limits.
So in other words, when
I say exists, I mean
exists and is finite.
So here, when I say that it
converges and I say the limit
exists, what I mean is that
it's a finite number.
And so that's indeed
what I said here.
The total area is finite.
And, similarly, over here.
I might add, however, that
there is another
part of this subject.
Which I'm skipping entirely.
Which is a little bit subtle.
Which is the following.
If f changes sign, there can
be some cancellation and
oscillation.
And then sometimes the limit
exists, but the total area, if
you counted it all positively,
is actually still infinite.
And we're going to
avoid that case.
We're we're just going to treat
these positive cases.
So don't worry about
that for now.
That's the next layer of
complexity which we're not
addressing in this class.
Another question.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is,
would this be OK on tests.
The answer is, absolutely yes.
I want to encourage
you to do this.
If you can think about
it correctly.
The subtle point is
just, you have to
plug in infinity correctly.
Namely, you have to realize
that this only
works if k is positive.
This is the step where you're
plugging in infinity.
And I'm letting you put this
infinity up here as an
endpoint value.
So in fact that's exactly
the theme.
The theme is dealing
with infinity here.
And I want you to be able
to deal with it.
That's my goal.
STUDENT: [INAUDIBLE]
PROFESSOR: OK, so another
question is, so let's be sure
here when the limit exists,
I say it has to be finite.
That means it's finite,
not infinite.
The limit can be 0.
It can also be - 1.
It can be anything.
Doesn't have to be a
positive number.
Other questions.
So we've had our
first example.
And now I just want to add one
physical interpretation here.
This is Example 1,
if you like.
And this is something that was
on your problem set, remember.
That we talked about the
probability, or the number, if
you like, the number of
particles on average that
decay in some radioactive
substance.
Say, in time between 0 and some
capital T. And then that
would be this integral, 0 to
capital T, some total quantity
times this integral here.
This is the typical kind
of radioactive decay
number that one gets.
Now, in the limit, so this is
some number of particles.
If the substance is radioactive,
then in the
limit, we have this.
Which is equal to the total
number of particles.
And that's something that's
going to be important for
normalizing and understanding.
How much does the whole
substance, how many moles do
we have of this stuff.
What is it.
And so this is a number that
is going to come up.
Now, I emphasize that this
notion of T going to infinity
is just an idealization.
We don't really believe that
we're going to wait forever
for this substance to decay.
Nevertheless, as theorists, we
write down this quantity.
And we use it.
All the time.
Furthermore, there's other good
reasons for using it, and
why physicists accept
it immediately.
Even though it's not really
completely physically
realistic ever to let
time go very, very
far into the future.
And the reason is, if you notice
this answer here, look
at how much simpler this number
is, 1 / k, than the
numbers that I got in the
intermediate stages here.
These are all ugly, the
limits are simple.
And this is a theme that
I've been trying to
emphasize all semester.
Namely, that the infinitesimal,
the things that
you get when you do
differentiation, are the
easier formulas.
The algebraic ones, the things
in the process of getting to
the limit, are the ugly ones.
These are the easy ones, these
are the hard ones.
So in fact, infinity is
basically easier than any
finite number.
And a lot of appealing formulas
come from those kinds
of calculations.
Another question.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is,
shouldn't the answer be a?
Well, the answer turns
out to be a / k.
Which means that when you set
up your arithmetic, and you
model this to a collection
of particles.
So you said it should be a.
But that's because you
made an assumption.
Which was that a was the total
number of particles.
But that's just false, right?
This is the total number
of particles.
So therefore, if you want to set
it up, you want set up so
that this number's the total
number of particles.
And that's how you set up a
model is, you do all the
calculations and you see what
it's coming out to be.
And that's why you need to do
this kind of calculation.
OK, so.
The main thing is, you shouldn't
make assumptions
about models.
You have to follow what the
calculations tell you.
They're not lying.
OK, so now.
We carried this out.
There's one other example which
we talked about earlier
in the class.
And I just wanted to
mention it again.
It's probably the most famous
after this one.
Namely, the integral from minus
infinity to infinity of
e^ - x ^2 dx.
Which turns out, amazingly, to
be able to be evaluated.
It turns out to be the
square root of pi.
So this one is also great.
This is the constant which
allows you to compute all
kinds of things in
probability.
So this is a key number
in probability.
It basically is the key to
understanding things like
standard deviation and basically
any other thing in
the subject of probability.
It's also what's driving these
polls that tell you within 4%
accuracy we know that
people are going to
vote this way or that.
So in order to interpret all of
those kinds of things, you
need to know this number.
And this number was only
calculated numerically
starting in the 1700s or so by
people who, actually, by one
guy whose name was de Moivre,
who was selling his services
to various royalty who were
running lotteries.
In those days they ran
lotteries, too.
And he was able to tell them
what the chances were of the
various games.
And he worked out this number.
He realized that this
was the pattern.
Although he didn't know that it
was the square root of pi,
he knew it to sufficient
accuracy that he could tell
them the correct answer to how
much money their lotteries
would make.
And of course we do this
nowadays, too.
In all kinds of ways.
Including slightly more legit
businesses like insurance.
So now, I I'm going to give
you some more examples.
And and the other examples are
much more close to the edge
between infinite and finite.
This distinction between
convergence and divergence.
And let me just, maybe I'll say
one more word about why we
care about this very gross issue
of whether something is
finite or infinite.
When you're talking about
something like this normal
curve here, there's an issue of
how far out you have to go
before you can ignore
the rest.
So we're going to ignore what's
called the tail here.
Somehow you want to know that
this is negligible.
And you want to know how
negligible it is.
And this is the job of a
mathematician, is to know what
finite region you have to
consider and which one you're
going to carefully calculate
numerically.
And then the rest, you're going
to have to take care of
by some theoretical reasoning.
You're going to have to know
that these tails are small
enough that they don't matter
in your finite calculation.
And so, we care very much
about the tails.
Because they're the only
thing that the
machine won't tell us.
So that's the part that
we have to know.
And these tails are also
something which are discussed
all the time in financial
mathematics.
They're very worried
about fat tails.
That is, unlikely events that
nevertheless happen sometimes.
And they get burned fairly
regularly with them.
As they have recently, with
the mortgage scandal.
So, these things are pretty
serious and they really are
spending a lot of
time on them.
Of course, there are lots of
other practical issues besides
just the mathematics.
But you've got to get
the math right, too.
So we're going to now talk about
some borderline cases
for these fat tails.
Just how fat do they have to be
before they become infinite
and overwhelm the
central bump.
So we'll save this for
just a second.
And what I'm saving up here is
the borderline case, which I'm
going to concentrate on, which
is this moderate rate, which
is x to powers.
Here's our next example.
I guess we'll call
this Example 3.
It's the integral from
1 to infinity dx / x.
That's the power p = 1.
And this turns out to be
a borderline case.
So it's worth carrying
out carefully.
Now, again I'm going to do
it by the slower method.
Rather than the shorthand
method.
But ultimately, you can
do it by the short
method if you'd like.
I break it up into an integral
that goes up to some
large number, n.
I see that its logarithm
function is the
antiderivative.
And so what I get is ln n
- ln 1, which is just 0.
So this is just log n.
In any case, it tends
to infinity
as n goes to infinity.
So the conclusion is, since the
limit is infinite, that
this thing diverges.
Now, I'm going to do this
systematically now with all
powers p, to see what happens.
I'll look at the integral.
Sorry, I'm going to have
to start at 1 here.
From 1 to infinity, dx
/ x ^p, and see what
happens with these.
And you'll see that p = 1 is
a borderline when I do this
calculation.
This time I'm going to do the
calculation the hard way.
But now you're going to have to
think and pay attention to
see what it is that I'm doing.
First of all, I'm going to
take the antiderivative.
And this is x ^ - p, so it's
- p + 1 / - p + 1.
That's the antiderivative of
the function 1 / x ^ - p.
And then I have to evaluate
that at 1 and infinity.
So now, I'll write this down.
But I'm going to be particularly
careful here.
I'll write it down.
It's infinity to the -
p + 1 / - p + 1 -,
so I plug in 1 here.
So I get 1 / - p + 1.
So this is what I'm getting.
Again, what you should be
thinking here is this is a
very large number
to this power.
Now, there are two cases.
There are two cases.
And they exactly
split at p = 1.
When p = 1, this number is 0.
This exponent is 0, and in fact
this expression doesn't
make any sense because the
denominator is also 0.
But for all of the
other values, the
denominator makes sense.
But what's going on is that
this is infinite when this
exponent is infinity to
a positive power.
And it's 0 when it's infinity
to a negative power.
So I'm going to say it
here, and you must
check this at home.
Because this is exactly what
I'm going to ask you
about on the exam.
This is it.
This type of thing, maybe with
a specific value of p here.
When p < 1, this thing
is infinite.
On the other hand, when p
> 1, this thing is 0.
So when p > 1, this
thing is 0.
It's just equal to 0.
And so the answer
is 1 / p - 1.
Because that's this number.
Minus the quantity
1 / - p + 1.
This is a finite number here.
Notice that the answer would
be weird if this thing went
away in the p < 1 case.
Then it would be a
negative number.
It would be a very strange
answer to this question.
So, in fact that's
not what happens.
What happens is that the answer
doesn't make sense.
It's infinite.
So let me just write this
down again, under here.
This is a test in a
particular case.
And here's the conclusion.
Ah.
No, I'm sorry.
I think I was going to write
it over on this board here.
So the conclusion is that the
integral from 1 to infinity dx
/ x^p diverges if p <= 1.
And converges if p > 1.
And in fact, we can actually
evaluate it.
It's equal to 1 / p - 1.
It's got a nice, clean
formula even.
Alright, now let
me remind you.
So I didn't spell the word
diverges right, did I?
Oh no, that's an r.
I guess that's right.
Diverges if p <= 1.
So really, I needed both of
these arguments, which are
sitting above it in
order to do it.
Because the second argument
didn't work at all when p = 1
because the formula for the
antiderivative is wrong.
The formula for the
antiderivative is given by the
ln function when p = 1.
So I had to do this
calculation too.
This is the borderline case,
between p > 1 and p < 1.
When p > 1, we got
convergence.
We could calculate
the integral.
When p < 1, when we got
divergence and we calculated
the integral over there.
And here in the borderline case,
we got a logarithm. and
we also got divergence.
So it failed at the edge.
Now, this takes care
of all the powers.
Now, there are a number of
different things that one can
deduce from this.
And let me carry them out.
So this is more or less
the second thing that
you'll want to do.
And I'm going to emphasize
maybe one aspect of it.
I guess we'll get rid of this.
But it's still the issue that
we're discussing here.
Is whether this area
is fat or thin.
I'll remind you of that.
So here's the next idea.
Something called limit
comparison.
Limit comparison is what you're
going to use when,
instead of being able actually
to calculate the number, you
don't yet know what
the number is.
But you can make a comparison to
something whose convergence
properties you already
understand.
Now, here's the statement.
If a function, f, is similar to
a function, asymptotically
the same as a function, g, as
x goes to infinity, I'll
remind you what that
means in a second.
Then the integral starting at
some point out to infinity of
f(x) dx, and the other
one, converge and
diverge at the same time.
So both, either, either
-- sorry, let's try
it the other way.
Either, both.
Either both converge,
or both diverge.
They behave exactly
the same way.
In terms of whether they're
infinite or not.
And, let me remind you what
this tilde means.
This thing means that f(
x) / g ( x) tends to 1.
So if you have a couple of
functions like that, then
their behavior is the same.
This is more or less obvious.
It's just because far enough
out, this is for
large a, if you like.
We're not paying any attention
to what happens.
It just has to do with the tail,
and after a while f ( x)
and g(x) are comparable
to each other.
So their integrals are
comparable to each other.
So let's just do a couple
of examples here.
If you take the integral from 0
to infinity dx / the square
root of x ^2 + 10, then I claim
that the square root of
x^2 + 10 resembles the square
root of x ^2, which is just x.
So this thing is going
to be like.
So now I'm going to have to
do one thing to you here.
Which is, I'm going to
change this to 1.
To infinity. dx /x And the
reason is that this x = 0 is
extraneous.
Doesn't have anything to
do with what's going
on with this problem.
This guy here, the piece of it
from, so we're going to ignore
the part integral from 0 to 1 dx
/ square root of x ^2 + 10,
which is finite anyway.
And unimportant.
Whereas, unfortunately, the
integral of dx will have a
singularity at x = 0.
So we can't make the
comparison there.
Anyway, this one is infinite.
So this is divergence.
Using what I knew from before.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is, why
did we switch from 0 to 1?
So I'm going to say a little
bit more about that later.
But let me just make
it a warning here.
Which is that this guy here is
infinite for other reasons.
Unrelated reasons.
The comparison that we are
trying to make is with the
tail as x goes to infinity.
So another way of saying this
is that I should stick an a
here and an a here and
stay away from 0.
So, say a = 1.
If I make these both 1,
that would be OK.
If I make them both 2,
that would be OK.
If I make them both 100,
that would be OK.
So let's leave it as
100 right now.
And it's acceptable.
I want you to stay away
from the origin here.
Because that's another
bad point.
And just talk about what's
happening with the tail.
So this is a tail, and I
also had a different
name for it up top.
Which is emphasizing this.
Which is limit comparison.
It's only what's happening at
the very end of the picture
that we're interested in.
So again, this is as
x goes to infinity.
That's the limit we're talking
about, the limiting behavior.
And we're trying not to pay
attention to what's happening
for small values of x.
So to be consistent, if I'm
going to do it up to 100 I'm
ignoring what's happening up
to the first 100 values.
In any case, this
guy diverged.
And let me give you
another example.
This one, you could
have computed.
This one you could have
computed, right?
Because it's a square root of
quadratic, so there's a trig
substitution that evaluates
this one.
The advantage of this limit
comparison method is, it makes
no difference whether you can
compute the thing or not.
You can still decide whether
it's finite or infinite,
fairly easily.
So let me give you an
example of that.
So here we have another
example.
We'll take the integral dx
/ square root of x^3 + 3.
Let's say, for the
sake of argument.
From 0 to infinity.
Let's leave off, let's make it
10 to infinity, whatever.
Now this one is problematic
for you.
You're not going to be able
to evaluate it, I promise.
So on the other hand 1 / the
square root of x^3 + 3 is
similar to 1 / the square root
of x ^3, which is 1 / x ^ 3/2.
So this thing is going to
resemble this integral here.
Which is convergent.
According to our rule.
So those are the, more or less
the main ingredients.
Let me just mention one other
integral, which was the one
that we had over here.
This one here.
If you look at this integral, of
course we can compute it so
we know the area is finite.
But the way that you would
actually carry this out, if
you didn't know the number and
you wanted to check that this
integral were finite, then you
would make the following
comparison.
This one is not so difficult.
First of all, you would write it
as twice the integral from
0 to infinity of e^ - x ^2 dx.
This is a new example here, and
we're just checking for
convergence only.
Not evaluation.
And now, I'm going to make a
comparison here, Rather than a
limit, comparison I'm actually
just going to make an ordinary
comparison.
That's because this thing
vanishes so fast. It's so
favorable that we can only put
something on top of it, we
can't get something underneath
it that exactly
balances with it.
In other words, this wiggle was
something which had the
same growth rate as the
function involved.
This thing just vanishes
incredibly fast. It's great.
It's too good for us,
for this comparison.
So instead what I'm going
to make is the following
comparison. e ^ - x
^2 < = e ^ - x.
At least for x >= 1.
When x > = 1, then x ^2 >=
x, and so - x ^2 < - x.
And so e^ - x^2 is
less than this.
So this is the reasoning
involved.
And so what we have here
is two pieces.
We have 2, the integral from
0 to 1, of e^ - x ^2.
That's just a finite part.
And then we have this other
part, which I'm going to
replace with the e ^ - x here.
2 times 1 to infinity
e ^ -x dx.
So this is, if you like,
this is ordinary
comparison of integrals.
It's something that we did way
at the beginning of the class.
Or much earlier on, when we were
dealing with integrals.
Which is that if you have a
larger integrand, then the
integral gets larger.
So we've replaced
the integral.
We've got the same integrand
on 0 to 1.
And we have a larger integrand
on, so this
one is larger integrand.
And this one we know
is finite.
This one is a convergent
integral.
So the whole business
is convergent.
But of course we replaced it
by a much larger thing.
So we're not getting the right
number out of this.
We're just showing that
it converges.
So these are the main
ingredients.
As I say, once the thing gets
really, really fast decaying,
it's relatively straightforward.
There's lots of room to show
that it converges.
Now, there's one last item of
business here which I have to
promise you.
Which I promised you, which had
to do with dealing with
this bottom piece here.
So I have to deal with what
happens when there's a
singularity.
This is known as an improper
integral of the second type.
And the idea of these examples
is the following.
You might have something
like this.
Something like this.
Or something like this.
These are typical sorts
of examples.
And before actually describing
what happens, I
just want to mention.
So first of all, the key point
here is you can just calculate
these things.
And plug in 0 and it works and
you'll get the right answer.
So you'll determine, you'll
figure out, that it turns out
that this one will converge,
this one will diverge, and
this one will diverge.
That's what will turn
out to happen.
However, I want to warn you that
you can fool yourself.
And so let me give you a
slightly different example.
Let's consider this
integral here.
The integral from -
1 to 1 dx / x ^2.
If you carry out this integral
without thinking, what will
happen is, you'll get the
antiderivative, which is - x ^
-1, evaluated at - 1 and 1.
And you plug it in.
And what do you get?
You get - 1 (1 ^ - 1) -,
uh-oh. (- ( -1) ^ - 1).
There's a lot of - 1's
in this problem.
OK, so that's - 1.
And this one, if you work it all
out, as I sometimes don't
get the signs right, but this
time I really paid attention.
It's - 1, I'm telling you
that's what it is.
So that comes out to be - 2.
Now, this is ridiculous.
This function here
looks like this.
It's positive, right?
1 / x ^2 is positive.
How exactly is it that the area
between - 1 and 1 came
out to be a negative number?
That can't be.
There was clearly something
wrong with this.
And this is the kind of thing
that you'll get regularly if
you don't pay attention to
convergence of integrals.
So what's going on here is
actually that this area in
here is infinite.
And this calculation that
I made is nonsense.
So it doesn't work.
This is wrong.
Because it's divergent.
Actually, when you get to
imaginary numbers, it'll turn
out that there's a way
of rescuing it.
But, still, it means something
totally different when that
integral is thought
to be at - 2.
So.
What I want you to do here, so
I think we'll have to finish
this up very briefly
next time.
We'll do these three
calculations and you'll see
that these two guys
are divergent
and this one converges.
And we'll do that next time.
