After the previous commentary had already been expressed in type, repeated meditations on the same topic led to a new demonstration of the theorem, which is indeed just as purely analytical as the previous one, but is based on completely different principles, and seems to be far preferable to it in terms of simplicity. The following pages are therefore dedicated to this third demonstration.
Consider the following function of the indeterminate
in which the coefficients
etc. are fixed real quantities. Let
be indeterminates, and consider
The factor
can obviously be removed from the numerator and denominator of the final formula, since it divides
. Finally, let
be a positive quantity, arbitrarily chosen but greater than the maximum of the following quantities:
where the signs of the quantities
etc. are excluded, i.e. the negatives, if any, have been changed to positives. These preparations being made, I say that
obtains a certain positive value when
for any real value assigned to
Proof. Let us set
Then it is clear that
I.
is composed of the parts
each of which, for any determined real value of
is easily seen to be positive. Hence,
necessarily takes a positive value. Similarly, it can be shown that
are positive, and thus
is necessarily a positive quantity.
II. For
the functions
respectively become
as can be easily proven by expanding. Thus for
, the value of the function
is derived to be
and thus it is a positive quantity. Q.E.D.
Moreover, from the same formulas, we infer that for
the value of the function
is
and therefore it is positive. Hence, we conclude that for no value of
which is simultaneously greater than
etc., is it possible to have
Theorem. Within the limits
and
and
and
there exist certain values of the indeterminates
for which
and
simultaneously.
Proof. Let us suppose that the theorem is not true. Then it is evident that for all values of the indeterminates within the assigned limits, the value of
must be a positive quantity, and therefore the value of
must always be finite. Let us consider the double integral
from
to
and from
to
which has a fully determined finite value. This value, which we denote by
must be the same whether the integration is first carried out with respect to
and then with respect to
or in the reverse order. However, we have indefinitely, considering
as constant,
as is easily confirmed by differentiation with respect to
. A constant need not be added, assuming that the integral begins at
since for
we have
. Therefore, since
clearly also vanishes for
the integral
from
to
becomes
with
remaining indefinite. It follows from this that
On the other hand, considering
as a constant, we have indefinitely
as is easily confirmed by differentiation with respect to
. Here too, a constant need not be added, assuming that the integral starts at
. Therefore, since the integral from
to
is carried out by what has been demonstrated in the previous article, it is
and therefore, by the theorem in the previous article, it is always a positive quantity for any real value of
. Hence
i.e., the value of the integral
from
to
is necessarily a positive quantity[1]. This is absurd, as we previously found that the same quantity is
Thus the assumption cannot hold, and the truth of the theorem is established.
The function
is transformed into
by the substitution
and likewise it is transformed into
by the substitution
Therefore, if for determined values of
and
, say for
,
, it simultaneously results in
(as demonstrated in the previous article), then
obtains the value
for both substitutions
Consequently, it is indefinitely divisible by
Whenever
is not equal to 0, nor
, these divisors are unequal. Thus,
is divisible by their product
Whenever either
, and hence
, or
, these factors are identical, namely
. It is therefore certain that the function
involves a real divisor of the second or first order. Since the same conclusion holds for the quotient,
can be completely resolved into such factors. Q.E.D.
Although we have fully dealt with the matter proposed in the preceding section, it will not be superfluous to add some further reasoning about art. 2. Starting from the assumption that
and
vanish for any values of the indeterminates
within the assigned limits, we have fallen into an inevitable contradiction, from which we concluded the falsity of the assumption itself. Therefore, this contradiction must cease if there are indeed values of
for which
and
simultaneously become
To illustrate this more clearly, we observe that for such values,
and consequently,
becomes infinite. Hence, it will no longer be permissible to treat the double integral
as an assignable quantity.
In general terms, denoting
as indefinite coordinates of points in space, the integral
represents the volume of a solid contained between five planes with equations:
and the surface with equation
considering those parts as negative where the
coordinates are negative. However, it is implicitly understood here that the sixth surface is continuous. When this condition ceases, and
becomes infinite, it is indeed possible that the concept lacks meaning. In such a case, it is impossible to speak about the integral
, and it is not surprising that analytical operations applied blindly to empty calculations lead to absurdities.
The integral
is a true integration, i.e., a summation, as long as within the limits over which it extends,
is everywhere a finite quantity. It becomes absurd, however, if
becomes infinite somewhere within those limits. For an integral like
which generally represents the area between the abscissa line and the curve with the ordinate
for the abscissa
when we evaluate it according to usual rules, often ignoring continuity, we are frequently entangled in contradictions. For example, assuming
the analysis provides an integral
by which the area is correctly defined as long as the curve maintains continuity. However, if it is interrupted at
and someone incorrectly inquires about the magnitude of the area from the negative abscissa to the positive one, the formula will yield an absurd answer, stating that it is negative. We will explore the meaning of these and similar paradoxes of analysis more extensively on another occasion.
Let me add a final observation. In the unrestricted statements of the questions, which may turn out to be absurd in certain cases, consulting analysis often leads to ambiguous answers. Thus, the value of the integral
from
to
and from
to
if the value of
can be extended, through analytical operations, and is easily obtained as
Of course, the integral can only have a definite value whenever
remains finite within the assigned limits. This value, given by the formula, is certainly satisfactory, but it is not yet fully determined through it, as
is a multiform function. It will be necessary to decide which values of the function should be used in the specific case, through other considerations (which are not difficult). On the other hand, whenever
becomes infinite somewhere within the assigned limits, the question of the value of the integral
is absurd. Despite this, if you insist on extracting an answer from the analysis, a variety of methods will lead you to the same result, in one way or another, each of which will be contained within the previously given general formula.
- ↑ As is now self-evident. However, the indefinite integral is easily shown to be
, and it can be shown elsewhere (since it is not immediately obvious which value from the infinitely many possible values of the multiform function
should be adopted for
), that the definite integral from
to
will be
or
. However, this is not necessary for our purpose.