### Video Transcript

The integral the integral from one to β of π to the power of negative one over π₯ divided by π₯ squared with respect to π₯ is convergent. What does it converge to?

Weβre given an integral where one of the limits of our integral is β. And weβre told that this integral is convergent. And we recall, for a constant π and a function π of π₯, we say that the integral from π to β of π of π₯ with respect to π₯ is the limit as π‘ approaches β of the integral from π to π‘ of π of π₯ with respect to π₯. And we say that the integral from π to β of π of π₯ is convergent if the limit as π‘ approaches β of the integral from π to π‘ of π of π₯ is convergent.

In the question, the lower limit of our integral is one. So weβll set π equal to one. And our integrand is equal to π to the power of negative one over π₯ divided by π₯ squared. So weβll set our function π of π₯ equal to this. So the definite integral given to us in the question is equal to the limit as π‘ approaches β of the integral from one to π‘ of π to the power of negative one over π₯ divided by π₯ squared with respect to π₯. And since weβre told that our integral is convergent, this limit must be convergent. We now want to evaluate this integral. However, we see that our intgrand π to the power of negative one over π₯ divided by π₯ squared is not in the standard form which we know how to integrate.

The easiest way to see how to evaluate this integral is to attempt to differentiate π to the power of negative one divided by π₯. And we can differentiate this by using the fact for any differentiable function π the derivative of π to the power of π of π₯ with respect to π₯ is equal to π prime of π₯ times π to the power of π of π₯.

In our case, we have that π of π₯ is equal to negative one divided by π₯. We can differentiate this using the power rule for differentiation. We multiply by the exponent of negative one and then reduce the exponent by one to give us an exponent of negative two. Giving us that our derivative is equal to negative one times negative one over π₯ squared times π to the power of negative one over π₯. Which we can simplify to just give us π to the power of negative one over π₯ divided by π₯ squared, which is just equal to our function π of π₯.

So if the derivative of π to the power of negative one over π₯ is equal to π of π₯, then, in particular, π to the power of negative one over π₯ is an antiderivative of π of π₯. And using the fact that π to the power of negative one over π₯ is an antiderivative of our function π of π₯, we can evaluate the integral inside of our limit. This gives us the limit as π‘ approaches β of π to the power of negative one over π₯ evaluated at the limits of our integral π₯ is equal to one and π₯ is equal to π‘.

So weβll clear some space and then evaluate this at the limits of our integral. This gives us the limit as π‘ approaches β of π to the power of negative one over π‘ minus π to the power of negative one divided by one. We can just evaluate negative one divided by one to be equal to one. So our second term inside of our limit is π to the power of negative one, which we can write as one divided by π. Now, to help us evaluate this limit, weβll use the fact that the limit of a sum is equal to the sum of the limits. This gives us the limit as π‘ approaches β of π to the power of negative one over π‘ plus the limit as π‘ approaches β of negative one divided by π.

We have that negative one divided by π is a constant with respect to π‘. Its value does not change when π‘ changes. Therefore, we can evaluate this limit to just be negative one divided by π. There are a few different ways to evaluate our first limit, the limit as π‘ approaches β of π to the power of negative one divided by π‘. Weβre going to use the fact that the exponential function is continuous. We know if a function π of π₯ is continuous when π₯ is equal to π and the limit as π₯ approaches β of a function π of π₯ is equal to π. Then we can evaluate the limit as π₯ approaches β of the composition π of π₯ and π of π₯ as just π is equal to π.

So by using this and the fact that the limit as π‘ approaches β of negative one divided by π‘ is equal to zero, we can evaluate the first limit as π to the zeroth power. And any number to the zeroth power is just equal to one. Therefore, our limit converged to one divided by one minus π. And this limit was equal to the integral from one to β of π to the power of negative one over π₯ divided by π₯ squared with respect to π₯, which must also be equal to one minus one over π.