Professor Dave here, let’s look at improper integrals.
In learning about integration, we’ve been focusing largely on indefinite integrals.
This is because these focus on the act of finding the antiderivative, which is the bulk
of the task.
When we want to evaluate them within some limits of integration, like with the definite
integrals we looked at, it’s just a little bit of algebra.
We evaluate the antiderivative at the upper limit, then the lower limit, and subtract
the latter from the former.
That gives us a number that represents the area under the curve within that finite interval.
But sometimes we will want to evaluate an integral over an interval that is infinitely large.
This will happen when one of the limits of integration is either positive or negative
infinity, or if the function has an infinite discontinuity within the finite interval of
a to b.
When this is the case, we call these improper integrals.
Let’s learn about these now.
Now one might initially assume that if the interval is infinite, the area under the curve
must also be infinite, but actually, this is not always the case.
We’ve seen countless times by now that calculus is all about doing something infinitely many
times and getting a finite result, kind of like the infinite series that sums to give
e, so we can indeed have a function that extends infinitely in a particular direction, and
yet spans a finite area.
Take for example the curve one over x squared.
Let’s say we want to integrate this starting at x equals one, and going to infinity.
To examine what happens as we get closer to infinity, let’s select an arbitrary upper
limit of integration and call it t.
We can treat this like a definite integral, and get the antiderivative.
X to the negative two will give us x to the negative one over negative one, which is negative
one over x.
Let’s evaluate this over our selected interval, and we get negative one over t minus negative
one over one.
These two negative signs cancel to give us positive one, and switching the order, we
can express this as one minus (one over t).
So if we take t to equal two, we get one minus one half, which is one half.
If we take t to equal ten, we get one minus one tenth, or nine tenths.
We can plug in larger and larger values for t, and we get closer and closer to one, and
if we plug in infinity for t, which means integrating over this whole portion of the
function all the way to infinity, then we get one as the answer.
That means, quite incredibly, that the area under the curve from x equals one all the
way to infinity, is one.
This gives us a simple way to evaluate improper integrals.
We can integrate f of x dx from a to infinity, and this will simply be equal to the integral
over the definite interval from a to t, in the limit of t approaching infinity.
This limit may or may not exist, which means that not all improper integrals can be evaluated,
but some indeed can.
This also works in the negative direction, if we are integrating from negative infinity to b.
We just integrate from t to b, and see what happens in the limit of t approaching negative
infinity.
If we are evaluating an improper integral and we do get a finite number as an answer,
then we say that the interval is convergent.
If the limit does not exist, meaning the area truly is infinite, we say that the interval
is divergent.
Unfortunately, it may be difficult to tell whether an integral is convergent or divergent
just by looking at it.
Say we try to do the same thing with one over x.
We can integrate this function from one to infinity by replacing infinity with t.
The integral of one over x is the natural log of x, so when we evaluate, we get the
natural log of t minus the natural log of one.
The natural log of one, or any log of one, must be zero, because anything to the zero
power gives us one.
So that term goes away, leaving us with the natural log of t.
As t gets larger, this term gets larger, and it does so without bound, all the way to infinity.
So this one didn’t work out as nicely as the first one, and this integral is indeed
divergent, even though the two functions look fairly similar.
From this we conclude that one over x squared gets small enough fast enough that the area
under the curve is finite, while one over x also gets small enough, as it too, goes
to zero as x approaches infinity, it just doesn’t do it fast enough, and we needed
calculus to tell us this fact.
Let’s try another one.
How about one over the quantity (one plus x squared).
This function looks like this, extending to positive and negative infinity in either direction,
and it’s perfectly symmetrical, so let’s just pick one half.
How about from zero to positive infinity.
So as we know, we change infinity to t, and we integrate.
If we consult our table of common integrals, this is actually a special one, as it fits
the form of one over x squared plus A squared, and in this case A equals one, so we get inverse
tangent of x.
We will then evaluate this at t, and then zero, and we get inverse tangent of T minus
inverse tangent of zero.
We can evaluate the inverse tangent of zero quite easily, it is simply the angle that
gives us a tangent value of zero.
As tangent is sine over cosine, when sine is zero, we get zero overall, and sine is
zero at zero, so inverse tangent of zero is zero.
That leaves us with the inverse tangent of t, which we must now examine as t approaches infinity.
For an inverse tangent to be getting larger, it must mean that the sine value is increasing,
while the cosine value is decreasing, because a larger and larger number being divided by
a smaller and smaller number must be approaching infinity.
Moving through quadrant one, the tangent value approaches infinity until we get to half pi,
where it is undefined.
That means that the inverse tangent of infinity is half pi.
So the area under this half of the curve is half pi.
To get the other half, we could do the exact same thing but integrate from T to zero, and
we would again get half pi, which makes sense as this is perfectly symmetrical, and half
pi plus half pi is pi, so the area under this entire curve, which spans from negative infinity
to positive infinity, is actually a finite number, pi, and this integral, no matter what
the interval, is indeed convergent.
Apart from being impressed that this integral from negative infinity to positive infinity
yields a finite number, we might also be quite astonished that this number is pi.
We are used to seeing pi when dealing with circles, and here there is no circle to be found.
Just another example of the stunning beauty and mystery that can be found in mathematics.
So integrating over an infinite interval is one way we can get an improper integral.
There is another way that involves finite intervals, and this will be when there is
some vertical asymptote within the interval.
This allows the curve to extend to infinity not in the horizontal direction, but in the
vertical direction, and we might not be too surprised by now to find out that this circumstance
can also yield a convergent integral with a finite value.
To evaluate these, it’s not much different, it’s just that instead of replacing infinity
with T, we replace either one of the limits of integration, if one of the limits is the
asymptote, or we replace some value in between with T, if the asymptote is in the middle
of the interval.
Let’s take the function one over (root x minus two), and integrate from two to five.
This function has a vertical asymptote at x equals two, so we can see that this will
be an improper integral.
So instead of integrating from two, we have to integrate from T, and then look at what
happens to the integral as T approaches two from the positive direction.
To integrate, let’s just use a simple substitution, changing x minus two into u.
That means that du will be equal to dx, and we have u to the negative one half du.
Integrating will give us u to the positive one half, over one half, or two root u.
We then change u back, and we have two root (x minus two).
Let’s evaluate this at five and T.
For five, we get two root three, and for t, we get two root (t minus two).
As T approaches two from the positive direction, what’s inside this radical remains positive
but gets closer and closer to zero, and in the limit of T approaching two, this term
will become zero.
So the answer is two root three, for the area under the curve for this interval, and the
integral is therefore convergent.
So we can see that evaluating improper integrals is not that much different than evaluating
regular definite integrals.
In fact, it’s exactly the same, it’s just that we see what happens when one of the limits
of integration starts moving towards some particular value, whether positive or negative
infinity, or some vertical asymptote.
It is very important to recognize when an integral is improper, because we must evaluate
it accordingly.
Evaluating an improper integral as though it is a definite integral, simply evaluating
the antiderivative at the limits of integration and subtracting, can indeed yield incorrect values.
So as you go about integrating, if something looks like it might have an asymptote of some
kind, check the graph of the function being integrated over the specified interval to
see if you need to adjust your approach accordingly.
With that understood, let’s check comprehension.