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Okay so now we'll talk a little bit about
the next type.

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00:00:05,010 --> 00:00:13,220
So that was improper integrals where I was
looking at an infinite infinite interval.

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00:00:13,220 --> 00:00:17,610
The other possibility is that the
integrand f

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of x is unbounded at either a or b.

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So a or b are the end points of the
interval where I want to integrate.

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And so,

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if I say it's unbounded at a, what I mean
is that the limit

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00:00:29,610 --> 00:00:35,030
as x goes down to a of f of x is plus or
minus infinity,

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so it either goes up to positive infinity
or it goes down to minus infinity,

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and then just the same thing at at the
upper end point of the limit.

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So if x is increasing towards b, if the
limit of f

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of x goes to either, so if the function
blows up and

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goes to infinity or it goes down to minus
infinity.

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And so were going to define this in
exactly

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the same way, so instead of having an
infinite

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value in the limit were just going to
define to be the limit as so for the lower

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endpoint as t goes down to a, the integral
from t to b of f of x,

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or the integral goes up to b, of the
integral from a to t of f of x.

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If those limits again exist and they're
finite, then are going to say that,

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oops didn't want that bit yet.
that this integral exists,

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and the value of this integral is equal to
the value of this limit, or this limit.

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And so this third point, is if you're
truly unlucky and it's unbounded at both

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endpoints, what you need to do is just
choose a point somewhere in the middle.

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And go both ways.

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So, what you'd be doing is splitting the
interval up into

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two intervals and then integrating over
each one of those separately.

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So, as a quick example of this guy,
imagine trying to integrate the function.

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1 over the square root of x dx from 0 to
1.

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And so the problem is as I move towards
zero, 1 over the

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00:02:16,930 --> 00:02:20,740
square root of x is going to get bigger
and bigger and bigger.

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And in the limit it's going to be
infinite.

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But I'm going to try and replace the zero
in the lower end point with

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t so t in this case has to be some number
greater than zero.

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So I'll integrate from t to one and then
it'll be easier to

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integrate, I can use my anti power rule,
if I just write it.

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1 over the square root of x has x to the
minus one half power.

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And so I'll go ahead and do this integral.
And then I'll take the limit of

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that as x goes down to 0.

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So evaluating my integral, I just have.

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2 x to the one half.

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So that's anti-derivative of x to the
minus one half.

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And you can verify that just by taking one
half times 2

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is 1.
Which is the coefficient here.

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X to the one half minus 1, so x to the
minus one half.

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And then I need to evaluate that at the
upper end

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point and subtract and evaluate it at the
lower end point.

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So that's going to give me 2 minus 2 times
the square root of t.

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And now I'm going to take the limit of

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this expression here as t deceases to
zero.

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And so, essentially the, the second term
that has the t

51
00:03:49,830 --> 00:03:51,910
in it, that's just going to get smaller,
and smaller and smaller.

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So, in the limit, that's just going to end
up being equal to two.

53
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So, even though this function takes on
arbitrarily

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large values in the interval where I'm
interested

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in integrating it, I can still come up
with a finite number For the area of it.

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Okay,

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and so this is this is a picture of what
it looks like.

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So in this lower square here from 0, 0 to
1, 1 square,

59
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that's half of the area, and then the area
where it's got this

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curved side here, all the way up to
infinity, so, it's difficult to

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draw something that goes all the way to
infinity so I stopped at six.

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but that has the same area so

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above the line, y equals 1, the yellow

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region has the same area as this unit
square.

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Here.

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And then we can also combine both types of
improper integrals.

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So for instance, suppose I wanted to try
to compute the integral from zero to

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infinity of 1 over the square root of the
absolute value of x

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minus 1.
So it has an infinity in the

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it has infinitely long interval.
And also

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when x is equal to 1, I'm going to end up
dividing by 0 and that's

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going to be a point that's in the interior

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of the interval where I'm trying to
integrate.

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00:05:21,860 --> 00:05:25,710
So the function is not only unbounded in
the positive

75
00:05:25,710 --> 00:05:28,920
x direction it's also unbounded in the
positive y direction.

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And so this is actually what it, what it's
going to look like.

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so at, at zero it has a point of
discontinuity, it's basically the limit

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of this function as I move towards one in
either direction is positive infinity.

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And so, basically my approach to doing
this

80
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is, I'm just going to split up the
interval.

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So I'm going to considering integrating
this function from

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zero to one, and then from one to two.

83
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And I'll use the second type of improper
integral there

84
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because that's going to help me handle
this unbounded y direction.

85
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And then I'll add on the integral from two
to positive infinity.

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And that'll be an indefinite

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yeah indefinite integral of the first
type.

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00:06:22,740 --> 00:06:27,370
So, this is just me writing that out, so
all I'm doing is

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splitting this up into the, so, it's the
integral from 0 to infinity.

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I'm going to think of that as, integral
from

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zero to one, integral from one to two.

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And then integral from two to infinity.

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But I'm going to replace all those with
the limit definitions.

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So if this integral exists,

95
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or for this integral to exist all three of
these limits have to exist and be finite.

96
00:06:57,308 --> 00:07:03,284
So lets just tackle them one at a time.

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So for the first one if I'm

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considering values in the interval 0, 1
then 1 minus x.

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So remember I use to have a absolute value
down there.

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1 minus x is going to be positive already
so I don't need the absolute value.

101
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So I've gotten rid of those

102
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and now this is just going to be 1 minus x
to the negative one half power.

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And when I find the anti-derivative of
that, have to remember to use

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the chain rule because I have a negative x
inside my function here.

105
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I end up finding out that that's equal to
two.

106
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So hopefully that's what we would expect,
because that's what we just did.

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So if you noticed this, this shape here is
just

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exactly the same one we had last time,
turned around.

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Then, I'm going to look at the, the next
bit.

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So it's actually symmetric about one.

111
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So it's not surprising that I, I end up
getting the same value, but

112
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to be to be more rigorous about it, we can
go through the steps again.

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So this time, if x is greater

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than 1, then x minus 1 did I write this
backwards the second time?

115
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So x minus 1 is going to be the way to get
a positive number.

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So if x is greater than 1, so for instance
if it's 2, 2 minus 1 is greater than 0.

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So I just flip the order around of these
guys

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and that let's me get rid of the absolute
value.

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And then I just take that anti-derivative.

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So this time, I don't get the minus sign,
because

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here I had a negative x underneath my
square root.

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Here I have a positive x underneath my
square root.

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So it's going to be positive.
And then I just end up,

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once I evaluate my limit at two and t.

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I get 2 minus 2 times the square root of t
minus 1.

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I'm taking the limit as t goes to one.

127
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So when t goes to 1 t minus 1 is

128
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going to 0 so the second term is getting
arbitrarily small.

129
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So in the limit I can just say that's
going to

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be zero and I just end up with this two
again.

131
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So, so far so good.

132
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But now when I try and integrate my third
integral, I have

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the same, the same integrand as the second
one so the first step is is pretty easy.

134
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And now I'm going to evaluate it at the
upper end point.

135
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And subtract, subtract this, evaluated at
the lower endpoint.

136
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And that gives me this expression here.
But now, I have a, a t.

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I'm, I'm no longer dividing by t.

138
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The, the power of the term containing the
t.

139
00:09:59,150 --> 00:10:02,109
So, t minus 1 is to a power greater than
zero.

140
00:10:03,190 --> 00:10:07,700
And so this is going to get bigger as t
gets bigger and in an unbounded way.

141
00:10:07,700 --> 00:10:11,730
So if t is going to infinity, then the

142
00:10:11,730 --> 00:10:14,390
square root of t is also going to
infinity.

143
00:10:16,510 --> 00:10:21,070
And so when I take the limit of this
expression.

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Oops, this expression, as t goes to
infinity.

145
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I get infinity, and so the conclusion I'm
going to

146
00:10:28,730 --> 00:10:31,170
make from this is that the integral from
zero to

147
00:10:31,170 --> 00:10:34,590
infinity of d x over the square root of

148
00:10:34,590 --> 00:10:38,160
the absolute value of x minus 1 does not
exist.

149
00:10:39,690 --> 00:10:41,580
Because when I split it up into these

150
00:10:41,580 --> 00:10:46,930
pieces one of them was not finite.
And so then in particular, that means that

151
00:10:46,930 --> 00:10:52,790
if I took the limit of this as the upper
end point goes to infinity.

152
00:10:52,790 --> 00:10:54,820
That's not going to be finite either.

153
00:10:54,820 --> 00:10:58,190
And our definition for an indefinite
integral existing

154
00:10:58,190 --> 00:11:00,490
required that the value has to be finite.

155
00:11:00,490 --> 00:11:05,200
So if this is not finite, this integral,
this indefinite integral does not exist.