# Lesson Video: Improper Integrals: Infinite Limits of Integration Mathematics • Higher Education

In this video, we will learn how to evaluate improper integrals where one or more of the endpoints approach infinity.

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### Video Transcript

Improper Integrals: Infinite Limits of Integration

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

Now, letβs start by considering the following function π of π₯ and letβs 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 doesnβt really seem to make sense. How can infinity be a bound when itβs not a number? And how can we find an area that goes on to infinity? Now, although this doesnβt 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. Itβs 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 thatβs this integral in our formula. We also require that the limit of this integral as π‘ tends to infinity exists too. So thatβs 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 weβre 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.

Letβs 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, weβre 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 thatβs negative one. And we mustnβt 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. Letβs suppose we had a function π of π₯, which looks something like this. And suppose weβre 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 letβs suppose itβs 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.

Letβs 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. Letβs 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. Itβs 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. Weβre 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, weβre 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. Weβre 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, weβre 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, weβll be combining the two formula weβve seen in this video so far in order to form a third formula. Weβll 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 weβre 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.

Letβs 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 letβs 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. Letβs choose π to be zero. However, you can pick any value of π that you like.

Letβs 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. Itβs negative one-half π to the power of negative π₯ squared.

Using this antiderivative, weβre 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, weβre 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. Letβs 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. Letβs 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.