Video: Determining Whether the Improper Integral of a Function Containing a Logarithmic Function Is Convergent or Divergent Using Integration by Substitution

Determine whether the integral ∫_(1)^(∞) (ln π‘₯)/π‘₯ dπ‘₯ is convergent or divergent.

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Video Transcript

Determine whether the integral, the integral from one to ∞, of the natural log of π‘₯ over π‘₯ with respect to π‘₯ is convergent or divergent.

This is an improper integral. Recalling an improper integral is an integral where one or both of the limits of integration are ∞. To deal with these kinds of integrals, we replace the infinite limit with a variable and then evaluate the integral taking the limit as that variable approaches ∞. Then, if we find this limit to exist, the integral is convergent. If the limit does not exist or it’s infinite, then the integral is divergent.

So let’s begin by replacing the infinite limit with a variable, which we’ll call 𝑑, which means that we now take the limit as 𝑑 approaches ∞ of this integral. Now, we need to integrate this function here, which we’ll do by substitution. If we choose the substitution 𝑒 equals the natural log of π‘₯, then we remember that the natural log of π‘₯ differentiates to one over π‘₯. So d𝑒 by dπ‘₯ is one over π‘₯. And this gives us that d𝑒 equals one over π‘₯ dπ‘₯. So now, we can substitute ln of π‘₯ with 𝑒 and the one over π‘₯ dπ‘₯ with d𝑒.

But remember, we also need to rewrite our limits in terms of 𝑒. The lower limit is one. So when π‘₯ equals one, 𝑒 equals the natural log of one. Because remember, we chose 𝑒 to be equal to the natural log of π‘₯. And then, we remember that the natural log of one is always zero. The upper limit is 𝑑. And when π‘₯ is 𝑑, 𝑒 is the natural log of 𝑑. So with the substitution of 𝑒 equals the natural log of π‘₯, these are our new limits of integration. So now let’s calculate this integral. To do this, we remember the power rule for integration, which tells us that to integrate a variable, as long as it’s not raised to the power of negative one, we just add one to the power and divide by the new power.

𝑒 on its own is the same as 𝑒 to the power of one. So 𝑒 integrates to 𝑒 squared over two. We evaluate this by substituting the upper limit into 𝑒 squared over two and subtracting the lower limit substituted into 𝑒 squared over two. But zero squared over two is just zero. So this is just the limit as 𝑑 approaches ∞ of the natural log of 𝑑 squared over two. Well, as 𝑑 approaches ∞, the natural log of 𝑑 also approaches ∞. So the natural log of 𝑑 squared over two also approaches ∞.

Remember, we said if the limit does not exist or is infinite, then the improper integral is divergent. And so, we can conclude that this improper integral is divergent.

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