Welcome to calculus. I'm professor Ghrist, and we're about to
begin lecture 27 on improper integrals. The fundamental theorem of
integral calculus is great, but it's not without its limitations. In this lesson, we'll consider what
happens when we encounter a difficulty with limits in a definite integral. The fundamental theorem of
integral calculus is great, but it does have its limitations. There are a few things that
you must be careful about. In particular, the fundamental
theorem has a hypothesis that f, the integrand,
is continuous on the interval from a to b. And within this statement
lies two dangers. The first is that of continuity. A discontinuous integrand
can cause problems. Here's an example. Consider the integral as x goes
from -1 to 1 of (1/(x squared))dx. If we simply apply
the fundamental theorem and take the antiderivative,
negative 1 over x, and evaluate that at 1, what do we get? -1. Then we subtract what we
get when we evaluate at -1. -1- 1= -2. Perfect. Except for the fact that our
integrand is 1 over x squared. All those terms are positive,
and if we go to the definition of the definite integral as a Riemann sum,
adding up the values, there's no way that we can add up positive
values to get a negative integral. The problem is this
integrand is not continuous. It's not even defined at x equals zero. The second hypothesis is that of having an interval from a to b,
a finite interval. Consider the integral as X
goes from negative infinity to positive infinity of
(2x/(1+x squared)) dx. Clearly, the antiderivative
is log of (1+x squared). What happens when we
evaluate this at the limits? You might think we get infinity minus
infinity, which is not defined. Or you might think, well,
this is an odd integrand over a symmetric domain, therefore
it must be zero, which is correct. These improper integrals are dangerous. In all cases,
we're going to use the technique of taking a limit to make sense of these integrals. There are two cases. The first we might call a blow-up. This is what happens when
you have an integrand that is not well defined at some point. Let's say at one of the endpoints, A. In this case, the way to evaluate
the limit is to integrate from some constant, t,
to the other endpoint. And then take a limit as T
goes to the singular input. In the other instance,
we might call it having a tail, where your integration
domain goes off to infinity. In this case, the appropriate thing to
do is integrate over a finite domain. Let's say up to a constant,
T, and then take the limit as T goes to infinity of
the finite domain integral. The following class of examples
are crucial to this subject. These are the p integrals, or the integral of ((1/x) to the p)dx. Let's consider first the example
of a tail singularity, where we integrate,
lets say from X equals one to infinity. Of course the value is going to depend
on P, but lets do all of them at once. If we integrate (x to the -p)dx,
that is easy enough. That's going to equal,
as long as p is not equal to 1, (x to the 1- p)/(1- p). We need to evaluate this at
the limit x goes from one to t. And then take the limit
as t goes to infinity. That is, we're taking the limit. This T goes to infinity of (T to the 1- p) / (1- p)- (1 / (1- p)). Now this limit is going
to be infinite sometimes. But when p is bigger than 1,
then the first term goes to 0 and we're left with 1 / (p- 1). If p is less than one,
then that first term dominates and goes to infinity, and
we say the integral diverges. In the particular case where p equals one, then this integral becomes the integral
of 1 over x, that is, log of x. And evaluating that at the limits and sending T to infinity yields,
again, infinity. And so we have a divergent integral
when p is less than or equal to one. Let's do it the other way and look at
a blow-up instead of a tail singularity. Consider the same integrand x to the -p,
but now integrated as x goes from zero to one. Well, doing the integral yields
the same anti-derivative, we just need to evaluate
our limits differently. So, evaluating as x goes from t to one,
then taking a limit as T goes to zero. This will break up into two cases
when p is not equal to one, or when p is equal to one. In these cases, again, we get sometimes
convergence, sometimes divergence. But note what is happening to the p's. When p is bigger then one,
we get a divergent integral. When p is less than one, then the (T to the 1-p)/(1-p) term drops out and we're left with an answer of 1/(1-p). Again, in the case where p equals one, log of T is not gonna be
convergent as T goes to zero. Let's summarize what we've
found with these p integrals. If we integrate along
a tail going to infinity, then the p integral converges when
p is strictly bigger than one. It diverges when p is less than or
equal to one. For a blow up singularity at zero,
these are reversed. And we get a convergent integral
when p is less than one, and a divergent integral when p is
bigger than or equal to one. Note, that at the particular value of one,
it's always divergent. No matter whether you're going from
zero to one or from one to infinity. Now you don't need to remember the actual
values of the convergent p integrals. But you do need to remember this chart,
this listing of when the integral converges and
when is diverges. And the reason you need to remember
this is because it will help you determine convergence or
divergence of other integrals. Integrals whose anti-derivatives
who may not be so easy to compute. Let's look at an example. Consider the integral
of dx/(square root of x squared + x) as x goes from zero to one. This is a finite domain,
however there is a singularity, or a blow up, at x equals zero. So how shall we proceed? I don't know the anti-derivative to this. It doesn't look like it's
going to be terribly easy. So, let us consider what the leading
order behavior for this integrand is. We're going to think in
terms of Taylor series. Now, this integrand doesn't have
a well-defined Taylor series at x equals zero, since the function is not
even defined, and it blows up. But notice that if we
factor out a square root of x from the denominator,
then we're left with 1/(square root of x+1) as a factor. Now let's rewrite that, thinking that we are going to be looking
at what happens as x is near zero. I can write this as (x to the -1/2)
times quantity (1+x) to the -1/2. And this then is helpful, why? Because the binomial series says that
whenever you have (1+x) to the alpha, as x is small, then this is of the form one plus something in big O of x. Now if we apply that to
the (1+x) to the -1/2 term, then we get the integral as x
goes zero to one of x to the -1/2 times quantity 1 + something in O(x). And we see that the leading order term in this integrand is x to the -1/2. So I'm going to split this
up into two integrals. The first is a p integral
with p equals 1/2. The second is an integral whose
precise form I haven't written down, but it's something that is in O(square
root of x), and indeed, is bounded. And I'm integrating it over
the domain from zero to one. The integral on the right
definitely converges. What about the integral on the left? Well remember,
I said you had to memorize some of these. Let's go back and
recall that when p equals one half, for a blow up p integral, it converges. Therefore, we know that this
entire integral converges. We may not know the value,
but we know it converges. Now, what happens if we take
that same integrand and integrate as x is going to infinity? For very large x, we need to consider that x squared term in
the denominator as the lead. Therefore, factoring out,
we get (1/(square root of x squared)) times (1/(square
root of 1+(1/x))). To simplify that a little bit, we see that we have again something to
which the binomial series applies. Namely, (1+(1/x)) to the -1/2. Expanding that out gives
(1 + something in O(1/x)). And now,
splitting this up into two integrals, we see that the leading order term
Is a p integral, with p equals one. And I don't think I need to remind
you that when p equals one, you always have divergence. Therefore, because one piece
of this integral diverges, the entire integral diverges. Same integrand,
different behavior as you go to infinity. There's one last thing we need to look at. And that is the problem of double limits. If you integrate something over
the entire real number line, you might be tempted to evaluate the limit as T goes to infinity of the integral
as x goes from negative T to T. Let's see what happens in this case. We already know that the anti-derivitave
is log of quantity (1 + x squared). And if we evaluate that at negative T and
T, well, we get something that cancels to zero. And so
it would seem as though our limit is zero. That is not correct, and
we have failed to be careful. What do we need to do? We need to take two independent limits. One, as say, s, is going to negative
infinity, the integral from s to zero. The other, as T, goes to positive
infinity of the integral from 0 to T. When we have two tails,
we need two limits. And I don't think it's too hard to see what the behavior of each
of these is going to be. If we consider what happens
when x gets very large, then the leading order term is 2/x. And what is left over
is 1/(1+(1/x)) squared. That means that using,
say, the geometric series, we see that the leading order term
is a p integral with p equals one. That connotes divergence. And this means that each of these
limiting integrals diverges. And hence the net integral does
not converge to zero, it diverges. In general, you'll want to remember
the p integrals and use them to tell you when an improper
integral converges or diverges. These improper integrals are dangerous. Indefinite integrals are not easy. Most students find that they have some
difficulty understanding what they're suppose to do with an indefinite integral. Be sure to do a lot of practice
problems in this lesson. In our next lesson,
we're going to step back and take a broader look at all of
the techniques that we have built. We'll apply them to a specific
class of trigonometric integrals.