Integrate.
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We've already seen how we can differentiate term by term.
Well can we integrate term by term?
Well yes, and here's a theorem.
Suppose I got some function f, and f of x is this power series, the
sum n goes from zero to infinity of a sub n x to the n.
And big R is the radius of convergence of this power series.
Well then, I can integrate this function term
by term, meaning that the integral of f
of x dx, x goes from zero to some parameter t, is the sum n goes from zero
to infinity of this, a sub n times t to the n plus one over n plus one.
And this is true for any value of t between minus r and r.
.
If you want to be a little more specific,
it turns out the radius of convergence of this
series is exactly the same is exactly the
same as the radius of convergence of this series.
Well, and if you're wondering where this term
comes from, right, that is exactly what you would get just
by integrating the term a sub n x to the n, right?
The integral of a sub n x to the n dx, x goes from zero to t,
well that just is a sub n times t to the n plus one over n plus one.
Armed with this result, we can do some pretty great stuff.
For example, we know that the sum n goes from zero to infinity of x
to the n, well, that's one over one minus x, as
long as the absolute value of x is less than one.
I mean, in other words, right, in this particular
instance, the radius of convergence, big R, is one.
Well, let's integrate that.
So, the integral, from x equals zero to t of one over one minus x d x.
Yeah, what is that interval? Well, I can
evaluate that by making u substitution, so let's set u equal to one
minus x, and in that case, d u is just minus d x.
I don't see a minus dx there, but I can artificially
create a minus dx with some carefully placed minus signs there.
Okay, now I've got a negative dx there.
That looks great. So, this is negative the integral, x goes
from zero to t of, I've just got a du over one minus x, but that's u.
So I want to know, how do I anti-differentiate 1 over u.
Well this is negative log the absolute value of U.
Then evaluate it at x equals t and x equals zero.
Now take the difference, so that's negative log of well what's U,
absolute value, one minus x, and we are evaluating this at x equals zero and at t.
So this is negative log of the absolute value of one minus t.
And then it would be subtracting this with x equals zero, but I'm subtracting
anegative, so I'm going to add the log of the absolute value of one minus zero.
Well, what's log of one?
Log of one is just zero, so I don't need to include this term.
So all told, if I integrate one over one minus x from zreo to t,
what I'm getting is negative the natural log, the absolute value of one minus t.
Now we can integrate the power series term by term.
Alright, so we just showed that negative log the absolute value of one minus t is
the integral x goes from zero to t of one over one minus x d x.
Now I'm going to replace this one over one minus x with a power series,
so this is equal to the integral, x goes from zero to t instead
of one over one minus x, the sum n goes from zero to infinity
of x to the n, because one over one minus x is equal to this.
Well,
as long as the absolute value of x is less than 1 d x.
Okay.
Now I'm integrating a power series, and by the theorem, I can do that term by term.
So this is the sum, n goes from zero to infinity of the integral
of x going from zero to t, of x to the n d x.
So now I have to be able to do this integration problem.
So this
the sum, n goes from zero to infinity of, what's this?
Well, I'm integrating x to the n d x from x equals zero to t.
So that's t to the n plus one over n plus one.
So there it is, right?
I found a series for this function, for negative the
natural log of the absoulte value of one minus t.
Here's a power series for that function,
and it's valid for which values of t?
Well as long as t is less than 1 in absolute value.
Before we get too excited, it's worth mentioning a fine point here.
Well one case when you'd like to use this formula is when t equals negative one.
Because in that case, what do you get?
Well, then you get negative the natural log
of the absolute value of 1 minus negative
1, is equal to what? Well, it's equal
to this series, the sum n goes from zero to infinity of, now, negative
one to the n plus one over n plus one. I can simplify this a bit.
Right, what is this?
Negative the natural log of one minus negative one, well this is just two.
So this is negative the natural log of 2 is
given by this series. Or is it?
Alright the issue here is that I'm going to plug in t equals negative one.
But strictly speaking, that's not one of the
values of t that I'm allowed to plug in.
If you want to dig deeper, there are some
other theorems that you might want to look into.
For example, it turns out that this is true.
Alright, it turns out that negative the natural
log of 2 is given by this alternating series.
But to justify that, you do a little bit more.
One way to do it is by invoking Abel's Theorem.
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