Question Video: Evaluating the Improper Integral of a Function with Infinite Limits of Integration by Using Integration by Substitution Mathematics

Video Transcript

The integral from 𝑒 to ∞ of one divided by 𝑥 times the natural logarithm squared of 𝑥 with respect to 𝑥 is convergent. What does it converge to?

In this question, we’re given a definite integral. And we’re told that this definite integral is convergent. We need to determine the value that it converges to. And to do this, we first need to notice something about our integral. The upper limit of integration is ∞. And when one of our limits of integration are positive or negative ∞, we call this an improper integral. So to evaluate our integral, we need to recall how we deal with improper integrals. We recall the integral from 𝑎 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 will be equal to the limit as 𝑡 approaches ∞ of the integral from 𝑎 to 𝑡 of 𝑓 of 𝑥 with respect to 𝑥.

And there’s a few things worth pointing out here. This is only true if this limit exists. This is what the question means when it says our integral is convergent. It means that this limit is convergent. Next, because our values of 𝑡 are approaching ∞ and 𝑎 is the lower limit of integration, we can assume that 𝑡 is bigger than 𝑎. So to answer this question, the first thing we’re going to want to do is set up this limit for the improper integral given to us in the question. We can see the lower limit of integration is 𝑒. So we’ll set our value of 𝑎 equal to 𝑒. This gives us the integral from 𝑒 to ∞ of one over 𝑥 times the natural logarithm squared of 𝑥 with respect to 𝑥 is equal to the limit as 𝑡 approaches ∞ of the integral from 𝑒 to 𝑡 of one over 𝑥 times the natural logarithm squared of 𝑥 with respect to 𝑥, provided that this limit exists.

However, the question actually tells us that this limit is convergent. We just need to find the value that this limit converges to. And to do this, we’re going to need to evaluate our definite integral. There’s a few different options on how we could do this. The easiest way is to use substitution. We’ll use the substitution 𝑢 is equal to the natural logarithm of 𝑥. The reason we do this is the derivative of the natural logarithm of 𝑥 with respect to 𝑥 is one over 𝑥. And we can see this appears in our integrand. So we start with our substitution 𝑢 is equal to the natural logarithm of 𝑥. We need to differentiate both sides with respect to 𝑥.

By doing this, we get d𝑢 by d𝑥 is equal to one over 𝑥. And of course, we know d𝑢 by d𝑥 is not a fraction. However, when we’re using integration by substitution, it can often help to treat it a little bit like a fraction. We can get the equivalent statement in terms of differentials d𝑢 is equal to one over 𝑥d𝑥. And there is one thing worth pointing out about our substitution. Remember that 𝑡 is approaching ∞, and so we can assume that 𝑡 is bigger than 𝑒. The reason we do this is because we want our integrand to be continuous on the entire interval of integration. In this case, we can see our integrand is the composition and product of continuous functions.

So it will be continuous across its entire domain. And we can just find the domain of this function. First, we’re not allowed to divide by zero, so 𝑥 is not allowed to be equal to zero. Next, we can’t take the natural logarithm of a number less than or equal to zero. And of course, the natural logarithm of one is equal to zero. And we can’t divide by this. However, we don’t need to worry about this because our values of 𝑥 start at 𝑥 is equal to 𝑒. And we know that our integrand is continuous for these values of 𝑥. So before we continue with our substitution, we now need to change the limits of integration. We need to substitute these values of 𝑥 into our expression for 𝑢.

We’ll start with the new upper limits of integration. We substitute 𝑥 is equal to 𝑡 into our expression for 𝑢. We get 𝑢 is equal to the natural logarithm of 𝑡. And for the lower limit of integration, we substitute 𝑥 is equal to 𝑒. We get 𝑢 is equal to the natural logarithm of 𝑒 which we know is equal to one. We’re now ready to apply our substitution 𝑢 is equal to the natural logarithm of 𝑥. First, we showed that the new lower limit of integration was one and the new upper limit of integration is the natural logarithm of 𝑡. Next, we want to replace one over 𝑥d𝑥 with d𝑢. And this just leaves our integrand as one over the natural logarithm squared of 𝑥.

But 𝑢 is the natural logarithm of 𝑥. So this is just one over 𝑢 squared. So this gives us the limit as 𝑡 approaches ∞ of the integral from one to the natural logarithm of 𝑡 of one over 𝑢 squared with respect to 𝑢. And remember, 𝑡 is bigger than 𝑒. So the natural logarithm of 𝑡 is bigger than one. So our integrand is continuous on the interval of integration and we can just use the power rule for integration. Of course, to do this, it might first help to rewrite our integrand as 𝑢 to the power of negative two. We want to add one to our exponent of 𝑢 and then divide by this new exponent.

Doing this and simplifying, we then get the limit as 𝑡 approaches ∞ of negative 𝑢 to the power of negative one evaluated at the limits of integration. 𝑢 is equal to one and 𝑢 is equal to the natural logarithm of 𝑡. Next, we’re going to want to evaluate this at the limits of integration. But we’ll, first, rewrite our antiderivative as negative one over 𝑢. Evaluating our antiderivative at the limits of integration, we get the limit as 𝑡 approaches ∞ of negative one divided by the natural logarithm of 𝑡 plus one divided by one. And at this point, we can just evaluate this limit directly. Our limit is as 𝑡 is approaching ∞. As 𝑡 approaches ∞, we can notice that the denominator of our first term is going to approach ∞. However, the numerator in this first term remains constant. So its denominator is growing without bound.

This means our first term limits to give us zero. Of course, our second term is just a constant. It doesn’t change as the value of 𝑡 changes. So its limit will just be equal to itself, one divided by one, which is of course just equal to one. Therefore, because this limit was convergent and equal to one, we can say our original improper integral must be equal to one. Therefore, we were able to show the integral from 𝑒 to ∞ of one divided by 𝑥 times the natural logarithm squared of 𝑥 with respect to 𝑥 converges to one.