### Video Transcript

Consider the integral from zero to
one of one divided by π₯ to the power of π with respect to π₯. Find all possible values of π for
which the integral is convergent. Evaluate the integral for those
values of π.

The question gives us a definite
integral, where our integrand is one divided by π₯ to the power π for some constant
π. The question wants us to find all
values of π for which this integral is convergent. Then it wants us to evaluate this
integral for all of the values of π for which the integral is convergent.

To answer this question, letβs
think about our integrand function one divided by π₯ to the power of π. For starters, if our integrand
function is continuous on the closed interval from zero to one, then the integral is
proper. And in particular, we could use the
fundamental theorem of calculus to evaluate this integral. However, if our integrand was not
continuous on the closed interval from zero to one, then the integral would be
improper. To deal with an improper integral,
weβll want to know all the points where π₯ to the power of negative π would be
discontinuous.

We see that π₯ to the power of
negative π is a rational function. In fact, when π is less than or
equal to zero, it will be a polynomial which is continuous across the whole real
numbers. So it will be a proper
integral. So weβre only interested in the
cases when π is greater than zero. So when π is greater than zero, we
have π₯ to the power of negative π is the rational function one divided by π₯ to
the power of π. So itβs continuous on the interval
from zero to one, where zero is not included, but is not continuous when π₯ is equal
to zero.

One way of seeing this is when π
is greater than zero, one divided by zero to the power of π gives us the
indeterminant form of one divided by zero. So our function is not defined when
π₯ is equal to zero. We also briefly discussed when π
is less than or equal to zero, π₯ to the power of negative π is actually a
polynomial. So it is continuous on the closed
interval from zero to one. And we can evaluate these integrals
by just using the fundamental theorem of calculus. In the case where π was greater
than zero, weβve shown that the only point of discontinuity on our interval was when
π₯ is equal to zero.

So we can check the convergence of
these integrals by using one of our rules for improper integrals. Which tells us the integral from
zero to one of π₯ to the power of negative 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₯ to
the power of negative π with respect to π₯. This is true if the limit exists
and is finite. And we remember this rule for
integrals is only true when our integrand is continuous on the interval from zero to
one, not including zero, and is discontinuous at the point when π₯ is equal to
zero.

So weβre now ready to check the
convergence of this integral. Weβll do it in parts. When π is less than or equal to
zero, weβll try to use the fundamental theorem of calculus. And when π is greater than zero,
weβll try to use our rule for improper integrals. Weβll need to check that the limit
exists and is finite.

So letβs clear some space and work
on each of these cases individually. Weβll start when π is less than or
equal to zero. This gives us the integral from
zero to one of one divided by π₯ to the power of π with respect to π₯. We rewrite our integrand as π₯ to
the power negative π. And then we recall that our
integrand is continuous on the closed interval from zero to one because π is less
than or equal to zero. We know that π is not equal to
one. So we can integrate this by adding
one to the exponent and dividing by this new exponent.

We then substitute in the limits of
our integral. We see the one to the power of
negative π plus one is just equal to one and zero to the power of negative π plus
one is just equal to zero. So this simplifies to just one
divided by negative π plus one. And weβve shown that, in the case
when π is less than or equal to zero, our integral is convergent.

Now, letβs do the cases when π is
greater than zero. We have that our integrand is
continuous on the interval from zero to one, not including zero, and is not
continuous when π₯ is equal to zero. So by using our rules for improper
integrals, we have the integral from zero to one of π₯ to the power of negative 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₯ to the power of negative π with respect to π₯.

It would be tempting at this point
to evaluate this integral by adding one to our exponent and then dividing by the new
exponent. But we have to be careful because
if π is equal to one, then weβre integrating π₯ to the power of negative one. And using this will give us an
indeterminate form. So weβre going to have to separate
this into two cases. Weβll start when π is not equal to
one. Since π is not equal to one, we
can add one to the exponent and then divide by this new exponent to find our
antiderivative.

Next, we evaluate this at the
limits of our integral, when π₯ is equal to π‘ and π₯ is equal to one. We see that our first term does not
change as the value of π‘ changes. So its limit is equal to
itself. And its numerator, one to the power
of negative π plus one, is just equal to one. So this gives us our new
expression. We now see that, inside our limit,
the denominator of negative π plus one does not vary as π‘ varies. The only part which varies is our
numerator, π‘ to the power of negative π plus one.

To evaluate this limit, we recall
two of our limit laws. If π is greater than zero, the
limit of π₯ to the πth power as π₯ approaches zero is equal to zero. And if π is less than zero, then
the limit of π₯ to the πth power as π₯ approaches zero does not exist. Using this, we see if negative π
plus one is greater than zero, then our numerator will converge and be equal to
zero. And if negative π plus one is less
than zero, then our numerator will not converge.

So weβve shown when π is greater
than one, the numerator of our limit does not converge and so our integral does not
converge. We must be careful with the
convergence part here. We might be tempted to say that it
converges when π is less than one. However, weβve already assumed that
π is greater than zero. So weβve shown that our integral
will converge when zero is less than π is less than one. In fact, since the numerator of our
limit is approaching zero, the limit is approaching zero. So we can evaluate this integral to
be equal to one divided by negative π plus one. And this is only when zero is less
than π is less than one.

So weβve now checked the
convergence of the integral when π is less than or equal to zero, when π is
between zero and one, or when π is greater than one. We now only need to check the case
when π is equal to one. So letβs clear some space and work
on the case when π is equal to one.

We have that our integrand π₯ to
the power of negative one is continuous on the interval from zero to one, not
including zero, and is not continuous when π₯ is equal to zero. So by using our rules for improper
integrals, we have that this integral is equal to the limit as π‘ approaches zero
from the right of the integral from π‘ to one of π₯ to the power of negative one
with respect to π₯ if this limit exists and is finite. We have the antiderivative of π₯ to
the power of negative one is the natural logarithm of the absolute value of π₯.

We then evaluate this at the bounds
of our integral. We see the natural logarithm of the
absolute value of one is a constant which is just equal zero. We know the limit as π₯ approaches
zero from the right of the natural logarithm of π₯ does not exist. So the limit as π‘ approaches zero
from the right of negative the natural logarithm of the absolute value π‘ also does
not exist. Therefore, since this limit does
not exist, our integral does not converge when π is equal to one.

Therefore, weβve shown, for all
values of π, our integral is only convergent when π is less than or equal to zero
or when zero is less than π is less than one. We can combine these into a single
inequality. So weβve shown that our integral
will converge when π is less than one. In fact, in both cases where our
integral converged, we got that we can evaluate our integral to be one divided by
negative π plus one. And by rearranging this
denominator, weβve shown when this integral converges, it evaluates to be equal to
one divided by one minus π.

Therefore, weβve shown that the
integral from zero to one of one divided by π₯ to the power π with respect to π₯
will converge when π is less than one. And it evaluates in these cases to
be one divided by one minus π.