Video: Finding the Unknown in a Rational Function That Makes the Improper Integral of the Function Convergent and Evaluating That Integral

Consider the integral ∫_(0)^(1) (1/π‘₯^(𝑝)) dπ‘₯. Find all possible values of 𝑝 for which the integral is convergent. Evaluate the integral for those values of 𝑝.

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Video Transcript

Consider the integral from zero to one of one divided by π‘₯ to the power of 𝑝 with respect to π‘₯. Find all possible values of 𝑝 for which the integral is convergent. Evaluate the integral for those values of 𝑝.

The question gives us a definite integral, where our integrand is one divided by π‘₯ to the power 𝑝 for some constant 𝑝. The question wants us to find all values of 𝑝 for which this integral is convergent. Then it wants us to evaluate this integral for all of the values of 𝑝 for which the integral is convergent.

To answer this question, let’s think about our integrand function one divided by π‘₯ to the power of 𝑝. For starters, if our integrand function is continuous on the closed interval from zero to one, then the integral is proper. And in particular, we could use the fundamental theorem of calculus to evaluate this integral. However, if our integrand was not continuous on the closed interval from zero to one, then the integral would be improper. To deal with an improper integral, we’ll want to know all the points where π‘₯ to the power of negative 𝑝 would be discontinuous.

We see that π‘₯ to the power of negative 𝑝 is a rational function. In fact, when 𝑝 is less than or equal to zero, it will be a polynomial which is continuous across the whole real numbers. So it will be a proper integral. So we’re only interested in the cases when 𝑝 is greater than zero. So when 𝑝 is greater than zero, we have π‘₯ to the power of negative 𝑝 is the rational function one divided by π‘₯ to the power of 𝑝. So it’s continuous on the interval from zero to one, where zero is not included, but is not continuous when π‘₯ is equal to zero.

One way of seeing this is when 𝑝 is greater than zero, one divided by zero to the power of 𝑝 gives us the indeterminant form of one divided by zero. So our function is not defined when π‘₯ is equal to zero. We also briefly discussed when 𝑝 is less than or equal to zero, π‘₯ to the power of negative 𝑝 is actually a polynomial. So it is continuous on the closed interval from zero to one. And we can evaluate these integrals by just using the fundamental theorem of calculus. In the case where 𝑝 was greater than zero, we’ve shown that the only point of discontinuity on our interval was when π‘₯ is equal to zero.

So we can check the convergence of these integrals by using one of our rules for improper integrals. Which tells us the integral from zero to one of π‘₯ to the power of negative 𝑝 with respect to π‘₯ is equal to the limit as 𝑑 approaches zero from the right of the integral from 𝑑 to one of π‘₯ to the power of negative 𝑝 with respect to π‘₯. This is true if the limit exists and is finite. And we remember this rule for integrals is only true when our integrand is continuous on the interval from zero to one, not including zero, and is discontinuous at the point when π‘₯ is equal to zero.

So we’re now ready to check the convergence of this integral. We’ll do it in parts. When 𝑝 is less than or equal to zero, we’ll try to use the fundamental theorem of calculus. And when 𝑝 is greater than zero, we’ll try to use our rule for improper integrals. We’ll need to check that the limit exists and is finite.

So let’s clear some space and work on each of these cases individually. We’ll start when 𝑝 is less than or equal to zero. This gives us the integral from zero to one of one divided by π‘₯ to the power of 𝑝 with respect to π‘₯. We rewrite our integrand as π‘₯ to the power negative 𝑝. And then we recall that our integrand is continuous on the closed interval from zero to one because 𝑝 is less than or equal to zero. We know that 𝑝 is not equal to one. So we can integrate this by adding one to the exponent and dividing by this new exponent.

We then substitute in the limits of our integral. We see the one to the power of negative 𝑝 plus one is just equal to one and zero to the power of negative 𝑝 plus one is just equal to zero. So this simplifies to just one divided by negative 𝑝 plus one. And we’ve shown that, in the case when 𝑝 is less than or equal to zero, our integral is convergent.

Now, let’s do the cases when 𝑝 is greater than zero. We have that our integrand is continuous on the interval from zero to one, not including zero, and is not continuous when π‘₯ is equal to zero. So by using our rules for improper integrals, we have the integral from zero to one of π‘₯ to the power of negative 𝑝 with respect to π‘₯ is equal to the limit as 𝑑 approaches zero from the right of the integral from 𝑑 to one of π‘₯ to the power of negative 𝑝 with respect to π‘₯.

It would be tempting at this point to evaluate this integral by adding one to our exponent and then dividing by the new exponent. But we have to be careful because if 𝑝 is equal to one, then we’re integrating π‘₯ to the power of negative one. And using this will give us an indeterminate form. So we’re going to have to separate this into two cases. We’ll start when 𝑝 is not equal to one. Since 𝑝 is not equal to one, we can add one to the exponent and then divide by this new exponent to find our antiderivative.

Next, we evaluate this at the limits of our integral, when π‘₯ is equal to 𝑑 and π‘₯ is equal to one. We see that our first term does not change as the value of 𝑑 changes. So its limit is equal to itself. And its numerator, one to the power of negative 𝑝 plus one, is just equal to one. So this gives us our new expression. We now see that, inside our limit, the denominator of negative 𝑝 plus one does not vary as 𝑑 varies. The only part which varies is our numerator, 𝑑 to the power of negative 𝑝 plus one.

To evaluate this limit, we recall two of our limit laws. If 𝑛 is greater than zero, the limit of π‘₯ to the 𝑛th power as π‘₯ approaches zero is equal to zero. And if 𝑛 is less than zero, then the limit of π‘₯ to the 𝑛th power as π‘₯ approaches zero does not exist. Using this, we see if negative 𝑝 plus one is greater than zero, then our numerator will converge and be equal to zero. And if negative 𝑝 plus one is less than zero, then our numerator will not converge.

So we’ve shown when 𝑝 is greater than one, the numerator of our limit does not converge and so our integral does not converge. We must be careful with the convergence part here. We might be tempted to say that it converges when 𝑝 is less than one. However, we’ve already assumed that 𝑝 is greater than zero. So we’ve shown that our integral will converge when zero is less than 𝑝 is less than one. In fact, since the numerator of our limit is approaching zero, the limit is approaching zero. So we can evaluate this integral to be equal to one divided by negative 𝑝 plus one. And this is only when zero is less than 𝑝 is less than one.

So we’ve now checked the convergence of the integral when 𝑝 is less than or equal to zero, when 𝑝 is between zero and one, or when 𝑝 is greater than one. We now only need to check the case when 𝑝 is equal to one. So let’s clear some space and work on the case when 𝑝 is equal to one.

We have that our integrand π‘₯ to the power of negative one is continuous on the interval from zero to one, not including zero, and is not continuous when π‘₯ is equal to zero. So by using our rules for improper integrals, we have that this integral is equal to the limit as 𝑑 approaches zero from the right of the integral from 𝑑 to one of π‘₯ to the power of negative one with respect to π‘₯ if this limit exists and is finite. We have the antiderivative of π‘₯ to the power of negative one is the natural logarithm of the absolute value of π‘₯.

We then evaluate this at the bounds of our integral. We see the natural logarithm of the absolute value of one is a constant which is just equal zero. We know the limit as π‘₯ approaches zero from the right of the natural logarithm of π‘₯ does not exist. So the limit as 𝑑 approaches zero from the right of negative the natural logarithm of the absolute value 𝑑 also does not exist. Therefore, since this limit does not exist, our integral does not converge when 𝑝 is equal to one.

Therefore, we’ve shown, for all values of 𝑝, our integral is only convergent when 𝑝 is less than or equal to zero or when zero is less than 𝑝 is less than one. We can combine these into a single inequality. So we’ve shown that our integral will converge when 𝑝 is less than one. In fact, in both cases where our integral converged, we got that we can evaluate our integral to be one divided by negative 𝑝 plus one. And by rearranging this denominator, we’ve shown when this integral converges, it evaluates to be equal to one divided by one minus 𝑝.

Therefore, we’ve shown that the integral from zero to one of one divided by π‘₯ to the power 𝑝 with respect to π‘₯ will converge when 𝑝 is less than one. And it evaluates in these cases to be one divided by one minus 𝑝.

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