So let us consider
another crucial example of improper integrals;
the integral of the power function.
So it's x powered minus a.
For the sake of simplicity,
firstly let us assume that a doesn't equal one.
I'm going to write this down.
It's our first case.
So what we should do here,
we should calculate our antiderivative.
In order to do so, we need to divide one
by power of ax plus one,
this results in a one minus a multiplied
by x powered one minus a.
Afterward, we need to
substitute our x with upper and lower limits.
So the question of convergence here actually
stems on the fact of
the existence of a limit at the upper point.
So we need to establish where
the limit of this function at
the plus infinity actually exists
and in what cases it doesn't.
In order to understand,
it's about kind of [inaudible].
Because once again we have
a constant multiplier and then we have power function.
So we need to decide whether the power function
has limit as a plus infinity or does not.
It's pretty much easy because as you do remember
the power function with
positive power actually approaches plus infinity.
It looks like this or like this.
The power function with negative values
actually looks like this,
and actually negative power.
So there is a simple one.
If the power of X is
positive then the limit doesn't exist.
The function approach is plus infinity at the infinity.
If the power actually is negative,
then the function approaches zero,
adds a plus infinity.
So we are going to write this results down.
If a is not the 1,
and if a minus 1,
that's 1 minus a is greater than 0,
then, the integral does not converge.
I'm going to symbolically
write there and just I close it.
If 1 minus a is actually negative as integral converges.
What for the case of a equal to 1.
We are going now to look at integral of 1 divided by X,
and [inaudible] you do know that the answer
is logarithm of the absolute value of X.
You need to substitute with upper and lower bounds,
and once again we need to understand what is
the limit of logarithm of plus infinity is.
As you well know,
it's plus infinity and thus the integral
does not converge. That's fine.
So we can simply add this case in
the scope for the not convergent cases.
Let us summarize our results here.
The integrals that we've talked about converges if A is
greater than 1 and does not
converge otherwise. That's fine.
But why do we need these special and pretty simple case?
When you choose into following idea,
sometimes it's hard or impossible
to calculate actual antiderivative or well,
it's easier to compare the function to some which
is in actually in the knowledge of the answer.
For example, just assumes that we're
speaking about improper integral of the function f,
and we have some function for example g. We do know
that both functions lies over zero,
they're both positive and
one function is always greater than that.
For example, let us assume that g is greater than f at
every point and the improper integral of g converges.
Then understandable,
the proper integral of f converges too
because along the way it can be not
convergent as it's area
under this curve approaches infinity.
But if the area under
f curve a right curve, approaches infinity,
then the area under the g curve also approaches
infinity because one actually include the other.
In that sense, integrals should also be non-convergent.
That's normally a called a comparator,
to prove out role of comparison in terms
of whether the integral converges on it.
Let us just think of some basic example here.
For example, the integral which is
X divided by 1 plus x squared.
Well, you can proceed
with actual calculation of the antiderivative,
and we've done it with
our magic borders where we
began off talking about anti derivatives.
But here, we are going to say something more general.
For example, if we take a look at
the function x divided by 1 plus x squared,
you've actually seen the antideritive of it.
We've started with it when we
were using our MagicTransfer board.
But whilst it's easy to
just derive an answer here and try to find limits,
when I go into rule with
some much more easier method
here using our rule of comparison.
So we have our function,
which is x divided by 1 plus x squared.
In fact, I'm going to say is that it is somehow
comparable with some function with known answer.
Right now, we know answer for all the function
which are a power function of x.
So maybe we can come up with some comparison with it.
So asymptotically speaking,
we are looking at the function which is somehow
close to 1 divided by x
because the denominator grows as x,
and denominate basically grows as x squared,
and they're equivalent in this manner.
So in order to understand how this integral converges,
we need to understand and remember how this converges.
As we previously, stated
Integral of the function
1 divided by x does not converge.
So arbitrarily speaking, we
need to come up with some inequality here,
for example, we need to say is
that since it does not converge,
the function we are comparing it
which should be greater as it needs.
So maybe we should write
something like 1 divided by 2 here.
But informally speaking,
equivalent functions are basically the same,
they are bounded by [inaudible].
So if one of them converges,
then the other one converges.
If one of them doesn't converge,
then the other doesn't converge.
Thus we can even generalize our comparison rule
into understanding
how equivalent functions actually look here.
So by using this very easy example,
we can, without
any calculations of additional antiderivatives,
just call the answer straightaway.
At very same time, let us just take a look
at the Gauss Distribution exponent of minus x squared.
Here, I'm going to try for schematic here for you,
and what we're going to look at,
we're going to look at the fact
whether this integral converges.
So we need to come up with something
which can be used as comparison for this function.
Let's just take a moment here and think about it.
Firstly, what we should understand,
we should understand that this function as we
do know whilst we were
talking about functions asymptotics,
it's extremely fast approaching
0 while it rises up to the infinite.
It is faster than any possible polynomial function,
it's faster than any power function.
So if we do use this function and we try to compare it
with some known polynomial function,
we already know the answer.
But as we pull down the functions,
that is tricky because we've actually spoke
about the convergence on
the segment from one to plus infinity,
and now we're talking about from 0 plus one to infinity.
As you all do understand,
0 is slightly more complex than one
towards functions one divided by x or x squared.
It's another tricky point for it.
So we're not going to compare it with the power function,
and we're going to compare it with
a widely known function that we've
actually calculated on the previous video.
Do you remember, we've worked with
a function which is called exponential
distribution or just basic exponents
of functions, of linear axis.
So let me just consider the following one.
Does this inequality hold?
Exponent of minus x squared is
greater or less than exponent of minus x.
Well, in order to understand it,
we need to compare those powers because we're looking at
exponents of obviously negative numbers.
Thus the greatest numbers allows
an exponent here, or well obviously,
x squared is greater than x if x
is greater or equal to one.
So for the segment from one
to plus infinity, this inequality holds,
and thus we can easily
call it urgent because we've actually
calculated the integral of
exponent of minus x as the previous feature,
and there was our answer,
so limits obviously exist.
So what for the actual
integral that look at which is not from one,
so plus infinity, which is from 0 to plus infinity?
That's easy because as we do know,
we can add areas under curves
because if you want to calculate the area of two figures,
you just need to summarize
the areas of two separate paths.
So we need to write the integral from 0 to 1,
plus the integral from 1 to plus infinity.
As for the second one,
we actually do know that it necessarily converges.
But as for the second one,
we do not even care about convergence
because it is an integral definite,
and more importantly, with
finite limits, with finite bounds.
More importantly, how function being integrated does not
have any discontinuations on the segments,
so it necessarily is integratable.
So as a result this,
also is a nice number,
and all integral existence convergence.
That's a really interesting thing
because as we previously stated,
we cannot calculate the antiderivative
for our exponent of minus x squared function,
but we can actually quite simply call it convergent.
So it makes sense that
all our tricky science here is extremely useful,
and can be used in understanding why do we actually
consider some basic example from probability theory,
and does not actually call
it a sensitive and all that stuff.
So at this very point,
we're actually in power of calculation
any possible definite integrals as with
proper and improper integrals using
our fundamental theorem of calculus.
Now, it's time to think about how the numerical methods
for calculation imply it is not
possible to calculate antiderivative works,
and what does it stand for,
and that's the theme of our following video.