Video Transcript
Determine whether the integral
between zero and one of one over 𝑥 with respect to 𝑥 is convergent or
divergent.
For this question, we have been
asked to evaluate a definite integral of the function one over 𝑥. This should be a familiar function
to us. And we clearly see that it has an
infinite discontinuity when 𝑥 is equal to zero, approaching positive infinity from
the right and negative infinity from the left. The standard technique that we
might use to evaluate a definite integral is the second part of the fundamental
theorem of calculus. But this requires that our
integrand 𝑓 is a continuous function on the closed interval between 𝑎 and 𝑏,
which are our limits of integration. We’ve just seen that one over 𝑥
has an infinite discontinuity when 𝑥 is equal to zero. And we clearly see that this is one
of our limits of integration. Hence, the continuity condition
here is not true. This means that what we’re dealing
with is an improper integral. And we must use a different
technique.
For a discontinuity occurring at
the lower limit of integration, the definition of an improper integral tells us the
following. If 𝑓 is continuous on the interval
which is open at 𝑎 and closed at 𝑏 and discontinuous at 𝑎. Then the integral between 𝑎 and 𝑏
of 𝑓 of 𝑥 with respect to 𝑥 is equal to the limit as 𝑡 approaches 𝑎 from the
right of the integral between 𝑡 and 𝑏 of 𝑓 of 𝑥 with respect to 𝑥. If this limit exists and is
finite. Don’t worry too much about the 𝑡
that we’ve introduced since this is just a dummy variable which helps us to evaluate
our limit. Let us now apply this to our
question. 𝑓 of 𝑥, our integrand, is one
over 𝑥. The lower limit of integration, 𝑎,
is zero. And the upper limit of integration,
𝑏, is one.
Our definition tells us that our
integral is equal to the limit as 𝑡 approaches zero from the right of the integral
between 𝑡 and one of one over 𝑥 with respect to 𝑥. Okay. Let’s now evaluate this. We know that the antiderivative of
one over 𝑥 is the natural log of the absolute value of 𝑥. To continue, we input our limits of
integration 𝑡 and one. We’re then left with the following
expression. The laws of limits allow us to
apply our limit individually to each of these terms. So let’s do so and see what happens
to them. Well, our first term has no 𝑡
dependents at all. So we can simply get rid of
this. We also know that the natural
logarithm of one is equal to zero. And this relationship might be seen
more clearly by taking the exponential of both sides. 𝑒 to the power of zero is of
course one.
Okay. What about this term? Taking a direct substitution
approach, we have the natural logarithm of zero, technically zero from the
right. In some sense, we can say that this
is equal to negative infinity. Since as 𝑡 approaches zero from
the right, the natural log of the absolute value of 𝑡 approaches negative
infinity. Inputting our two values back into
our expression gives us the following. Our limit is equal to zero minus
negative infinity, which is positive infinity.
Now, saying that a limit is equal
to infinity does give us information about that limit. But it’s a particular way of
expressing that the limit does not exist since infinity is not a number. Since our limit does not exist, we
conclude that our limit does not have a numerical and finite evaluation. In cases such as this, we say that
the integral is divergent. And so we have arrived at the
answer to our question.
