### Video Transcript

The integral from zero to one of π
multiplied by the natural logarithm of π with respect to π 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 π is continuous on the closed interval from π to π, and capital πΉ is
an antiderivative of our function lowercase π. Then, the integral from π to π of
lowercase π of π with respect to π is equal to our antiderivative function
evaluated at π minus our antiderivative function evaluated at π. But this is only valid if our
integrand, π multiplied by the natural logarithm of π, is continuous on the closed
interval from zero to one.

However, we see that if we plug π
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 is equal to zero. If our integrand is not defined
when π is equal to zero, it canβt 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 is equal to zero. Or we can notice that our function
π multiplied by the natural logarithm of π is continuous on the interval from zero
to one, not including zero. This is because π is continuous on
this interval and the natural logarithm of π 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 π is a continuous function on
the interval from π to π which does not include π. And it is discontinuous at the
point π. Then, we can introduce a dummy
variable π‘. Which tells us the integral from π
to π of π of π with respect to π is equal to the limit as π‘ approaches π from
the right of the integral from π‘ to π of π of π with respect to π.

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

Weβre now ready to start evaluating
this integral. However, we canβt integrate π
natural logarithm of π directly since itβs not in a standard form. So, weβre going to need to perform
some manipulation. We could attempt to simplify this
by using a substitution. And this would work. However, itβs simpler to use
integration by parts. We recall that integration by parts
tells us that the integral of π’π£ prime with respect to π is equal to π’π£ minus
the integral of π’ prime π£ with respect to π.

Now, we need to decide which
function to set as π’ and which function to set as π£ prime. Weβre 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 π’ to be equal to a logarithmic function. We can do this in this case. So, weβll set π’ equal to the
natural logarithm of π. And that leaves π£ prime to be
equal to π.

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

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

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 π‘ multiplied by π‘ squared over two. And then, we can simplify the
integrand inside our limit by canceling the shared factor of π in the numerator and
the denominator. This gives us the integral from π‘
to one of π over two with respect to π.

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 π‘
is approaching zero from the right, we have that π‘ is always positive. Therefore, the natural logarithm of
π‘ and π‘ 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 over
two is π squared over four, which we get by adding one to the exponent and dividing
by this new exponent. This gives us the limit as π‘
approaches zero from the right of negative one multiplied by π squared over four
evaluated between the limits of π is equal to π‘ and π is equal to one. Evaluating at the limits of our
integral gives us the limit as π‘ approaches zero from the right of negative one
squared over four plus π‘ squared over four.

We have that negative one squared
over four does not vary based on the value of π‘. So, the limit as π‘ approaches zero
from the right of this term is equal to negative a quarter. However, as π‘ 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 natural logarithm of π with respect to π converges
to negative a quarter.