So next section is improper integrals.
An improper integral is just above the,
let's
see, I guess I should go to my slide.
and an improper integral comes in two
types.
So, you might be interested in integrating
a function over an unbounded interval
so that would be an interval where at
least one of the end points is infinite.
Or the function
could be unbounded, so I might be
integrating something like 1
over x and one of the end points would be
zero.
So, the limit as my function approaches
zero is going to be infinite.
And so these are going to be the
conditions
where we can make sense of integrals like
that.
So we'll start by considering the first
type, so I want to integrate a function f
of x over either the inter, interval minus
infinity b or the interval a to infinity.
And if I want to integrate a function over
the entire
real line so minus infinity to infinity I
just think
of it as I pick some point in the middle
and go up to there and then finish the
rest.
The improper integral of f of x over a
infinity Exists if and only if the limit,
so
basically all I'm going to do, I want to
have an infinite, infinity up in the top
here.
But if I actually plugged in an infinity,
most of the time something's going to go
wrong.
So I want to consider this as a limit.
I'll evaluate this integral using a and t,
and then take
the limit of the resulting function as t
goes to infinity.
If that limit exists and it's finite, then
we say
that the improper integral of f of x over
(a, infinity)
exists.
So now it's getting a bit trickier, so our
definition of existing means the limit
exists and it's finite.
And so in particular, that means that if
this
integral, you know, as I take the limit as
t goes to infinity, if I get infinity,
Then
f of x is not integrable over this this
interval.
And then we do exactly the same thing
except now I have a t on the
bottom so when I want the, the lower limit
to be able to go to minus infinity.
I'm just going to put t in the bottom and
take the limit as t goes to minus
infinity.
And the condition has to be the same.
For this to be well-defined, I need this
limit to
exist, and I need it to be a finite value.
And so when those limits exist and they're
finite, then I'm going
to define the improper integral of f of x
dx over a infinity.
To be the value of this limit, so the, the
limit s t
goes to infinity integral from a to t of f
of x d x.
And then the, sort of the complementary
piece for
the lower limit, so I'm going to define
the improper
integral from minus infinity to b of f of
x d x.
To be the value of this limit, as t goes
to minus
infinite, the integral of t to b of f of x
d x.
So you can also add and subtract improper
integrals, so all I'm doing here.
Is I'm considering b greater than a so
this is a normal
way of setting up an integral if I think
of if I think
of an interval I have a lower end point of
a and an
upper end point of b so b has to be
greater than a.
And so if I intergrate.
f of x, from a to infinity.
The integral of f of x from b to infinity
has to sort of fit inside of that.
So what I've done, is I've said is a to
infity.
and then b to infinity.
And if I do this subtraction, all I
want to think about
is, I'm taking a to infinity, infinnity is
way over there.
And b to infinity and I'm just sort of
chopping of
the tail at b, and so probably just as you
would
expect what's left is just going to be the
integral of f of x from a to
b, and then I can do exactly the same
thing with the other tail.
So if I have the integral from minus
infinity up
to b, so remember b is bigger than a, so
this integral, minus infinity up to a of f
of
x, that's if you imagine what the area
looks like,
that's just going to fit in the first, the
first area.
And so what I did.
When I do this subtraction, essentially
all I'm
doing is just chopping off the tail at
a, and again I'm left with the integral
from a to b of f of x dx.
Okay.
And now I want to be able to also
make sensitive integral over the entire
real line.
So like I mentioned before, the integral
from minus infinity to infinity of
f of x, this is going to exist if and only
if there's a point a such that both.
The integral from minus infinity up to a,
and the integral from
a to infinity of f of x exists so in
particular both those
need to be finite.
And then if that integral exists We're
just going to define its value.
So I, I'm just splitting up the real line
out of point
a so I have numbers bigger than a and
numbers smaller than a.
The value of this integral is just
going to be the sum
of these two integrals that we know how to
compute already.
And so I just wrote down the, the
definitions down below, and the reason
we need to be able to split this thing in
the
middle, you know, why, why not just define
our limit like this
so i take the limit as t goes to infinity
of
the integral from minus t to t of f of x
dx.
So that, that seems like it should work
its going to get the whole real line but
the problem we are running into, is I
could do something like this, so what,
what I'm doing here just imagining a line
with
a slop of 45 degrees so for positive
values of
x that's going to be above the x axis, for
negative
values of x it's going to be below the x
axis.
So if I integrate that from t to t, I'm
always going to get zero.
But if I split it up anywhere in the
middle, I
would end up getting a minus infinity and
a plus infinity.
And kind of like we saw in the first
lecture when we're looking
at this series of 1 minus 1 plus 1 minus 1
you know.
If I have something that I can express as
infinity
minus infinity, I can't really make any
sense out of that.
Another good way to imagine what's
happening
with this is, I could also get an intergal
from minus infinity to
infinity by putting a 2t up here but then
that would mean that.
All of these integrals here, instead of
integrating
to zero, they would integrate to some,
let's see.
Well this one would depend on t, but
if you're clever about the choice, you can
make
these all integrate to some non-zero
value, so as you
take the limit, then you can get any
answer you want to.
So if I, if I defined it this way, I would
have the limit from minus t to
t being 0, but then I'm not going to have
the property, so I wanted to be
able to do addition and subtraction with
my indefinite integrals.
And so now If I try to do that, if, if I
want to add
two together, I need to be able to break
it up somewhere in the middle.
And that's gona give me, so if I take a
being zero for
instance, I'm going to get minus infinity
plus infinity and that's not defined.
On the other hand, if you know for some
reason, so for instance
if you know that f of x belongs to a
family of function.
So I'm thinking of probability density
functions, where this integral does exist.
Then you're free to evaluate it however
you like.
So if you know that this integral exist
basically what that's
telling you is if you evaluated correctly
and get an answer.
That's going to be the correct answer.
So you can evaluate it however you like,
you'll get the same answer.
So you can evaluate it however you like.
So this formula if it was useful, you
could use it to evaluate an improper
integral.
Okay, so now This is probably going to be
a little bit strange for some of you.
But, I want to, for my example, show that
for an
integer value of alpha, that's greater
than or equal to zero,
this integral, so the integral from zero
to infinity, of x
to the alpha, times e to the minus x
squared exists.
And so what's going to happen with, with
this particular
argument if I'm pretty sure the way you
would
actually evaluate this, if you wanted to
get if
you wanted to get a number for an answer.
You need to use integration by parts.
And when you use integration by parts,
you're going to
end up with an integral where alpha is 1
smaller.
And so if alpha
was, say, 10, you would need to use
integration by parts ten times.
but I want to get a result that's valid
for any alpha that's greater than zero.
And so this is kind of a strange
situation.
Well remember, I said this integral exists
if the limit is finite.
So I can show that the limit is finite,
without actually finding out what that
limit is specifically.
But if I can say whatever limit it
is, it's less than a finite number, then I
can say that this integral exists.
This indefinite integral exists.
So, what I want to show, is that if I
integrate this function from zero to t.
And then I take the limit as t goes to
infinity.
I have to get a finite number.
And so another sort of shorthand for
finite
is you just write strictly less than
infinity.
So that means some real number.
some finite real number.
So, we can recall form the, from the first
lecture when I was talking about limits.
If p of x is a poloynomial and e to the x
is the exponential function.
Then if I take the limit of this as x goes
to infinity I get 0.
So in particular, I have this minus x
squared here.
So if I have
e to a negative power I can also write
that as 1 over e to that power.
So I can rewrite the integrand here as x
to the alpha Divided by e to the x
squared.
And so, by using those rules of limits,
you can also show that this is
going to the limit of this as x goes to
infinity is also going to be 0.
And so, you might
wonder why this plus 2 is up there,
because there's no plus 2 in the intregran
And the answer is you know I just put it
there because I want it
there.
So it, it'll pop out later on.
And so often when you're trying to work
these things out, so I know
how to solve this problem already, so I
can put the plus 2 there.
Otherwise I wouldn't have put the plus
2 there, I would have done more work.
I would have realized, hey, it would be
really
handy if there's another x squared in this
equation.
So then I would have gone backwards to
this line.
I would have added 2 to this exponent.
But the point is, it's not it's not meant
to be equal to this.
It's basically, what's going to end up
happening is that
I can now Write this as something times x
squared and
have my limit here, or, my integrand here.
And so
this, the, if, if I consider a sequence as
the x goes to
infinity, this is going to be equal to 0.
And so because
I know that limits equal to 0, I can now
use the
definition to say that there is a value M
so capital
M such that, so this is my, my integrand
again this value is
less than 1 whenever x is greater than M,
so this is basically
this epsilon delta argument except you
know, for
the, for the, for the sequence that is
going to infinity.
So the epsilon is 1 and then that
means because I know the limit exists,
there must
be some value such that when I'm beyond
that value I'm within epsilon of what I
want
it to be and so I'm going to choose 1, I
can choose any number I want to by the
definition of the limit but I'm going to
choose 1 because its easy to work with.
So generally when
you are free to choose a number, if you
choose anything other
than zero or one, you better have a really
good reason for it.
So now it should be clear why I wanted
this x
squared here, so if this quantity has to
be less 1
one then I can just divide both sides by x
squared
and so I'm considering the limit as x goes
to infinity.
you know, at some point this M is going to
be
a number bigger than zero.
And x, you know, I'm looking at values of
x greater than that M.
So x is not going to be 0.
So I can divide by x as many times as I
want to.
So I'm going to divide by x twice.
So my alpha over 2 becomes an alpha.
And my 1 becomes a 1 over x squared.
And so the important thing that I've done
here is I've taken the integrand and I've
bounded it by one over x squared.
So that means for every value of x greater
than M, this thing is going to be less
than one over x squared.
And what that's going to let me do.
Is now if I consider values of t that are
greater than M, I'm
going to split my integral so my integral
zero to t into a piece
from zero to M
[COUGH]
and another piece from M up to t.
And now what can I say about the, the
first term here?
Though it's not pretty, but 0 is a
constant and M is a
constant, so whatever, you know whatever
number this is it's going to be constant.
And so, t is the only thing that's
changing now,
this is the, what I'm taking to be the
limit.
So I can just rewrite this as a constant.
It's no longer really important.
If it's constant, it's finite.
And I'll I'm trying to do, is show that
this
thing is not going to be infinite.
And also I have said that this integrand
can be bounded by 1 over x squared.
So if I were to replace this with 1
over x squared, like I've done in the
second line.
It means it's everywhere for t bigger then
M, x squared is bigger than that.
So c, this thing, plus the integral from m
to t.
You remember one of the properties of the
integral was if I
had to functions and one of those
functions was always bigger than
the other function, If I integrate both of
those functions over the
same integral, then whichever function is
always bigger, it's integral is also
bigger.
So that allows me to bound this integral
by some constant,
and a func, and the integral of one over x
squared.
And now one over x squares,
that's something that is actually easy to
integrate.
So that just becomes minus 1 over x so 1
over x
squared, I can think of that as x to the
minus 2 power
and I can use my anti-power rule and that
just means its going
to come from, its the derivative of minus
x to the minus 1.
So I get minus x to the minus 1 here
evaluated at from M to t.
So again
I can rewrite that now as a constant minus
this thing here.
So this is just to summarize what I've
done on the other slide.
I now have bounded this integral from zero
to t of this very unhappy looking
integrand, by this expression over here
that's a much simpler expression of t.
And so I have c plus 1 over M so these are
just constant.
And now remember what I was trying to do
is show that the limit
of this guy over here as t goes to
infinity is approaching a finite value.
And so 1 over t, for t is positive.
So, this can never be, this can
never be bigger than c plus 1 over M.
So, I can make an upper bound of c.
So, that's the integral of my integrand
from 0 to M.
That's a constant value, so, it's going to
be finite.
And then M is a really big number, and I
have 1 over
M, so I have to add just a little bit more
to my constant.
And that gets me an upper bound, for no
matter what value of t
I choose, this is it's going to be smaller
c plus 1 over M.
So that means I can say that the limit as
t goes to infinity Of the
integral from 0 to t of x to the alpha, e
to the minus x squared.
That's going to be less than c plus 1 over
M.
And that's going to be a finite number.
So I write less than infinity.
And so without actually evaluating this
integral, and finding out
the, the number, I was able to show that,
let's see.
That this indefinite integral, from 0 to
infinity, of x
to the alpha, e to the minus x squared,
dx.
This exists.