### Video Transcript

Improper Integrals: Infinite Limits
of Integration

In this video, weβll learn how to
evaluate improper integrals, where one or more of the endpoints approaches
infinity. Weβll be looking at some examples
of how we can solve integrals of this form.

Now, letβs start by considering the
following function π of π₯ and letβs consider an integral of this function π of π₯
with an infinite bound, so the integral from π to infinity of π of π₯ with respect
to π₯. And we can let π be some point on
our π₯-axis, say here. Then, this integral represents the
area between π of π₯ and the π₯-axis from π to infinity. Now, this doesnβt really seem to
make sense. How can infinity be a bound when
itβs not a number? And how can we find an area that
goes on to infinity? Now, although this doesnβt seem to
make sense, it is in fact possible to evaluate certain integrals of this form. And we do this using limits.

The formula we use to solve this
integral tells us that this integral is equal to the limit as π‘ goes to infinity of
the integral from π to π‘ of π of π₯ with respect to π₯, for π‘ greater than
π. Itβs important to note that this
only works in certain cases. Firstly, we require that the
integral from π to π‘ of π of π₯ with respect to π₯ exists. So thatβs this integral in our
formula. We also require that the limit of
this integral as π‘ tends to infinity exists too. So thatβs this limit in our
formula. Now, we can say that if both the
integral and the limit exist, then we can say that our integral with an infinite
limit, which is the integral weβre going to find, converges. And if either the integral does not
exist or the limit does not exist, then we can say that our integral with infinite
limit diverges. Just to recap, we can say that our
integral from π to infinity of π of π₯ with respect to π₯ converges if it exists
and is equal to some real constant π. And the integral from π to
infinity of π of π₯ with respect to π₯ diverges if it does not exist.

Letβs now look at an example of an
integral of this form.

The integral from one to infinity
of one over π₯ squared with respect to π₯ is convergent. What does it converge to?

In this question, weβre told that
this integral is convergent. And this tells us that this
integral does evaluate to some constant. We can find what this constant is
by using the following formula, which tells us that the integral from π to infinity
of π of π₯ with respect to π₯ is equal to the limit as π‘ goes to infinity of the
integral from π to π‘ of π of π₯ with respect to π₯. Now, if you look at our integral,
we can see the π is equal to one and π of π₯ is equal to one over π₯ squared. We can substitute these two into
our formula. And we obtain that the integral is
equal to the limit as π‘ tends to infinity of the integral from one to π‘ of one
over π₯ squared with respect to π₯.

We can start by finding this
integral. We can rewrite one over π₯ squared
as π₯ to the power of negative two. And then, we can use the power rule
for integration. We increase the power by one and
then divide by the new power. Increasing the power by one gives
us π₯ to the power of negative one. And then, we divide by the new
power. So thatβs negative one. And we mustnβt forget our bounds of
π‘ and one. Now, dividing by negative one is
the same as multiplying by negative one. And π₯ to the power of negative one
is one over π₯. So we can rewrite π₯ to the power
of negative one over negative one as negative one over π₯.

Now, we can substitute in our
bounds. Since π‘ is our upper bound, when
we substitute it in, the sign remains the same. We get negative one over π‘. However, since one is our lower
bound, we must change the sign when we substitute it in, giving us minus negative
one over one. Simplifying this, we can see that
our integral is equal to one minus one over π‘. And therefore, we can substitute
this into our limit. And now, we have that the integral
from one to infinity of one over π₯ squared with respect to π₯ is equal to the limit
as π‘ tends to infinity of one minus one over π‘.

Now, we can split this limit up
using limit rules. We have that the limit of a
difference of function, so π of π₯ minus π of π₯, is equal to the difference of
the limits of the function, so the limit of π of π₯ minus the limit of π of
π₯. We can apply this to our limit to
say that it is equal to the limit as π‘ tends to infinity of one minus the limit as
π‘ tends to infinity of one over π‘. Now, since one has no π‘
dependence, the limit as π‘ tends to infinity of one is simply one. And we can consider the limit as π‘
tends to infinity of one over π‘. As π‘ tends to infinity, π‘ is
getting larger and larger and larger. Therefore, the reciprocal of π‘ or
one over π‘ is getting closer and closer to zero. And so, we can say that the limit
as π‘ tends to infinity of one over π‘ is equal to zero. And so here, we reach our solution,
which is that the integral from one to infinity of one over π₯ squared with respect
to π₯ converges to β and is therefore equal to β one.

Next, we can take this formula for
evaluating integrals with infinite limit one step further. We will be editing it slightly so
we can evaluate integrals of a slightly different form. Letβs suppose we had a function π
of π₯, which looks something like this. And suppose weβre asked to find the
integral from negative infinity to π of π of π₯. Now, π can be any point on our
π₯-axis. But letβs suppose itβs here. Now, this integral is asking us to
find the area between π of π₯ and the π₯-axis from π all the way to negative
infinity. Now, this is very similar to the
integrals to going to positive infinity. Sometimes, its integral may
converge, and sometimes, it may diverge. In the cases where it converges, we
can say that it is equal to the limit as π‘ tends to negative infinity of the
integral from π‘ to π of π of π₯ with respect to π₯ for π‘ is less than π.

Now, this formula works very, very
similar to the one we saw earlier. The main difference is that instead
it is our lower bound that is infinite and it is negative infinity. Otherwise, this formula works very,
very similar to the previous one.

Letβs now look at an example of how
it works.

The integral from negative infinity
to zero of two to the power of π with respect to π is convergent. What does it converge to?

Now, we know a formula to evaluate
integrals of this form. It tells us that the integral from
negative infinity to π of π of π₯ with respect to π₯ is equal to the limit as π‘
tends to negative infinity of the integral from π‘ to π of π of π₯ with respect to
π₯. In our case, we can see that π is
equal to zero. We are in fact integrating with
respect to π instead of respect to π₯. Therefore, our function π of π₯,
or in our case, π of π, is two to the power of π. We can substitute these values into
our formula. We now have that the integral from
negative infinity to zero of two to the power of π with respect to π is equal to
the limit as π‘ tends to negative infinity of the integral from π‘ to zero of two to
the power of π with respect to π. Letβs start by finding this
integral.

We can consider the
antiderivative. If we try differentiating two to
the power of π with respect to π, then we obtain the natural logarithm of two
multiplied by two to the power of π. Since the natural logarithm of two
is a constant, we can divide by it on both sides and we can put it inside the
differential. And we obtain that the derivative
of two to the power of π of the natural logarithm of two is equal to two to the
power of π. And so, here, we found the
antiderivative of two to the power of π. Itβs two to the power of π over
the natural logarithm of two.

And we can use this to say that the
integral from π‘ to zero of two to the power of π with respect to π is equal to
two to the power of π over the natural logarithm of two from π‘ to zero. Since zero is the upper bound, when
we substitute it in, the sign remains the same, giving us two to the power of zero
over the natural logarithm of two. And since π‘ is the lower bound,
when we substitute it in, we must change the sign, giving us negative two to the
power of π‘ over the natural logarithm of two. Since anything to the power of zero
is one, we can rewrite to use the power of zero as one.

Therefore, we evaluated that this
integral is equal to one over the natural logarithm of two minus two to the power of
π‘ over the natural logarithm of two. And we can substitute this value
for our integral back into our limit. Weβre able to use the properties of
limits in order to split this limit up. We have for the limit of a
difference of functions is equal to the difference of the limits of the functions,
giving us that our limit is equal to the limit as π‘ tends to negative infinity of
one over the natural logarithm of two minus the limit as π‘ tends to negative
infinity of two to the power of π‘ over the natural logarithm of two.

Now, in our first limit here, weβre
taking the limit of one over the natural logarithm of two, which has no π‘
dependence. Therefore, this limit is simply
equal to one over the natural logarithm of two. Weβre able to apply another limit
property to the second limit. And that is that the limit of a
quotient of functions is equal to the quotient of the limit of the functions. Now in the denominator here, weβre
taking the limit of a constant function. Therefore, the denominator is
simply equal to the natural logarithm of two. In the numerator, we have the limit
as π‘ tends to negative infinity of two to the power of π‘. Here, π‘ is getting more and more
negative. Therefore, two to the power of π‘
will be getting closer and closer to zero. And so, we can say that the limit
as π‘ tends to negative infinity of two to the power of π‘ must be zero. Here, we reach our solution which
is that the integral from negative infinity to zero of two to the power of π with
respect to π converges to β and is therefore equal to β one over the natural
logarithm of two.

Next, weβll be combining the two
formula weβve seen in this video so far in order to form a third formula. Weβll be using the property of
integration that tells us how to integrate a function over adjacent integrals, which
tells us that the integral from π to π of π of π₯ with respect to π₯ is equal to
the integral from π to π of π of π₯ with respect to π₯ plus the integral from π
to π of π of π₯ with respect to π₯. What this formula tells us is that
if weβre trying to find the area between some π of π₯ and the π₯-axis between two
points π and π, then this area will be equal to the sums of the areas between π
of π₯ and the π₯-axis between π and π and π and π.

Letβs consider the integral from
negative infinity to infinity of π of π₯ with respect to π₯. If we have some π of π₯, as shown
here, then this integral represents the area between π of π₯ and the π₯-axis, from
negative infinity to positive infinity. And using the property of integrals
over adjacent intervals, we can pick any value π on the π₯-axis. And our integral from negative
infinity to positive infinity will be equal to the sum of the integrals from
negative infinity to π and from π to positive infinity, giving us this following
formula, which is that the integral from negative infinity to positive infinity of
π of π₯ with respect to π₯ is equal to the integral from negative infinity to π of
π of π₯ with respect to π₯ plus the integral from π to positive infinity of π of
π₯ with respect to π₯.

Now, we know formulas for
evaluating both of these integrals. We can use them here to say that
the integral from negative infinity to positive infinity of π of π₯ with respect to
π₯ is equal to the limit as π tends to negative infinity of the integral from π to
π of π of π₯ with respect to π₯ plus the limit as π‘ tends to infinity of the
integral from π to π‘ of π of π₯ with respect to π₯. And here, π must be less than π
and π‘ must be greater than π. Here, we have changed the π‘ from
the first limit to an π to avoid confusion since one of the limits is tending to
negative infinity and the other to positive infinity. And letβs also note that π can be
any real number. We can use this formula to help us
solve the following problem.

The integral between negative
infinity and positive infinity of π₯ multiplied by π to the power of negative π₯
squared with respect to π₯ is convergent. What does it converge to?

We have a formula which can help us
to evaluate integrals of this form. The formula tells us that the
integral from negative infinity to positive infinity of π of π₯ with respect to π₯
is equal to the limit as π tends to negative infinity of the integral from π to π
of π of π₯ with respect to π₯ plus the limit as π‘ tends to positive infinity of
the integral from π to π‘ of π of π₯ with respect to π₯, where π can be any real
number. Now, in our case, π of π₯ is equal
to π₯ multiplied by π to the power of negative π₯ squared. So we can substitute this into our
formula. Now, we can pick π to be any real
number. Letβs choose π to be zero. However, you can pick any value of
π that you like.

Letβs now try to solve our
integrals. And we can do this by trying to
find the antiderivative of π₯ multiplied by π to the negative π₯ squared. We notice that the derivative of π
to the power of negative π₯ squared is equal to negative two π₯π to the power of
negative π₯ squared. Now, we can divide both sides of
this equation by the constant term negative two. And we obtain this. On the left-hand side, we have a
constant term multiplying a derivative. Using the derivative rule for
scalar multiplication, we can move the negative one-half inside the derivative. And so, we found our antiderivative
of π₯ multiplied by π to the power of negative π₯ squared. Itβs negative one-half π to the
power of negative π₯ squared.

Using this antiderivative, weβre
able to evaluate the integrals inside our limits. Next, we can substitute in our
upper and lower bounds. We obtained the limit as π tends
to negative infinity of negative one-half multiplied by π to the power of negative
zero squared minus negative one-half multiplied by π to the power of negative π
squared. Plus the limit as π‘ tends to
positive infinity of negative one-half multiplied by π to the power of negative π‘
squared minus negative one-half multiplied by π to the power of negative zero
squared. Since π to the power of negative
zero squared is simply π to the power of zero, which is just one, the negative
one-half multiplied by π to the power of negative zero squared terms is simply
negative one-half. And now, we can sort out the double
negative signs which will give us a positive sign.

Next, we can use one of the
properties of limits, which tells us that the limit of a sum of functions is equal
to the sum of the limits of their functions. And so, weβre able to split our
limits up like this. Here, we have two limits of
constant terms. So these limits are simply equal to
that constant term. The first one being negative
one-half and the second just being one-half. Therefore, when we add these two
terms together, they will cancel out with one another. Now, we just need to worry about
these two remaining limits. Letβs consider the one on the right
first.

As π‘ gets larger and larger and
larger, negative π‘ squared will be getting larger and larger but in the negative
direction. Therefore, π to the power of
negative π‘ squared will be getting closer and closer to zero. And so, we can say that the limit
as π‘ tends to infinity of negative one-half π to the power of negative π‘ squared
is equal to zero. Similarly, as π gets larger and
larger in the negative direction, π squared will be getting larger and larger in
the positive direction. And so, negative π squared will be
getting larger in the negative direction. And when this happens, π to the
power of negative π squared gets closer and closer to zero. Therefore, this limit is also equal
to zero.

Since the negative one-half and the
positive one-half canceled out with one another and these two limits are both equal
to zero, therefore, we can say that the integral from negative infinity to positive
infinity of π₯ multiplied by π to the power of negative π₯ squared with respect to
π₯ converges to β and is therefore equal to β zero.

We have now seen a variety of
examples of how to evaluate limits with infinite bounds. Letβs recap some key points of this
video.

Key Points

The integral from π to infinity of
π of π₯ with respect to π₯ is equal to the limit as π‘ tends to infinity of the
integral from π to π‘ of π of π₯ with respect to π₯, where π‘ is greater than
π. The integral from negative infinity
to π of π of π₯ with respect to π₯ is equal to the limit as π‘ tends to negative
infinity of the integral from π‘ to π of π of π₯ with respect to π₯, where π‘ is
less than π. And the integral from negative
infinity to positive infinity of π of π₯ with respect to π₯ is equal to the limit
as π tends to negative infinity of the integral from π to π of π of π₯ with
respect to π₯ plus the limit as π‘ tends to positive infinity of the integral from
π to π‘ of π of π₯ with respect to π₯, where π is any real number. And an integral with an infinite
bound converges if the limit exists and diverges, if it does not.