# Question Video: Determining Whether the Improper Integral of a Function Containing a Logarithmic Function Is Convergent or Divergent Using Integration by Substitution Mathematics • Higher Education

Determine whether the integral β«_(1)^(β) (ln π₯)/π₯ dπ₯ is convergent or divergent.

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### Video Transcript

Determine whether the integral, the integral from one to β, 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 β. 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 β. Then, if we find this limit to exist, the integral is convergent. If the limit does not exist or itβs infinite, then the integral is divergent.

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

But remember, we also need to rewrite our limits in terms of π’. The lower limit is one. So when π₯ equals one, π’ equals the natural log of one. Because remember, we chose π’ 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 π‘. And when π₯ is π‘, π’ is the natural log of π‘. So with the substitution of π’ equals the natural log of π₯, these are our new limits of integration. So now letβ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 itβs not raised to the power of negative one, we just add one to the power and divide by the new power.

π’ on its own is the same as π’ to the power of one. So π’ integrates to π’ squared over two. We evaluate this by substituting the upper limit into π’ squared over two and subtracting the lower limit substituted into π’ squared over two. But zero squared over two is just zero. So this is just the limit as π‘ approaches β of the natural log of π‘ squared over two. Well, as π‘ approaches β, the natural log of π‘ also approaches β. So the natural log of π‘ squared over two also approaches β.

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.