Video: Improper Integrals: Discontinuous Integrands

In this video, we will learn how to evaluate improper integrals where the integrand has a vertical asymptote.

17:38

Video Transcript

Improper integrals: discontinuous integrands.

In this video, we’ll 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 we’ll call 𝑓 of π‘₯, has a discontinuity on the closed interval between π‘Ž and 𝑏, where π‘Ž and 𝑏 are the limits of integration. This is the case of a discontinuous integrand, which we’ll 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 we’ll see in a moment, we cannot use standard techniques to evaluate this integral. And we’ll 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 𝑓 is a continuous function on the closed interval between π‘Ž and 𝑏, then the integral between π‘Ž and 𝑏 of lowercase 𝑓 of π‘₯ with respect to π‘₯ is equal to uppercase 𝐹 of 𝑏 minus uppercase 𝐹 of π‘Ž. And this is where uppercase 𝐹 is any antiderivative of lowercase 𝑓.

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. Let’s 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, let’s 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 we’ll reexpress in this form. And then we’ll input our limits of integration, negative one and one. We’re 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 𝑓 is continuous on the interval which is closed at π‘Ž and open at 𝑏 and has a discontinuity at 𝑏. Then the integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ is equal to the limit as 𝑑 approaches 𝑏 from the left of the integral between π‘Ž and 𝑑 of 𝑓 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 𝑓 is continuous on the interval which is open at π‘Ž and closed at 𝑏 and has a discontinuity at π‘Ž. Then the integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ is equal to the limit as 𝑑 approaches π‘Ž from the right of the integral between 𝑑 and 𝑏 of 𝑓 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 𝑑 into both of our definitions. And 𝑑 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 𝑓 has a discontinuity at 𝑐, where 𝑐 is greater than π‘Ž and less than 𝑏, and both the integral between π‘Ž and 𝑐 of 𝑓 of π‘₯ with respect to π‘₯ along with the integral between 𝑐 and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ are convergent. Then we define that the integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ is equal to the integral between π‘Ž and 𝑐 of 𝑓 of π‘₯ with respect to π‘₯, added to the integral between 𝑐 and 𝑏 of 𝑓 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 we’re 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 we’ve outlined here. It’s worth noting that this definition is only valid if both of the two smaller integrals that we’re 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 we’ve 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 it’s 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.

It’s 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, there’s 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 we’ll 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 𝑓 is a continuous function on the closed interval between π‘Ž and 𝑏, which are our limits of integration. We’ve 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 we’re 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 𝑓 is continuous on the interval which is open at π‘Ž and closed at 𝑏 and discontinuous at π‘Ž. Then the integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ is equal to the limit as 𝑑 approaches π‘Ž from the right of the integral between 𝑑 and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. If this limit exists and is finite. Don’t worry too much about the 𝑑 that we’ve introduced since this is just a dummy variable which helps us to evaluate our limit. Let us now apply this to our question. 𝑓 of π‘₯, our integrand, is one over π‘₯. The lower limit of integration, π‘Ž, is zero. And the upper limit of integration, 𝑏, is one.

Our definition tells us that our integral is equal to the limit as 𝑑 approaches zero from the right of the integral between 𝑑 and one of one over π‘₯ with respect to π‘₯. Okay. Let’s 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 𝑑 and one. We’re then left with the following expression. The laws of limits allow us to apply our limit individually to each of these terms. So let’s do so and see what happens to them. Well, our first term has no 𝑑 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. 𝑒 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 𝑑 approaches zero from the right, the natural log of the absolute value of 𝑑 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 it’s 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. Let’s look at one of these now.

The integral between zero and πœ‹ over two of cos πœƒ divided by the square root of sin πœƒ with respect to πœƒ 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 we’re looking at an improper integral. Now, since we don’t see any infinities as our limits of integration, it’s also implied that our integrand has a discontinuity on the closed interval between zero and πœ‹ over two, which are the limits of the integral. Let’s look for this discontinuity. Since the integrand that we’ve been given is a quotient, we should try and find where its denominator is equal to zero. The square root of sin πœƒ is zero, where sin πœƒ itself is also equal to zero. This occurs when πœƒ is equal to zero plus π‘›πœ‹, which is of course just π‘›πœ‹. And 𝑛 here is an integer. So here we see that our denominator is equal to zero when πœƒ 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, let’s now check what happens to our numerator when πœƒ is equal to zero. Our numerator is cos πœƒ. So this becomes cos of zero, which is equal to one. We do this to check that when πœƒ 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œƒ is equal to zero, our function is equal to one over zero, which is the case of an infinite discontinuity. Okay. We’ve 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 𝑓 is a continuous function on the interval which is open at π‘Ž and closed at 𝑏 and discontinuous at π‘Ž. Then the integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ is equal to the limit as 𝑑 approaches π‘Ž from the right of the integral between 𝑑 and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. If this limit exists and is finite.

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

Now that we’ve done the legwork on our integral, let’s use this result to move forward with our question. We use the antiderivative that we’ve just found for our integral. And we get the following. We now input our limits of integration 𝑑 and πœ‹ over two. And we’re left with the following limit. We can now take a direct substitution approach to our limit. And we can evaluate since sin of πœ‹ over two is equal to one. And sin of zero is equal to zero. We’re 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 we’re 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, we’ve 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, we’ve been given an improper integral with a discontinuous integrand. To move forward, we’ll 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, let’s 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 we’ll call 𝑓 of π‘₯, for the type of discontinuity that we expect to see when π‘₯ equals four. Trying to evaluate 𝑓 of four, we’re left with four over zero, which implies that we’re looking at an infinite discontinuity, as opposed to if it’d been zero over zero, which would imply a removable discontinuity.

So we’ve ignored the discontinuity at π‘₯ equals negative four since it does not interact with our limits of integration. We’ve 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, 𝑐, into the sum of two smaller integrals. Let us apply this to our question, where we have our integrand 𝑓 of π‘₯, our lower limit π‘Ž, our upper limit 𝑏, and our discontinuity 𝑐. 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. We’ll therefore need to check our two smaller integrals for convergence. However, before we do this, the technique that we’ll be using to solve these two integrals will be a 𝑒-substitution. And so we might as well get this out of the way first. The substitution that we’ll be using is that 𝑒 is equal to π‘₯ squared minus 16. Differentiating with respect to π‘₯, we get that d𝑒 by dπ‘₯ is equal to two π‘₯. And an equivalent statement for this is that d𝑒 is equal to two π‘₯ dπ‘₯. It turns out that a more useful equation for us is that half d𝑒 is equal to π‘₯ dπ‘₯.

Okay. Since we’re working with definite integrals, we should also pay attention to our limits of integration. When π‘₯ is equal to zero, 𝑒 is equal to negative 16. When π‘₯ is four, 𝑒 is zero. And when π‘₯ is five, 𝑒 is nine. So let us now perform our substitutions, replacing π‘₯ squared minus 16 with 𝑒, π‘₯ dπ‘₯ with a half d𝑒, and our limits of integration as we’ve 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, it’s 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, we’re 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. We’ll 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 𝑒 is the natural logarithm of the absolute value of 𝑒. We’ll 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, we’re 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 we’re summing, are convergent. Since the first one we checked turned out to be divergent, we don’t 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.

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