Video: Determining Whether the Integrals of Polynomial Functions Multiplied by Natural Logarithm Functions Are Convergent or Divergent and Finding Their Values

The integral ∫_(0)^(1) π‘Ÿ ln π‘Ÿ dπ‘Ÿ is convergent. What does it converge to?

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Video Transcript

The integral from zero to one of π‘Ÿ multiplied by the natural logarithm of π‘Ÿ with respect to π‘Ÿ is convergent. What does it converge to?

We might be tempted to try and evaluate this integral by using the fundamental theorem of calculus. Which tells us if the function lowercase 𝑓 is continuous on the closed interval from π‘Ž to 𝑏, and capital 𝐹 is an antiderivative of our function lowercase 𝑓. Then, the integral from π‘Ž to 𝑏 of lowercase 𝑓 of π‘Ÿ with respect to π‘Ÿ is equal to our antiderivative function evaluated at 𝑏 minus our antiderivative function evaluated at π‘Ž. But this is only valid if our integrand, π‘Ÿ multiplied by the natural logarithm of π‘Ÿ, is continuous on the closed interval from zero to one.

However, we see that if we plug π‘Ÿ is equal to zero into our integrand, we get zero multiplied by the natural logarithm of zero, which is not defined. So, our function is not defined when π‘Ÿ is equal to zero. If our integrand is not defined when π‘Ÿ is equal to zero, it can’t be continuous on the closed interval from zero to one.

There are several different ways of approaching this. We could analyze what type of discontinuity we have when π‘Ÿ is equal to zero. Or we can notice that our function π‘Ÿ multiplied by the natural logarithm of π‘Ÿ is continuous on the interval from zero to one, not including zero. This is because π‘Ÿ is continuous on this interval and the natural logarithm of π‘Ÿ is also continuous on this interval. So, their product will be continuous on this interval.

We can then use our definition for an improper integral. If 𝑓 is a continuous function on the interval from π‘Ž to 𝑏 which does not include π‘Ž. And it is discontinuous at the point π‘Ž. Then, we can introduce a dummy variable 𝑑. Which tells us the integral from π‘Ž to 𝑏 of 𝑓 of π‘Ÿ with respect to π‘Ÿ is equal to the limit as 𝑑 approaches π‘Ž from the right of the integral from 𝑑 to 𝑏 of 𝑓 of π‘Ÿ with respect to π‘Ÿ.

And we’ve already shown that this applies to our integrand function π‘Ÿ natural logarithm of π‘Ÿ. So, we can use this to see that the integral from zero to one of π‘Ÿ natural logarithm of π‘Ÿ with respect to π‘Ÿ is equal to the limit as 𝑑 approaches zero from the right of the integral from 𝑑 to one of π‘Ÿ natural logarithm of π‘Ÿ with respect to π‘Ÿ.

We’re now ready to start evaluating this integral. However, we can’t integrate π‘Ÿ natural logarithm of π‘Ÿ directly since it’s not in a standard form. So, we’re going to need to perform some manipulation. We could attempt to simplify this by using a substitution. And this would work. However, it’s simpler to use integration by parts. We recall that integration by parts tells us that the integral of 𝑒𝑣 prime with respect to π‘Ÿ is equal to 𝑒𝑣 minus the integral of 𝑒 prime 𝑣 with respect to π‘Ÿ.

Now, we need to decide which function to set as 𝑒 and which function to set as 𝑣 prime. We’re going to use the method of L.I.A.T.E. The L in L.I.A.T.E stands for logarithm. And it tells us: if possible, we should attempt to set 𝑒 to be equal to a logarithmic function. We can do this in this case. So, we’ll set 𝑒 equal to the natural logarithm of π‘Ÿ. And that leaves 𝑣 prime to be equal to π‘Ÿ.

We differentiate both sides of our equation with respect to π‘Ÿ. This gives us that 𝑒 prime is equal to the derivative of the natural logarithm of π‘Ÿ, which is equal to one divided by π‘Ÿ. If 𝑣 prime is equal to π‘Ÿ, then 𝑣 is equal to the antiderivative of π‘Ÿ. We get this by adding one to the exponent and then dividing by the new exponent. This gives us that 𝑣 is equal to π‘Ÿ squared over two.

So, by using integration by parts, we have that our integral is equal to the limit as 𝑑 approaches zero from the right of the natural logarithm of π‘Ÿ multiplied by π‘Ÿ squared over two evaluated at the limits of 𝑑 and one. Minus the integral from 𝑑 to one of one over π‘Ÿ multiplied by π‘Ÿ squared over two with respect to π‘Ÿ.

Evaluating the limits of the first term in our limit gives us the natural logarithm of one multiplied by one squared over two minus the natural logarithm of 𝑑 multiplied by 𝑑 squared over two. And then, we can simplify the integrand inside our limit by canceling the shared factor of π‘Ÿ in the numerator and the denominator. This gives us the integral from 𝑑 to one of π‘Ÿ over two with respect to π‘Ÿ.

We see that the first term inside of our limit is a constant. And the natural logarithm of one is equal to zero. So, our first term is equal to zero. Similarly, since our limit is as 𝑑 is approaching zero from the right, we have that 𝑑 is always positive. Therefore, the natural logarithm of 𝑑 and 𝑑 squared over two are both approaching zero. So, our second term in our limit is equal to zero. Therefore, we only need to evaluate the integral in our final term.

Well, the antiderivative of π‘Ÿ over two is π‘Ÿ squared over four, which we get by adding one to the exponent and dividing by this new exponent. This gives us the limit as 𝑑 approaches zero from the right of negative one multiplied by π‘Ÿ squared over four evaluated between the limits of π‘Ÿ is equal to 𝑑 and π‘Ÿ is equal to one. Evaluating at the limits of our integral gives us the limit as 𝑑 approaches zero from the right of negative one squared over four plus 𝑑 squared over four.

We have that negative one squared over four does not vary based on the value of 𝑑. So, the limit as 𝑑 approaches zero from the right of this term is equal to negative a quarter. However, as 𝑑 is approaching zero from the right, the numerator of our second term is approaching zero, but the denominator remains constant. So, our second term approaches zero. Therefore, we have shown that the integral from zero to one of π‘Ÿ natural logarithm of π‘Ÿ with respect to π‘Ÿ converges to negative a quarter.

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