Okay so now we'll talk a little bit about
the next type.
So that was improper integrals where I was
looking at an infinite infinite interval.
The other possibility is that the
integrand f
of x is unbounded at either a or b.
So a or b are the end points of the
interval where I want to integrate.
And so,
if I say it's unbounded at a, what I mean
is that the limit
as x goes down to a of f of x is plus or
minus infinity,
so it either goes up to positive infinity
or it goes down to minus infinity,
and then just the same thing at at the
upper end point of the limit.
So if x is increasing towards b, if the
limit of f
of x goes to either, so if the function
blows up and
goes to infinity or it goes down to minus
infinity.
And so were going to define this in
exactly
the same way, so instead of having an
infinite
value in the limit were just going to
define to be the limit as so for the lower
endpoint as t goes down to a, the integral
from t to b of f of x,
or the integral goes up to b, of the
integral from a to t of f of x.
If those limits again exist and they're
finite, then are going to say that,
oops didn't want that bit yet.
that this integral exists,
and the value of this integral is equal to
the value of this limit, or this limit.
And so this third point, is if you're
truly unlucky and it's unbounded at both
endpoints, what you need to do is just
choose a point somewhere in the middle.
And go both ways.
So, what you'd be doing is splitting the
interval up into
two intervals and then integrating over
each one of those separately.
So, as a quick example of this guy,
imagine trying to integrate the function.
1 over the square root of x dx from 0 to
1.
And so the problem is as I move towards
zero, 1 over the
square root of x is going to get bigger
and bigger and bigger.
And in the limit it's going to be
infinite.
But I'm going to try and replace the zero
in the lower end point with
t so t in this case has to be some number
greater than zero.
So I'll integrate from t to one and then
it'll be easier to
integrate, I can use my anti power rule,
if I just write it.
1 over the square root of x has x to the
minus one half power.
And so I'll go ahead and do this integral.
And then I'll take the limit of
that as x goes down to 0.
So evaluating my integral, I just have.
2 x to the one half.
So that's anti-derivative of x to the
minus one half.
And you can verify that just by taking one
half times 2
is 1.
Which is the coefficient here.
X to the one half minus 1, so x to the
minus one half.
And then I need to evaluate that at the
upper end
point and subtract and evaluate it at the
lower end point.
So that's going to give me 2 minus 2 times
the square root of t.
And now I'm going to take the limit of
this expression here as t deceases to
zero.
And so, essentially the, the second term
that has the t
in it, that's just going to get smaller,
and smaller and smaller.
So, in the limit, that's just going to end
up being equal to two.
So, even though this function takes on
arbitrarily
large values in the interval where I'm
interested
in integrating it, I can still come up
with a finite number For the area of it.
Okay,
and so this is this is a picture of what
it looks like.
So in this lower square here from 0, 0 to
1, 1 square,
that's half of the area, and then the area
where it's got this
curved side here, all the way up to
infinity, so, it's difficult to
draw something that goes all the way to
infinity so I stopped at six.
but that has the same area so
above the line, y equals 1, the yellow
region has the same area as this unit
square.
Here.
And then we can also combine both types of
improper integrals.
So for instance, suppose I wanted to try
to compute the integral from zero to
infinity of 1 over the square root of the
absolute value of x
minus 1.
So it has an infinity in the
it has infinitely long interval.
And also
when x is equal to 1, I'm going to end up
dividing by 0 and that's
going to be a point that's in the interior
of the interval where I'm trying to
integrate.
So the function is not only unbounded in
the positive
x direction it's also unbounded in the
positive y direction.
And so this is actually what it, what it's
going to look like.
so at, at zero it has a point of
discontinuity, it's basically the limit
of this function as I move towards one in
either direction is positive infinity.
And so, basically my approach to doing
this
is, I'm just going to split up the
interval.
So I'm going to considering integrating
this function from
zero to one, and then from one to two.
And I'll use the second type of improper
integral there
because that's going to help me handle
this unbounded y direction.
And then I'll add on the integral from two
to positive infinity.
And that'll be an indefinite
yeah indefinite integral of the first
type.
So, this is just me writing that out, so
all I'm doing is
splitting this up into the, so, it's the
integral from 0 to infinity.
I'm going to think of that as, integral
from
zero to one, integral from one to two.
And then integral from two to infinity.
But I'm going to replace all those with
the limit definitions.
So if this integral exists,
or for this integral to exist all three of
these limits have to exist and be finite.
So lets just tackle them one at a time.
So for the first one if I'm
considering values in the interval 0, 1
then 1 minus x.
So remember I use to have a absolute value
down there.
1 minus x is going to be positive already
so I don't need the absolute value.
So I've gotten rid of those
and now this is just going to be 1 minus x
to the negative one half power.
And when I find the anti-derivative of
that, have to remember to use
the chain rule because I have a negative x
inside my function here.
I end up finding out that that's equal to
two.
So hopefully that's what we would expect,
because that's what we just did.
So if you noticed this, this shape here is
just
exactly the same one we had last time,
turned around.
Then, I'm going to look at the, the next
bit.
So it's actually symmetric about one.
So it's not surprising that I, I end up
getting the same value, but
to be to be more rigorous about it, we can
go through the steps again.
So this time, if x is greater
than 1, then x minus 1 did I write this
backwards the second time?
So x minus 1 is going to be the way to get
a positive number.
So if x is greater than 1, so for instance
if it's 2, 2 minus 1 is greater than 0.
So I just flip the order around of these
guys
and that let's me get rid of the absolute
value.
And then I just take that anti-derivative.
So this time, I don't get the minus sign,
because
here I had a negative x underneath my
square root.
Here I have a positive x underneath my
square root.
So it's going to be positive.
And then I just end up,
once I evaluate my limit at two and t.
I get 2 minus 2 times the square root of t
minus 1.
I'm taking the limit as t goes to one.
So when t goes to 1 t minus 1 is
going to 0 so the second term is getting
arbitrarily small.
So in the limit I can just say that's
going to
be zero and I just end up with this two
again.
So, so far so good.
But now when I try and integrate my third
integral, I have
the same, the same integrand as the second
one so the first step is is pretty easy.
And now I'm going to evaluate it at the
upper end point.
And subtract, subtract this, evaluated at
the lower endpoint.
And that gives me this expression here.
But now, I have a, a t.
I'm, I'm no longer dividing by t.
The, the power of the term containing the
t.
So, t minus 1 is to a power greater than
zero.
And so this is going to get bigger as t
gets bigger and in an unbounded way.
So if t is going to infinity, then the
square root of t is also going to
infinity.
And so when I take the limit of this
expression.
Oops, this expression, as t goes to
infinity.
I get infinity, and so the conclusion I'm
going to
make from this is that the integral from
zero to
infinity of d x over the square root of
the absolute value of x minus 1 does not
exist.
Because when I split it up into these
pieces one of them was not finite.
And so then in particular, that means that
if I took the limit of this as the upper
end point goes to infinity.
That's not going to be finite either.
And our definition for an indefinite
integral existing
required that the value has to be finite.
So if this is not finite, this integral,
this indefinite integral does not exist.