Welcome back to recitation.
In this video, I'd like us to do the following problem.
I want us to show that the function, if I integrate x to
the n, e to the minus x from 1 to infinity, that that
actually converges for any value of n, and I want us to
show this without using integration by parts n times.
So I'd like you to figure out a way to show that this
integral converges.
And again, what we mean by converges, is for a fixed
value here, say, some b, if I let, as the limit is b goes to
infinity, that that sequence of values converges to a
finite number.
That's what we mean, again, when we
say an integral converges.
Ultimately, we want to show this is finite.
So show this is finite, without using integration by
parts n times.
So I'll give you a little while to work on it, and then
I'll be back, and I will show you how I did it.
OK.
Welcome back.
Well, I want to show you how we can show that this integral
actually convergences for any n.
And I don't want to have to use integration by parts and
kill off powers of x in order to do that.
So again, the integral is from 1 to infinity of x to the n, e
to the minus x dx.
And if you recall, Professor Jerison was showing, in the
lecture video, that if you can show even from, not 1 to
infinity, but from very far out to infinity, that that
integral converges, from 1 to whatever the far out value is,
this integral is finite.
OK?
Doesn't blow up.
It's going to be potentially a very big number, but it is
going to be a finite number.
There are no places where we run into
trouble with this function.
It's a continuous function from 1 to infinity.
So we can go very far out and say, OK, from very far out to
infinity, the integral converges.
And then that's going to be enough.
So we did see that kind of technique earlier.
But I just want to remind you, that's what we're going to do.
Now, you might have thought about this problem and said,
well, I know that x to the n is much smaller than e to the
x for values of x very large.
So you might have said--
I think you used this notation also in the lecture. x to the
n is much smaller than e to the x for large x.
So what if we tried to do a comparison
with those two functions?
We're going to see, that's not quite enough.
But let's say x is very large, and let's look at a
comparison.
If I say, the integral from, say, some very large r to
infinity of x to the n, e to the minus x dx, it's certainly
going to be much smaller than the integral from big r to
infinity of e to the x times e to the minus x dx.
And you think, well, you know, that's a
pretty good first step.
I'm doing all right.
But what happens here?
What's e to the x times e to the minus x?
It's e to the x plus negative x, so it's e to
the 0, so it's 1.
So this is integrating the constant 1.
Well, the constant 1 from r to infinity, think of that.
It's the line y equals 1 from r to infinity.
It's an arbitrarily long rectangle.
That's got a lot of area.
It's got infinite area.
So this integral diverges.
That doesn't mean this one diverges, right?
Because this one is smaller than that one.
So this one here could still converge, even though this
integral diverged.
Again, let me remind you.
Why does this integral diverge?
Because this is actually equal to 1.
If I integrate 1 from r to infinity, I
get something infinite.
So you might have started with that, but that's not quite
good enough.
Right?
What is going to be good enough, is if I pick any
constant in front of this n--
I can put any constant in front of this n I
want, bigger than 1.
And that's going to help us out.
So I'm going to pick the constant 2, because it's going
to be easy, and it's a nice fixed number.
If I--
so this is how you do it correctly, or one strategy to
do it correctly.
If I take x to the 2n, instead of just x to the n, for any n,
there's some r big enough so that x to the 2n is much less
than e to the x for all x bigger than sum, bigger than
or equal to sum r.
So if I go far enough out in x-values, x to the 2n is much
smaller than e to the x.
So let's say that value is capital R, and
then we're much smaller.
This is not very formal, but it's getting closer to a
formal kind of thing.
Then that means x to the n is much smaller than e
to the x over 2.
Right?
And that's going to be the key.
Let's anticipate why that is.
The problem with the substitution of e to the x was
that e to the x times e to the minus x gave you 1.
But if I use e to the x over 2, I'm going to end up with
some function, e to the minus something, and that's going to
be good, because that's going to converge.
So if you tried e to the x first, and you saw you didn't
get a good function, you didn't have to stop there.
You could say, well, I was close.
The problem is, I cancelled off all the e to the minus
power, which is what I want to keep around.
If I want to keep some of that around, then I have to have a
little less power of e to the x there.
I have to have, I can't just have e to the x.
I should have something like square root of e to the x.
So e to the x over 2.
That's going to help us out.
So this is, this is kind of, as you're working on this type
of problem, this is some of the thought process you want
to go through.
So what do we see here?
We have x to the n is much less than e to the x over 2
for x bigger than or equal to r.
So now let's do a comparison with our
new comparative function.
So I'm going to come over, and this will be our last line to
finish this off.
So now we're integrating from r to infinity x to the n, e to
the minus x dx.
And we know that's going to be much less than integral from r
to infinity, e to the x over 2, e to the minus x dx.
And now let's figure out what this is.
e to the minus x plus x over 2 is e to the minus x over 2.
And the good news is, this is, we know this converges.
I'll check, and I'll show you, remind you that it converges.
But we know this converges.
And the reason is, because e to the minus x is a function
that decays so fast as it goes to 0.
That's really why you get the convergence.
Actually, you could even compare this to x to the minus
2 right away.
And you could get something like, you know this decays
faster than x to the minus 2, and we know x
to the minus 2 converges.
So you could even compare it to that.
You could do a second comparison in here.
But I'll actually calculate this, just to remind
us how we do this.
So e to the minus x over 2.
If I want to find an antiderivative, I'm going to
guess and check.
I know I'm going to have an e to the minus x over 2 again,
and then I need to be able to kill off a negative 1/2.
So I should put a negative 2 in front.
Let's double check.
This is basically a substitution problem.
An easy substitution.
So e to the minus x over 2.
Its derivative is negative 1/2 x, and then itself again.
The negative 1/2 times negative 2 gives me a 1.
So again, I'm just, it's an easy substitution problem, but
I always want to check.
So I evaluate that from r to infinity.
Well, the point is, e to the minus infinity--
this is 0.
As x goes to infinity, this quantity goes to 0.
So I get 0 minus a negative 2, so plus 2, e to the
minus r over 2.
For some fixed, big r.
Well, that's finite.
That's a finite number.
So we come back here.
This integral converges.
That integral was this integral.
And this integral, then, is bigger than this one.
So this one converges.
Now, this had a lot of pieces to it, so I'm going to remind
us sort of what was happening.
So let's go back to the original function, and I'll
just take us back through one more time.
OK.
So the original problem was, show that this integral from 1
to infinity of x to the n e to the minus x dx converges.
And I reminded you that you knew from lecture that if I
could show that it converged for some very large number
down here to infinity, that was sufficient, because this
function is continuous from 1 to infinity.
I don't have to worry about places where I might get
infinite area in a finite interval.
I'm always going to have finite area
and a finite interval.
So if I start at some big R to infinity and that converges,
then I'm good.
Because from 1 to r, that'll be finite.
So then the point is, you want to compare.
And I mentioned, a comparison that doesn't quite work, but
is a good first test. Because you know x to the n is much
less than e to the x for a sufficiently large x.
You might think to compare it to that, but the problem again
was when we do that substitution, we actually get
an integral that diverges.
But a divergent integral bigger than something doesn't
mean this one diverges.
If the divergence was on, the inequality was the other way,
then you could show this one diverged.
But we actually show convergence this way.
Or we're trying to show convergence in this direction,
so we need to say, if this one converges,
then that one converges.
This diverging doesn't tell us anything about this.
So then we say, OK.
This one didn't work, but it almost worked.
So what if I figure out a way to compare x to the n to a
slightly smaller thing than e to the x?
And that slightly smaller thing is e to the x over 2.
It's not really slightly smaller.
But the smaller function is e to the x over 2.
OK?
And this is a way to think about how that works.
Is that for any power of x to the 2n, I can still get it
smaller than e to the x for some
sufficiently large number.
OK?
And you don't even have to think about these R's
as being the same.
I can change them, I can make this bigger.
This doesn't compare to this problem, also.
So don't be confused by those two R's.
OK.
So then we found something we wanted to
compare x to the n with.
Then we come back over here, and we actually see that we
get a good comparison, because we're able to see that the
integral on the right hand side converges.
This integral is bigger than this integral.
So this integral converges, so we know
this integral converges.
And then, the integral from 1 to r of this is finite, so the
integral from 1 to infinity of this converges.
And that's sort of the strategy for doing these types
of
problems. OK.
That's where I'll stop.