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.