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Integrate.

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We've already seen how we can
differentiate term by term.

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Well can we integrate term by term?

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Well yes, and here's a theorem.

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Suppose I got some function f, and f of x
is this power series, the

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sum n goes from zero to infinity of a sub
n x to the n.

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And big R is the radius of convergence of
this power series.

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Well then, I can integrate this function
term

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by term, meaning that the integral of f

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of x dx, x goes from zero to some
parameter t, is the sum n goes from zero

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to infinity of this, a sub n times t to
the n plus one over n plus one.

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And this is true for any value of t
between minus r and r.

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.

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If you want to be a little more specific,

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it turns out the radius of convergence of
this

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series is exactly the same is exactly the

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same as the radius of convergence of this
series.

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Well, and if you're wondering where this
term

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comes from, right, that is exactly what
you would get just

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by integrating the term a sub n x to the
n, right?

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The integral of a sub n x to the n dx, x
goes from zero to t,

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well that just is a sub n times t to the n
plus one over n plus one.

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Armed with this result, we can do some
pretty great stuff.

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For example, we know that the sum n goes
from zero to infinity of x

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to the n, well, that's one over one minus
x, as

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long as the absolute value of x is less
than one.

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I mean, in other words, right, in this
particular

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instance, the radius of convergence, big
R, is one.

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Well, let's integrate that.

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So, the integral, from x equals zero to t
of one over one minus x d x.

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Yeah, what is that interval?
Well, I can

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evaluate that by making u substitution, so
let's set u equal to one

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minus x, and in that case, d u is just
minus d x.

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I don't see a minus dx there, but I can
artificially

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create a minus dx with some carefully
placed minus signs there.

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Okay, now I've got a negative dx there.

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That looks great.
So, this is negative the integral, x goes

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from zero to t of, I've just got a du over
one minus x, but that's u.

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So I want to know, how do I
anti-differentiate 1 over u.

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Well this is negative log the absolute
value of U.

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Then evaluate it at x equals t and x
equals zero.

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Now take the difference, so that's
negative log of well what's U,

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absolute value, one minus x, and we are
evaluating this at x equals zero and at t.

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So this is negative log of the absolute
value of one minus t.

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And then it would be subtracting this with
x equals zero, but I'm subtracting

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anegative, so I'm going to add the log of
the absolute value of one minus zero.

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Well, what's log of one?

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Log of one is just zero, so I don't need
to include this term.

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So all told, if I integrate one over one
minus x from zreo to t,

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what I'm getting is negative the natural
log, the absolute value of one minus t.

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Now we can integrate the power series term
by term.

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Alright, so we just showed that negative
log the absolute value of one minus t is

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the integral x goes from zero to t of one
over one minus x d x.

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Now I'm going to replace this one over one
minus x with a power series,

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so this is equal to the integral, x goes
from zero to t instead

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of one over one minus x, the sum n goes
from zero to infinity

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of x to the n, because one over one minus
x is equal to this.

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Well,

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as long as the absolute value of x is less
than 1 d x.

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Okay.

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Now I'm integrating a power series, and by
the theorem, I can do that term by term.

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So this is the sum, n goes from zero to
infinity of the integral

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of x going from zero to t, of x to the n d
x.

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So now I have to be able to do this
integration problem.

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So this

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the sum, n goes from zero to infinity of,
what's this?

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Well, I'm integrating x to the n d x from
x equals zero to t.

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So that's t to the n plus one over n plus
one.

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So there it is, right?

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I found a series for this function, for
negative the

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natural log of the absoulte value of one
minus t.

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Here's a power series for that function,

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and it's valid for which values of t?

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Well as long as t is less than 1 in
absolute value.

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Before we get too excited, it's worth
mentioning a fine point here.

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Well one case when you'd like to use this
formula is when t equals negative one.

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Because in that case, what do you get?

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Well, then you get negative the natural
log

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of the absolute value of 1 minus negative

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1, is equal to what?
Well, it's equal

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to this series, the sum n goes from zero
to infinity of, now, negative

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one to the n plus one over n plus one.
I can simplify this a bit.

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Right, what is this?

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Negative the natural log of one minus
negative one, well this is just two.

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So this is negative the natural log of 2
is

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given by this series.
Or is it?

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Alright the issue here is that I'm
going to plug in t equals negative one.

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But strictly speaking, that's not one of
the

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values of t that I'm allowed to plug in.

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If you want to dig deeper, there are some

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other theorems that you might want to look
into.

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For example, it turns out that this is
true.

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Alright, it turns out that negative the
natural

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log of 2 is given by this alternating
series.

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But to justify that, you do a little bit
more.

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One way to do it is by invoking Abel's
Theorem.

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