### Video Transcript

Improper integrals: discontinuous
integrands.

In this video, weβll practise
evaluating improper integrals, where the integrand has a vertical asymptote. First, let us recall what makes an
integral improper. The first case is when one or both
of the limits of integration are positive or negative infinity. So we could have the upper, the
lower, or both of these limits being an infinity. The second case is when the
integrand, which weβll call π of π₯, has a discontinuity on the closed interval
between π and π, where π and π are the limits of integration. This is the case of a discontinuous
integrand, which weβll be focusing on in this video. An example of this would be the
integral between negative one and one of one over π₯ squared with respect to π₯.

An important point worth noting is
that this type of integral may not appear to be improper at first glance since there
are no signs of obvious infinities appearing in the integral limits. However, as weβll see in a moment,
we cannot use standard techniques to evaluate this integral. And weβll need some additional
tools. So looking at our example integral,
the standard tool we might use to evaluate it is the second part of the fundamental
theorem of calculus. This tells us that if lowercase π
is a continuous function on the closed interval between π and π, then the integral
between π and π of lowercase π of π₯ with respect to π₯ is equal to uppercase πΉ
of π minus uppercase πΉ of π. And this is where uppercase πΉ is
any antiderivative of lowercase π.

Now, if we were to sketch one over
π₯ squared, clearly, we can see that it is not continuous on the closed interval
between negative one and one, since it has an infinite discontinuity when π₯ is
equal to zero. Letβs suppose that we forgot the
continuity condition of the fundamental theorem of calculus. And we try to evaluate our example
integral using this technique anyway. What would happen? Well, letβs see. We first reexpress one over π₯
squared as π₯ to the power of minus two. This allows us to more clearly see
our antiderivative, which is negative π₯ to the power of minus one, which could also
be expressed as negative one over π₯.

Okay, so weβll reexpress in this
form. And then weβll input our limits of
integration, negative one and one. Weβre then left with the following
expression, which simplifies to negative one minus one, which is, of course,
negative two. This is a very strange result
since, looking at our sketch, the area under the curve appears to be entirely above
the π₯-axis. So it should be positive. Yet our integral has evaluated to a
negative number. With this integral and others like
it, the infinite discontinuity is causing us problems. And in fact, it may not even be
possible to find a numerical evaluation for this integral at all, certainly not
using this technique.

Let us introduce some new
definitions, which will allow us to evaluate some integrals where an infinite
discontinuity exists. If π is continuous on the interval
which is closed at π and open at π and has a discontinuity at π. Then the integral between π and π
of π of π₯ with respect to π₯ is equal to the limit as π‘ approaches π from the
left of the integral between π and π‘ of π of π₯ with respect to π₯. And this is if the limit exists and
is finite. Here is an example of what this
might look like. To go with this, if π is
continuous on the interval which is open at π and closed at π and has a
discontinuity at π. Then the integral between π and π
of π of π₯ with respect to π₯ is equal to the limit as π‘ approaches π from the
right of the integral between π‘ and π of π of π₯ with respect to π₯. Again if the limit exists and is
finite. And here is the corresponding
visual example.

A few points to note here. Firstly, we have introduced π‘ into
both of our definitions. And π‘ is simply a dummy variable,
which helps us to evaluate our limits. Also, in our first definition, we
have used a left sided limit. This is because the interval of
integration is entirely on the left side of our upper limit. And we approach our discontinuity
from the left. In our second definition, we have
used a right sided limit since the interval of integration is entirely on the right
side of our lower limit. And of course, we approach the
discontinuity from the right. Now, the final statement in both
cases says that this is only true if the limits exist and are finite.

When describing this, we say that
the improper integral is convergent if the corresponding limit exists. And of course, it has to be finite
to exist and is divergent if the limit does not exist. Okay. So we have a definition for the
cases where an infinite discontinuity exists at one of the limits of
integration. But what if it exists between the
limits? An example of this would be the one
over π₯ squared case that we were considering before. Well, if π has a discontinuity at
π, where π is greater than π and less than π, and both the integral between π
and π of π of π₯ with respect to π₯ along with the integral between π and π of
π of π₯ with respect to π₯ are convergent. Then we define that the integral
between π and π of π of π₯ with respect to π₯ is equal to the integral between π
and π of π of π₯ with respect to π₯, added to the integral between π and π of π
of π₯ with respect to π₯. Now, this last line should be
familiar to us, since the technique of splitting an integral is used in many other
areas of calculus.

What weβre really doing here is
spitting our integral into the sum of two smaller integrals, one with the infinite
discontinuity at the upper limit, as shown here, and two with the infinite
discontinuity at the lower limit. We can then evaluate these two
smaller integrals using the limits that weβve outlined here. Itβs worth noting that this
definition is only valid if both of the two smaller integrals that weβre summing are
convergent. And then, we say that our main
integral is also convergent. If either of the two smaller
integrals are divergent, then so is our main improper integral.

Given what weβve learned, our steps
of solving problems will then be the following. We first check our integrand for
values of π₯ where discontinuities exist. For something like one over π₯
squared, this is fairly obvious. For other functions, we might need
to put in a bit more work. A general rule is that itβs common
for infinite discontinuities to occur in functions involving quotients. Once we have found our
discontinuities, we think about where this value of π₯ occurs with respect to the
limits of integration. For one over π₯ squared, our
discontinuity is when π₯ is equal to zero. This can be at the lower limit of
integration, the upper limit of integration, between the limits of integration, or
outside the limits of integration.

Itβs worth noting that, actually,
this final integral would not be considered improper at all, since the discontinuity
does not occur either between or on the limits of integration. In other words, thereβs no
discontinuity on the closed interval between one and two. The position of the discontinuity
with respect to the limits of integration dictates the method and the techniques
that weβll use to solve the integral.

Let us now take a look at an
example.

Determine whether the integral
between zero and one of one over π₯ with respect to π₯ is convergent or
divergent.

For this question, we have been
asked to evaluate a definite integral of the function one over π₯. This should be a familiar function
to us. And we clearly see that it has an
infinite discontinuity when π₯ is equal to zero, approaching positive infinity from
the right and negative infinity from the left. The standard technique that we
might use to evaluate a definite integral is the second part of the fundamental
theorem of calculus. But this requires that our
integrand π is a continuous function on the closed interval between π and π,
which are our limits of integration. Weβve just seen that one over π₯
has an infinite discontinuity when π₯ is equal to zero. And we clearly see that this is one
of our limits of integration. Hence, the continuity condition
here is not true. This means that what weβre dealing
with is an improper integral. And we must use a different
technique.

For a discontinuity occurring at
the lower limit of integration, the definition of an improper integral tells us the
following. If π is continuous on the interval
which is open at π and closed at π and discontinuous at π. Then the integral between π and π
of π of π₯ with respect to π₯ is equal to the limit as π‘ approaches π from the
right of the integral between π‘ and π of π of π₯ with respect to π₯. If this limit exists and is
finite. Donβt worry too much about the π‘
that weβve introduced since this is just a dummy variable which helps us to evaluate
our limit. Let us now apply this to our
question. π of π₯, our integrand, is one
over π₯. The lower limit of integration, π,
is zero. And the upper limit of integration,
π, is one.

Our definition tells us that our
integral is equal to the limit as π‘ approaches zero from the right of the integral
between π‘ and one of one over π₯ with respect to π₯. Okay. Letβs now evaluate this. We know that the antiderivative of
one over π₯ is the natural log of the absolute value of π₯. To continue, we input our limits of
integration π‘ and one. Weβre then left with the following
expression. The laws of limits allow us to
apply our limit individually to each of these terms. So letβs do so and see what happens
to them. Well, our first term has no π‘
dependents at all. So we can simply get rid of
this. We also know that the natural
logarithm of one is equal to zero. And this relationship might be seen
more clearly by taking the exponential of both sides. π to the power of zero is of
course one.

Okay. What about this term? Taking a direct substitution
approach, we have the natural logarithm of zero, technically zero from the
right. In some sense, we can say that this
is equal to negative infinity. Since as π‘ approaches zero from
the right, the natural log of the absolute value of π‘ approaches negative
infinity. Inputting our two values back into
our expression gives us the following. Our limit is equal to zero minus
negative infinity, which is positive infinity.

Now, saying that a limit is equal
to infinity does give us information about that limit. But itβs a particular way of
expressing that the limit does not exist since infinity is not a number. Since our limit does not exist, we
conclude that our limit does not have a numerical and finite evaluation. In cases such as this, we say that
the integral is divergent. And so we have arrived at the
answer to our question.

Okay. With the conclusion of this
example, we note that, so far, only functions involving powers of π₯ have been
seen. However, these techniques can also
be used to evaluate discontinuous integrands involving other functions, such as
trigonometrics or exponentials. Letβs look at one of these now.

The integral between zero and π
over two of cos π divided by the square root of sin π with respect to π is
convergent. What does it converge to?

Straight away, we can notice that
the question has told us our integral is convergent. This gives us a hint that weβre
looking at an improper integral. Now, since we donβt see any
infinities as our limits of integration, itβs also implied that our integrand has a
discontinuity on the closed interval between zero and π over two, which are the
limits of the integral. Letβs look for this
discontinuity. Since the integrand that weβve been
given is a quotient, we should try and find where its denominator is equal to
zero. The square root of sin π is zero,
where sin π itself is also equal to zero. This occurs when π is equal to
zero plus ππ, which is of course just ππ. And π here is an integer. So here we see that our denominator
is equal to zero when π is equal to any of the following. The only one of these values which
occurs over the interval of our integration is zero itself.

Okay, for due diligence, letβs now
check what happens to our numerator when π is equal to zero. Our numerator is cos π. So this becomes cos of zero, which
is equal to one. We do this to check that when π
equals zero, our function is not equal to the indeterminate form of zero over
zero. Since we would then be looking at a
removable discontinuity. In actual fact, when π is equal to
zero, our function is equal to one over zero, which is the case of an infinite
discontinuity. Okay. Weβve now confirmed that we have an
improper integral. And we have an infinite
discontinuity at the lower limit of integration. The definition of an improper
integral specifically for this case tells us that if π is a continuous function on
the interval which is open at π and closed at π and discontinuous at π. Then the integral between π and π
of π of π₯ with respect to π₯ is equal to the limit as π‘ approaches π from the
right of the integral between π‘ and π of π of π₯ with respect to π₯. If this limit exists and is
finite.

Letβs now apply this to our
question. The integral that weβve been given
is equal to the limit as π‘ approaches zero from the right of the same integral, but
now between the limits of π‘ and π over two. Now, at this point, we might notice
that our integral is not entirely trivial. And weβll need to put in some work
to evaluate it. To move forward, weβre gonna be
using a π’-substitution, specifically π’ is equal to sin π. From this, we differentiate to find
that dπ’ by dπ is equal to cos π. An equivalent statement to this is
that dπ’ is equal to cos π dπ. We now do the following with our
substitution. We replace sin π with π’. And we replace cos π dπ with
dπ’. Weβre then left with the integral
of one over the square root of π’ with respect to π’. Another way of expressing this
integrand is π’ to the power of negative a half. Evaluating this, we get two π’ to
the power of a half, which, is of course, two times the square root of π’. We can now go back to our original
substitution and replace our π’ with sin π. We therefore have that our integral
is equal to two times the square root of sin π, of course, plus the constant of
integration.

Now that weβve done the legwork on
our integral, letβs use this result to move forward with our question. We use the antiderivative that
weβve just found for our integral. And we get the following. We now input our limits of
integration π‘ and π over two. And weβre left with the following
limit. We can now take a direct
substitution approach to our limit. And we can evaluate since sin of π
over two is equal to one. And sin of zero is equal to
zero. Weβre then left with two times the
square of one minus two times the square root of zero. Of course, the square root of one
is just one. And this entire term is just
zero. The answer that weβre left with is
therefore two. Upon reaching this result, we have
answered our question. We used our definition to find that
the integral given in the question converges to a value of two.

So far, weβve seen examples where
the discontinuity exists at one of the limits of integration. But what if it exists between
them?

Is the integral between zero and
five of π₯ divided by π₯ squared minus 16 with respect to π₯ convergent? If so, what does it converge
to?

For this question, weβve been given
an improper integral with a discontinuous integrand. To move forward, weβll need to see
where these discontinuities occur and to see how they relate to our limits of
integration. Since our integrand has been given
in the form of a quotient, we can find where its discontinuities occur by finding
where its denominator, π₯ squared minus 16, is equal to zero. Solving this equation, we find that
our denominator is equal to zero when π₯ is equal to positive or negative four. Okay, letβs see how these values
relate to our limits of integration, zero and five. Our first value, negative four, is
not equal to either of the limits of integration, nor does it lie between them. The value of four, however, does
lie between our limits of integration. For due diligence, we can check our
integrand, which weβll call π of π₯, for the type of discontinuity that we expect
to see when π₯ equals four. Trying to evaluate π of four,
weβre left with four over zero, which implies that weβre looking at an infinite
discontinuity, as opposed to if itβd been zero over zero, which would imply a
removable discontinuity.

So weβve ignored the discontinuity
at π₯ equals negative four since it does not interact with our limits of
integration. Weβve also confirmed that the
question has given us an improper integral with a discontinuity occurring between
the limits of integration, which is when π₯ equals four. The definition of an improper
integral gives us the following tool, which allows us to split an integral at the
discontinuity, π, into the sum of two smaller integrals. Let us apply this to our question,
where we have our integrand π of π₯, our lower limit π, our upper limit π, and
our discontinuity π. Using our definition, we can say
that our original integral is equal to the sum of two smaller integrals, which are
adjacent to each other at the discontinuity when π₯ is equal to four. We can see this in the fact that
the discontinuity occurs at the upper limit of our first integral and the lower
limit of our second integral.

At this point, we should note that
this statement is only valid and our original integral is only convergent if both of
the two smaller integrals themselves are convergent. Weβll therefore need to check our
two smaller integrals for convergence. However, before we do this, the
technique that weβll be using to solve these two integrals will be a
π’-substitution. And so we might as well get this
out of the way first. The substitution that weβll be
using is that π’ is equal to π₯ squared minus 16. Differentiating with respect to π₯,
we get that dπ’ by dπ₯ is equal to two π₯. And an equivalent statement for
this is that dπ’ is equal to two π₯ dπ₯. It turns out that a more useful
equation for us is that half dπ’ is equal to π₯ dπ₯.

Okay. Since weβre working with definite
integrals, we should also pay attention to our limits of integration. When π₯ is equal to zero, π’ is
equal to negative 16. When π₯ is four, π’ is zero. And when π₯ is five, π’ is
nine. So let us now perform our
substitutions, replacing π₯ squared minus 16 with π’, π₯ dπ₯ with a half dπ’, and
our limits of integration as weβve just found. After performing these
substitutions, we also might as well take this factor of a half outside of the
integrals.

Okay. At this stage, itβs relevant to
give a quick but very important side note. When working with an improper
integral which contains a discontinuity between the limits of integration, we should
always split the integral at the discontinuity before performing any
substitutions. We do this because performing
substitutions first can sometimes cause problems by removing the discontinuity. Although this sounds great, in
practice, it can lead to some problems. To move forward, weβre gonna need
to check our integrals for convergence. We note that since our integrals
have the discontinuity at the upper and lower limits, respectively, before the
substitution, the same is true after the substitution. Weβll begin with the integral with
a discontinuity at the upper limit.

The definition of an improper
integral gives us the following tool to deal with this case. Applying this to our first term
gives us the following result. To move forward, we use the fact
that the antiderivative of one over π’ is the natural logarithm of the absolute
value of π’. Weβll get rid of these definitions
to clear some room. We input our limits of
integration. We can then take a shortcut by
noticing that if we try to evaluate this first term, we would be left with a
negative infinity. Since our other term, which is the
natural logarithm of the absolute value of negative 16, is finite, weβre forced to
conclude that our original limit will evaluate to a negative infinity. This is another way of saying that
the limit does not exist. Since the limit that defines the
first of our smaller integrals does not exist, we say it is divergent. And in fact, this divergence
cascades all the way back to our original integral.

Recall that our definition told us
that the original relationship we used is only valid if both of the two smaller
integrals, which weβre summing, are convergent. Since the first one we checked
turned out to be divergent, we donβt actually need to check the second one to
conclude that our original integral is also divergent itself. With no further work, we can say
that the integral given by the question is not convergent, but rather divergent.

To conclude, here are the key
points of this video. An improper integral with a
discontinuous integrand occurs in the following way. The definition of an improper
integral gives us the tools to deal with discontinuities at the upper limit, the
lower limit, and between the limits of integration. And finally, if the corresponding
limit which defines an improper integral exists and, of course, is finite, then we
say that that integral is convergent. And if the limit does not exist, we
say that it is divergent.