Translation:Disquisitiones generales circa seriem infinitam ...

The series which we propose to investigate in this treatise can be regarded as a function of four quantities $\alpha ,$ $\beta ,$ $\gamma ,$ $x,$ which we shall call its elements. We will distinguish these by their order, with the first element being $\alpha ,$ the second $\beta ,$ the third $\gamma ,$ and the fourth $x.$ It is clear that the first and second elements can be interchanged: therefore, if for the sake of brevity we denote our series by the symbol $F(\alpha ,\beta ,\gamma ,x),$ then we shall have $F(\alpha ,\beta ,\gamma ,x)=F(\beta ,\alpha ,\gamma ,x).$

By assigning definite values to the elements $\alpha ,$ $\beta ,$ $\gamma ,$ our series becomes a function of a single variable $x,$ which is clearly cut off after the $(1-\alpha {)}^{\text{th}}$ or $(1-\beta {)}^{\text{th}}$ term if $\alpha -1$ or $\beta -1$ is a negative integer, but in other cases it extends indefinitely. In the former case, the series yields a rational algebraic function, but in the latter case, it usually yields a transcendental function. The third element $\gamma$ must neither be a negative integer nor equal to zero, so that we do not have infinitely large terms.

The coefficients of the powers $x^m,$ $x^{m+1}$ in our series are as Template:C and therefore they approach equality as the value of $m$ increases. So, if a definite value is also assigned to the fourth element $x,$ the convergence or divergence of the series will depend on the nature of this value. Indeed, whenever a real value, positive or negative but less than unity, is assigned to $x,$ the series, while not convergent immediately from the beginning, will nevertheless converge after a certain interval and will lead to a sum which is finite and determinate. The same will occur for an imaginary value of $x$ of the form $a+b\sqrt{-1},$ whenever $aa+bb<1.$ On the other hand, for a real value of $x$ greater than unity, or for an imaginary value of the form $a+b\sqrt{-1}$ with $aa+bb>1,$ the series will diverge, perhaps not immediately, but after a certain interval, so that it is meaningless to speak of its "sum". Finally, for the value $x=1$ (or more generally for a value of the form $a+b\sqrt{-1}$ with $aa+bb=1$), the convergence or divergence of the series will depend on the nature of the elements $\alpha ,$ $\beta ,$ $\gamma ,$ as we will discuss, with particular attention to the sum of the series for $x=1,$ in the third section.

It is therefore clear that, to the extent that our function is defined as the sum of a definite series, our investigation must, by its nature, be restricted to those cases where the series actually converges, and hence it is meaningless to ask for the value of the series for values of $x$ which are greater than unity. Furthermore, from the fourth section onwards, we will construct our function on the basis of a deeper principle, which permits the most general application.

Differentiation of our series, considering only the fourth element $x$ as the variable, leads to a similar function, since it is clear that Template:C The same applies to repeated differentiation.

It will be worth our while to include here certain functions that can be reduced to our series and whose use is very common in analysis.

The preceding functions are algebraic or transcendental depending upon logarithms and the circle. However, we do not undertake our general investigation for the sake of these functions, but rather to advance the theory of higher transcendental functions, of which our series encompasses a vast range. Among these, amid countless others, are the coefficients which arise in the expansion of the function $(aa+bb-2ab\cos \phi {)}^{-n}$ into a series in terms of the cosines of the angles $\phi ,2\phi ,3\phi ,$ etc., about which we will speak particularly on another occasion. However, those coefficients can be reduced to the form of our series in several ways. Namely, setting Template:C we have "firstly", Template:C For if we view $aa+bb-2ab\cos \phi$ as the product of $a-br$ and $a-b{r}^{-1}$ (where $r$ denotes the quantity Template:Nowrap then $\mathrm{\Omega }$ is equal to the product of $a^{-2n}$ with Template:C and Template:C Since this must be identical to Template:C the values given above are obtained automatically.

Secondly, we have Template:C These values are easily derived from Template:C

Thirdly, Template:C

Finally fourthly, Template:C

These values and those following are easily derived from Template:C Template:Rule
Template:C

We say that a function is contiguous with $F^{\prime }(\alpha ,\beta ,\gamma ,x),$ if it is obtained from the latter by increasing or decreasing the first, second, or third element by unity, with the remaining three elements being held constant. Thus the primary function $F(\alpha .\beta ,\gamma ,x)$ produces six contiguous ones, any two of which are related to the primary function by a very simple linear equation. These equations, fifteen in number, are given below. For the sake of brevity we have omitted the fourth element, which is always understood to be $=x,$ and we have denoted the primary function simply by $F.$

[1]	$0=(\gamma -2\alpha -(\beta -\alpha )x)F+\alpha (1-x)F(\alpha +1,\beta ,\gamma )-(\gamma -\alpha )F(\alpha -1,\beta ,\gamma )$
[2]	$0=(\beta -\alpha )F+\alpha F(\alpha +1,\beta ,\gamma )-\beta F(\alpha ,\beta +1,\gamma )$
[3]	$0=(\gamma -\alpha -\beta )F+\alpha (1-x)F(\alpha +1,\beta ,\gamma )-(\gamma -\beta )F(\alpha ,\beta -1,\gamma )$
[4]	$0=\gamma (\alpha -(\gamma -\beta )x)F-\alpha \gamma (1-x)F(\alpha +1,\beta ,\gamma )+(\gamma -\alpha )(\gamma -\beta )xF(\alpha ,\beta ,\gamma +1)$
[5]	$0=(\gamma -\alpha -1)F+\alpha F(\alpha +1,\beta ,\gamma )-(\gamma -1)F(\alpha ,\beta ,\gamma -1)$
[6]	$0=(\gamma -\alpha -\beta )F-(\gamma -\alpha )F(\alpha -1,\beta ,\gamma )+\beta (1-x)F(\alpha ,\beta +1,\gamma )$
[7]	$0=(\beta -\alpha )(1-x)F-(\gamma -\alpha )F(\alpha -1,\beta ,\gamma )+(\gamma -\beta )F(\alpha ,\beta -1,\gamma )$
[8]	$0=\gamma (1-x)F-\gamma F(\alpha -1,\beta ,\gamma )+(\gamma -\beta )xF(\alpha ,\beta ,\gamma +1)$
[9]	$0=(\alpha -1-(\gamma -\beta -1)x)F+(\gamma -\alpha )F(\alpha -1,\beta ,\gamma )-(\gamma -1)(1-x)F(\alpha ,\beta ,\gamma -1)$
[10]	$0=(\gamma -2\beta +(\beta -\alpha )x)F+\beta (1-x)F(\alpha ,\beta +1,\gamma )-(\gamma -\beta )F(\alpha ,\beta -1,\gamma )$
[11]	$0=\gamma (\beta -(\gamma -\alpha )x)F-\beta \gamma (1-x)F(\alpha ,\beta +1,\gamma )-(\gamma -\alpha )(\gamma -\beta )F(\alpha ,\beta ,\gamma +1)$
[12]	$0=(\gamma -\beta -1)F+\beta F(\alpha ,\beta +1,\gamma )-(\gamma -1)F(\alpha ,\beta ,\gamma -1)$
[13]	$0=\gamma (1-x)F-\gamma F(\alpha ,\beta -1,\gamma )+(\gamma -\alpha )xF(\alpha ,\beta ,\gamma +1)$
[14]	$0=(\beta -1-(\gamma -\alpha -1)x)F+(\gamma -\beta )F(\alpha ,\beta -1,\gamma )-(\gamma -1)(1-x)F(\alpha ,\beta ,\gamma -1)$
[15]	$0=\gamma (\gamma -1-(2\gamma -\alpha -\beta -1)x)F+(\gamma -\alpha )(\gamma -\beta )xF(\alpha ,\beta ,\gamma +1)-\gamma (\gamma -1)(1-x)F(\alpha ,\beta ,\gamma -1)$

Now here is the proof of these formulas. If we set Template:C then the coefficient of $x^m$ will be as follows:

in $F....$	$..\alpha (\beta +m-1)M$
in $F(\alpha ,\beta -1,\gamma )$	$..\alpha (\beta -1)M$
in $F(\alpha +1,\beta ,\gamma )$	$..(\alpha +m)(\beta +m-1)M$
in $F(\alpha ,\beta ,\gamma -1)$	$..\frac{\alpha (\beta +m-1)(\gamma +m-1)M}{\gamma -1}$

Moreover, the coefficient of $x^{m-1}$ in $F(\alpha +1,\beta ,\gamma )$, or the coefficient of $x^m$ in $xF(\alpha +1,\beta ,\gamma ),$ is Template:C Hence, the truth of formulas 5 and 3 is immediately apparent. Formula 12 arises from 5 by swapping $\alpha$ and $\beta ,$ and from these two, elimination yields 2. Similarly, by the same permutation, formula 6 arises from 3; combining 6 and 12 yields 9, permuting yields 14, and combining these gives 7. Finally, from 2 and 6, 1 is derived, and then by permutation, 10. Formula 8 can be derived in a similar manner to formulas 5 and 3 above, from the consideration of coefficients (if desired, all 15 formulas could be derived in a similar way), or more elegantly from the known equations, as follows.

By changing the element $\alpha$ to $\alpha -1$ and $\gamma$ to $\gamma +1$ in formula 5, we obtain Template:C On the other hand, by changing only $\gamma$ to $\gamma +1$ in formula 9, we get Template:C Subtracting these formulas immediately yields 8, and hence by permutation, 13. From 1 and 8, 4 follows, and then by permutating, 11. Finally, 15 is deduced from 8 and 9.

If ${\alpha }^{\prime }-\alpha ,$ ${\beta }^{\prime }-\beta ,$ ${\gamma }^{\prime }-\gamma ,$ and ${\alpha }^{\prime \prime }-\alpha ,$ ${\beta }^{\prime \prime }-\beta ,$ ${\gamma }^{\prime \prime }-\gamma$ are all integers (positive or negative), one can go from the function $F(\alpha ,\beta ,\gamma )$ to the function $F({\alpha }^{\prime },{\beta }^{\prime },{\gamma }^{\prime })$, and likewise from there to the function $F({\alpha }^{\prime \prime },{\beta }^{\prime \prime },{\gamma }^{\prime \prime })$ through a series of similar functions, such that each one is contiguous to the preceding and succeeding ones. This is achieved by changing one element, e.g. $\alpha$, by one unit repeatedly, until one reaches $F({\alpha }^{\prime },\beta ,\gamma ),$ and then changing the second element until one reaches $F({\alpha }^{\prime },{\beta }^{\prime },\gamma ),$ and finally changing the third element until one reaches $F({\alpha }^{\prime },{\beta }^{\prime },{\gamma }^{\prime }),$ and so on to $F({\alpha }^{\prime \prime },{\beta }^{\prime \prime },{\gamma }^{\prime \prime }).$ Since linear equations exist, according to art. 7, between the first, second, and third functions, and generally between any three consecutive functions in this series, it is easily understood that linear equations between the functions $F(\alpha ,\beta ,\gamma ),$ $F({\alpha }^{\prime },{\beta }^{\prime },{\gamma }^{\prime }),$ $F({\alpha }^{\prime \prime },{\beta }^{\prime \prime },{\gamma }^{\prime \prime }),$ and so forth can be deduced by elimination. Thus, generally speaking, from two functions whose first three elements differ by integers, any other function with the same property can be obtained, provided that the fourth element remains the same. For what remains, it suffices to establish this remarkable truth generally; we shall not dwell on the shortcuts by which the operations required for this purpose can be made as brief as possible.

Suppose that we are given e.g. the functions Template:C between which a linear relation must be found. We can connect them through the following contiguous functions: Template:C Thus we have five linear equations (from formulas 6, 13, 5 of art. 7): Template:Ordered list Eliminating $F^{\prime }$ from I and II yields Template:Ordered list Eliminating $F^{\prime \prime }$ from this and III yields Template:Ordered list Eliminating $F^{\prime \prime \prime \prime \prime }$ from IV and V yields Template:Ordered list Finally, eliminating $F^{\prime \prime \prime \prime }$ from this and VII yields Template:Ordered list

If we wanted to exhaust all relations among triplets of functions $F(\alpha ,\beta ,\gamma ),$ $F(\alpha +\lambda ,\beta +\mu ,\gamma +\nu ),$ $F(\alpha +{\lambda }^{\prime },\beta +{\mu }^{\prime },\gamma +{\nu }^{\prime }),$ where $\lambda ,$ $\mu ,$ $\nu ,$ ${\lambda }^{\prime },$ ${\mu }^{\prime },$ ${\nu }^{\prime }$ are either $=0,$ $=+1,$ or $=-1,$ then the number of formulas would increase to 325. Such a collection would not be useless; but it will suffice to present only a few here. These can be easily demonstrated, either from the formulas in art. 7 or, if one prefers, in the same manner as in art. 8.

[16]	$F(\alpha ,\beta ,\gamma )-F(\alpha ,\beta ,\gamma -1)$	$=-\frac{\alpha \beta x}{\gamma (\gamma -1)}F(\alpha +1,\beta +1,\gamma +1)$
[17]	$F(\alpha ,\beta +1,\gamma )-F(\alpha ,\beta ,\gamma )$	$=\frac{\alpha x}{\gamma }F(\alpha +1,\beta +1,\gamma +1)$
[18]	$F(\alpha +1,\beta ,\gamma )-F(\alpha ,\beta ,\gamma )$	$=\frac{\beta x}{\gamma }F(\alpha +1,\beta +1,\gamma +1)$
[19]	$F(\alpha ,\beta +1,\gamma +1)-F(\alpha ,\beta ,\gamma )$	$=\frac{\alpha (\gamma -\beta )x}{\gamma (\gamma +1)}F(\alpha +1,\beta +1,\gamma +2)$
[20]	$F(\alpha +1,\beta ,\gamma +1)-F(\alpha ,\beta ,\gamma )$	$=\frac{\beta (\gamma -\alpha )x}{\gamma (\gamma +1)}F(\alpha +1,\beta +1,\gamma +2)$
[21]	$F(\alpha -1,\beta +1,\gamma )-F(\alpha ,\beta ,\gamma )$	$=\frac{(\alpha -\beta -1)x}{\gamma }F(\alpha ,\beta +1,\gamma +1)$
[22]	$F(\alpha +1,\beta -1,\gamma )-F(\alpha ,\beta ,\gamma )$	$=\frac{(\beta -\alpha -1)x}{\gamma }F(\alpha +1,\beta ,\gamma +1)$
[23]	$F(\alpha -1,\beta +1,\gamma )-F(\alpha +1,\beta -1,\gamma )$	$=\frac{(\alpha -\beta )x}{\gamma }F(\alpha +1,\beta +1,\gamma +1)$

Template:Rule
Template:C

Denoting Template:C by $G(\alpha ,\beta ,\gamma ,x),$ we have Template:C and thus, dividing equation 19 by $F(\alpha ,\beta +1,\gamma +1,x),$ Template:C or

[24]	$G(\alpha ,\beta ,\gamma ,x)=\frac{1}{1-\frac{\alpha (\gamma -\beta )}{\gamma (\gamma +1)}xG(\beta +1,\alpha ,\gamma +1,x)}$

and since similarly Template:C etc., we obtain the following continued fraction for $G(\alpha ,\beta ,\gamma ,x),$

[25]	$\frac{F(\alpha ,\beta +1,\gamma +1,x)}{F(\alpha ,\beta ,\gamma ,x)}=\frac{1}{1-\frac{ax}{1-\frac{bx}{1-\frac{cx}{1-\frac{dx}{1-\dots }}}}}$

where Template:C etc., where the law of the progression is obvious.

Moreover, from equations 17, 18, 21, 22, we have

[26]	$\frac{F(\alpha ,\beta +1,\gamma ,x)}{F(\alpha ,\beta ,\gamma ,x)}=\frac{1}{1-\frac{\alpha x}{\gamma }G(\beta +1,\alpha ,\gamma ,x)},$
[27]	$\frac{F(\alpha +1,\beta ,\gamma ,x)}{F(\alpha ,\beta ,\gamma ,x)}=\frac{1}{1-\frac{\beta x}{\gamma }G(\alpha +1,\beta ,\gamma ,x)},$
[28]	$\frac{F(\alpha -1,\beta +1,\gamma ,x)}{F(\alpha ,\beta ,\gamma ,x)}=\frac{1}{1-\frac{(\alpha -\beta -1)x}{\gamma }G(\beta +1,\alpha -1,\gamma ,x)},$
[29]	$\frac{F(\alpha +1,\beta -1,\gamma ,x)}{F(\alpha ,\beta ,\gamma ,x)}=\frac{1}{1-\frac{(\beta -\alpha -1)x}{\gamma }G(\alpha +1,\beta -1,\gamma ,x)},$

from which, substituting the values of the function $G$ into the continued fractions, an equal number of new continued fractions emerge.

Finally, it is clear that the continued fraction in formula 25 automatically terminates if any of the numbers $\alpha ,$ $\beta ,$ $\gamma -\alpha ,$ $\gamma -\beta$ is a negative integer, and otherwise it runs to infinity.

The continued fractions in the previous article are of the utmost importance, and it can be asserted that hardly any continued fractions progressing according to a known law have so far been extracted by analysts, which are not special cases of ours. Especially notable is the case where we set $\beta =0$ in formula 25, so that $F(\alpha ,\beta ,\gamma ,x)=1,$ and therefore, writing $\gamma -1$ instead of $\gamma ,$ Template:C where Template:C etc.

It will be worth our while to include some special cases here. Setting $t=1,$ $\beta =1,$ it follows from formula I of art. 5 that

[31]	$(1+u{)}^n=\frac{1}{1-\frac{nu}{1+\frac{\frac{n+1}{2}u}{1-\frac{\frac{n-1}{2.3}u}{1+\frac{\frac{2(n+2)}{3.4}u}{1-\frac{\frac{2(n-2)}{4.5}u}{1+\dots }}}}}}$

From formulas VI and VII of art. 5, we have

[32]	$\log (1+t)=\frac{t}{1+\frac{{\frac{1}{2}}_{}t}{1+\frac{{\frac{1}{6}}_{}t}{1+\frac{{\frac{2}{6}}_{}t}{1+\frac{{\frac{2}{10}}_{}t}{1+\frac{{\frac{3}{10}}_{}t}{1+\frac{{\frac{3}{14}}_{}t}{1^{}+\dots }}}}}}}$
[33]	$\log \frac{1+t}{1-t}=\frac{2t}{1-\frac{\frac{1}{3}tt}{1-\frac{{\frac{2.2}{3.5}}_{}tt}{1-\frac{{\frac{3.3}{5.7}}_{}tt}{1-\frac{{\frac{4.4}{7.9}}_{}tt}{1^{}-\dots }}}}}$

Changing the sign $-$ to $+$ here yields the continued fraction for $\mathrm{arc.tang}t.$

Furthermore, we have

[34]	$e^t=\frac{1}{1-\frac{t}{1+\frac{\frac{1}{3}t}{1-\frac{\frac{1}{6}t}{1+\frac{\frac{1}{6}t}{1-\frac{\frac{1}{10}t}{1+\frac{\frac{1}{10}t}{1-\dots }}}}}}}$
[35]	$t=\frac{\sin t\cos t}{1-\frac{\frac{1.2}{1.3}{\sin t}^2}{1-\frac{\frac{1.2}{3.5}{\sin t}^2}{1-\frac{\frac{3.4}{5.7}{\sin t}^2}{1-\frac{\frac{3.4}{7.9}{\sin t}^2}{1-\frac{\frac{5.6}{9.11}{\sin t}^2}{1-\dots }}}}}}$

Setting $\alpha =3,$ $\gamma =\frac{5}{2},$ the continued fraction presented in art. 90 of Theoria motus corporum coelestium follows automatically from formula 30. Two other continued fractions are also proposed there, the development of which we thought to supply here. Setting Template:C then (l.c.) $x-\xi =\frac{x}{1+\frac{2x}{35Q}}=\frac{xQ}{Q+\frac{2}{35}x},$ hence Template:C which is the first formula; the second is derived as follows. Setting Template:C we have, by formula 25, Template:C

Hence Template:C or by swapping the first and second elements, Template:C

However, by equation 21, we have Template:C from which it follows that $Q=R-\frac{4}{7}x,$ and substituting this value into the formula above yields Template:C which is the second formula.

Setting $\alpha =\frac{m}{n},$ $x=-\gamma nt,$ in formula 30 yields, for an infinitely large value of $\gamma ,$

[36]	$F(\frac{m}{n},1,\gamma ,-\gamma nt)=1-mt+m(m+n)tt-m(m+n)(m+2n)t^3+\dots$
	$=\frac{1}{1+\frac{mt}{1+\frac{nt}{1+\frac{(m+n)t}{1+\frac{2nt}{1+\frac{(m+2n)t}{1+3nt\dots }}}}}}$

Template:Rule
Template:C

Whenever the elements $\alpha ,$ $\beta ,$ $\gamma$ are all positive quantities, all coefficients of powers of the fourth element $x$ become positive: and whenever one or another of those elements is negative, at least from some power $x^m$ onwards all coefficients will have the same sign, provided that $m$ is taken greater than the absolute value of the most negative element. It is clear from this that the sum of the series for $x=1$ cannot be finite unless the coefficients decrease to infinity after a certain term, or, to speak in the manner of analysts, unless the coefficient of the term $x^{\mathrm{\infty }}$ is $=0.$ Indeed, for the benefit of those who favor the rigorous methods of the ancient geometers, we will show with all rigor that

first, the coefficients (since the series is not terminated) increase to infinity indefinitely whenever $\alpha +\beta -\gamma -1$ is a positive quantity.

second, the coefficients converge continually towards a finite limit whenever $\alpha +\beta -\gamma -1=0.$

third, the coefficients decrease to infinity indefinitely whenever $\alpha +\beta -\gamma -1$ is a negative quantity.

fourth, the sum of our series for $x=1,$ notwithstanding convergence in the third case, is infinite whenever $\alpha +\beta -\gamma$ is a positive quantity or $=0.$

fifth, the sum is truly finite whenever $\alpha +\beta -\gamma$ is a negative quantity.

We will apply this general discussion to the infinite series $M,$ $M^{\prime },$ $M^{″},$ $M^{‴},$ etc., which is formed so that the quotients $\frac{M^{\prime }}{M},$ $\frac{M^{″}}{M^{\prime }},$ $\frac{M^{‴}}{M^{″}},$ etc. resp. are the values of the fraction Template:C for $t=m,$ $t=m+1,$ $t=m+2$, etc. For brevity, we will denote the numerator of this fraction by $P$ and the denominator by $p$. Furthermore, we assume that $P$ and $p$ are not identical, or equivalently that the differences $A-a,$ $B-b,$ $C-c,$ etc., do not all vanish simultaneously.

I. Whenever the first of the differences $A-a,$ $B-b,$ $C-c,$ etc., which does not vanish is positive, some limit $l$ can be assigned, beyond which the values of the functions $P$ and $p$ will always be positive and $P>p$. It is evident that this occurs when $l$ is taken as the largest real root of the equation $p(P-p)=0;$ if this equation has no real roots at all, then this property holds for all values of $t$. Therefore, in the series $\frac{M^{\prime }}{M},$ $\frac{M^{″}}{M^{\prime }},$ $\frac{M^{‴}}{M^{″}},$ etc., at least after a certain interval (if not from the beginning), all terms will be positive and greater than unity. Consequently, if none of them tends to zero or infinity, it is clear that

the series $M,$ $M^{\prime },$ $M^{″},$ $M^{‴},$ etc., if not from the beginning, then at least after a certain interval, will have all its terms affected by the same sign and continually increasing.

By the same reasoning, if the first of the differences $A-a,$ $B-b,$ $C-c,$ etc. which does not vanish is negative, then the series $M,$ $M^{\prime },$ $M^{″},$ $M^{‴},$ etc., will, if not from the beginning, then at least after a certain interval, have all its terms affected by the same sign and continually decreasing.

II. Now, if the coefficients $A,$ $a$ are unequal, the terms of the series $M,$ $M^{\prime },$ $M^{″},$ $M^{‴},$ etc., will either increase or decrease to infinity, depending on whether the difference $A-a$ is positive or negative: we demonstrate this as follows. If $A-a$ is positive, let an integer $h$ be chosen so that $h(A-a)>1,$ and let $\frac{M^h}{m}=N,$ $\frac{M{\text{'}}^h}{m+1}=N^{\prime },$ $\frac{M^{\prime }{\text{'}}^h}{m+2}=N^{″},$ $\frac{M^{″}{\text{'}}^h}{m+3}=N^{‴}$, etc., and also $t{P}^h=Q,$ $(t+1)p^h=q.$ Then it is clear that $\frac{N^{\prime }}{N},$ $\frac{N^{″}}{N^{\prime }},$ $\frac{N^{‴}}{N^{″}},$ etc., are values of the fraction $\frac{Q}{q}$ when $t=m,$ $t=m+1,$ $t=m+2$, etc., while $Q,$ $q$ themselves are algebraic functions of the form Template:C Therefore, since by hypothesis the difference $hA-(ha+1)$ is positive, the terms of the series $N,$ $N^{\prime },$ $N^{″},$ $N^{‴},$ etc. will, if not from the beginning, then after a certain interval, continually increase (by I). Hence the terms of the series $mN,$ $(m+1)N^{\prime },$ $(m+2)N^{″},$ $(m+3)N^{‴},$ etc., will necessarily increase beyond all limits, and therefore the terms of the series $M,$ $M^{\prime },$ $M^{″},$ $M^{‴}$, etc., whose exponents are equal to $h$, will do so as well. Q.E.D.

If $A-a$ is negative, then the integer $h$ must be chosen so that $h(a-A)$ is greater than $1$, and similar reasoning leads to the conclusion that the terms of the series Template:C will continually decrease after a certain interval. Therefore, the terms of the series $M^h,$ $M{\text{'}}^h,$ $M^{\prime }{\text{'}}^h,$ etc., and consequently also the terms of the series $M,$ $M^{\prime },$ $M^{″},$ $M^{‴}$, etc., will necessarily tend to infinity. Q.E.S.

III. On the other hand, if the coefficients $A,$ $a$ are equal, then the terms of the series $M,$ $M^{\prime },$ $M^{″},$ $M^{‴},$ etc., converge continually to a finite limit: we demonstrate this as follows. First, let us suppose that the terms of the series increase continually after a certain interval, or equivalently that the first of the differences $B-b,$ $C-c,$ etc. which does not vanish is positive. Let $h$ be an integer such that $h+b-B$ becomes a positive quantity. Set Template:C and $(tt-1{)}^{h}P=Q,$ $t^{2h}p=q,$ such that $\frac{N^{\prime }}{N},$ $\frac{N^{″}}{N^{\prime }}$, etc., are values of the fraction $\frac{Q}{q}$ when $t=m,$ $t=m+1$, etc. Therefore, since we have Template:C and since $B-h-b$ is a negative quantity by hypothesis, the terms of the series $N,$ $N^{\prime },$ $N^{″},$ $N^{‴},$ etc., will decrease continually after a certain interval. Therefore, the corresponding terms of the series $M,$ $M^{\prime },$ $M^{″},$ $M^{‴},$ etc. which are always smaller, while also increasing continually, must converge to a finite limit. Q.E.D.

If the terms of the series $M,$ $M^{\prime },$ $M^{″},$ $M^{‴},$ etc., decrease continually after a certain interval, an integer $h$ must be chosen such that $h+B-b$ becomes a positive quantity. It then becomes evident from entirely similar reasoning that the terms of the series Template:C increase continually after a certain interval. Therefore, the corresponding terms of the series $M,$ $M^{\prime },$ $M^{″},$ etc., which are always greater, while also decreasing continually, must converge to a finite limit. Q.E.S.

IV. Lastly, concerning the sum of the series whose terms are $M,$ $M^{\prime },$ $M^{″},$ $M^{‴},$ etc. in the case where these terms decrease indefinitely, let us first suppose that $A-a$ falls between $0$ and $-1$, meaning that $A+1-a$ is either a positive quantity or $=0.$ Let $h$ be a positive integer, chosen arbitrarily in the case where $A+1-a$ is positive, or so that it makes the quantity $h+m+A+B-b$ positive in the case where $A+1-a=0.$ Then we will have Template:C where either $A+1-m-h-(a-m-h)$ is positive, or, if it equals $=0,$ then at least $B-A(m+h-1)-(b-a(m+h))$ will be positive. Hence (by I), a value $l$ can be assigned to the quantity $t$, which, once exceeded, will ensure that the values of the fraction $\frac{P(t-(m+h-1))}{p(t-(m+h))}$ will always be positive and greater than unity. Let $n$ be an integer greater than $l$ and also greater than $h,$ and let the terms of the series $M,$ $M^{\prime },$ $M^{″},$ $M^{‴},$ etc., corresponding to the values $t=m+n,$ $t=m+n+1,$ $t=m+n+2,$ etc., be denoted by $N,$ $N^{\prime },$ $N^{″},$ $N^{‴},$ etc. Then Template:C will be positive quantities greater than one, so that Template:C Consequently, the sum of the series $N+N^{\prime }+N^{″}+N^{‴}+\dots$ will be greater than the sum of the series Template:C no matter how many terms are included. However, as the number of terms increases indefinitely, the latter series exceeds all limits, as the sum of the series $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots$ is known to be infinite and remains infinite even if the terms $1+\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{n-1-h}$ are removed from the beginning. Hence, the sum of the series $N+N^{\prime }+N^{″}+N^{‴}+\dots ,$ and consequently the sum of $M+M^{\prime }+M^{″}+M^{‴}+\dots$ of which it is a part, increases beyond all limits.

V. However, when $A-a$ is a negative quantity that is absolutely greater than one, the sum of the series $M+M^{\prime }+M^{″}+M^{‴}+\dots$ will certainly be finite when continued indefinitely. Indeed, let $h$ be a positive quantity less than $a-A-1.$ Then similar reasoning shows that the quantity $t$ can be assigned a value $l,$ beyond which the fraction $\frac{Pt}{p(t-h-1)}$ always has positive values less than unity. Now, if we take an integer $n$ greater than $l,$ $m,$ $h+1,$ and let the terms of the series $M,$ $M^{\prime },$ $M^{″},$ $M^{‴},$ etc., corresponding to the values $t=n,$ $t=n+1,$ $t=n+2,$ etc., be denoted by $N,$ $N^{\prime },$ $N^{″},$ etc., then Template:C Consequently, the sum of the series $N+N^{\prime }+N^{″}+\dots ,$ no matter how many terms are included, is less than the product of $N$ with the sum of the same number of terms of the series Template:C However, the sum of this series can be easily found for any number of terms. In particular,

The first term	$=\frac{n-1}{h}-\frac{n-h-1}{h}$
The sum of the first two terms	$=\frac{n-1}{h}-\frac{(n-h-1)(n-h)}{hn}$
The sum of the first three terms	$=\frac{n-1}{h}-\frac{(n-h-1)(n-h)(n-h+1)}{hn(n+1)}$ etc.

and since the second part (by II) forms a series which decreases beyond all limits, the sum must be $=\frac{n-1}{h}.$ Hence $N+N^{\prime }+N^{″}+\dots ,$ when continued infinitely, will always remain less than $\frac{N(n-1)}{h},$ and thus $M+M^{\prime }+M^{″}+\dots$ will certainly converge to a finite sum. Q.E.D.

VI. In order to apply those general assertions concerning the series $M,$ $M^{\prime },$ $M^{″},$ etc. to the coefficients of the powers $x^m,$ $x^{m+1},$ $x^{m+2},$ etc. in the series $F(\alpha ,\beta ,\gamma ,x),$ it is necessary to set $\lambda =2,$ $A=\alpha +\beta ,$ $B=\alpha \beta ,$ $a=\gamma +1,$ $b=\gamma ,$ from which the five assertions in the preceding article follow automatically.

Therefore, investigations of the nature of the sum of the series $F(\alpha ,\beta ,\gamma ,1)$ are naturally restricted to the case where $\gamma -\alpha -\beta$ is a positive quantity, in which case the sum will always be a finite quantity. However, we must begin with the following observation. If, after a certain term, the coefficients of the series $1+ax+bxx+c{x}^3+\dots =S$ decrease beyond all limits, then the product Template:C must $=0$ when for $x=1,$ even if the sum of the series $S$ becomes infinitely large. For since the sum of two terms is $=a,$ the sum of three is $=b,$ the sum of four is $=c$, etc., the limit of the sum when continued indefinitely will be $=0.$ Therefore, whenever $\gamma -\alpha -\beta$ is a positive quantity, we must have $(1-x)F(\alpha ,\beta ,\gamma -1,x)=0$ for $x=1,$ and hence, by equation 15 of art. 7,

	Template:Nowrap
[37]	$F(\alpha ,\beta ,\gamma ,1)=\frac{(\gamma -\alpha )(\gamma -\beta )}{\gamma (\gamma -\alpha -\beta )}F(\alpha ,\beta ,\gamma +1,1)$

Thus, similarly, we have Template:C and so on, where $k$ denotes an arbitrary positive integer, Template:C with $(\gamma -\alpha )(\gamma +1-\alpha )(\gamma +2-\alpha )\dots (\gamma +k-1-\alpha )$
and $(\gamma -\beta )(\gamma +1-\beta )(\gamma +2-\beta )\dots (\gamma +k-1-\beta ),$
divided by the product of
$\gamma (\gamma +1)(\gamma +2)\dots (\gamma +k-1)$
with $(\gamma -\alpha -\beta )(\gamma +1-\alpha -\beta )(\gamma +2-\alpha -\beta )\dots (\gamma +k-1-\alpha -\beta ).$

We now introduce the notation

[38]	$\mathrm{\Pi }(k,z)=\frac{1\cdot 2\cdot 3\dots \dots k}{(z+1)(z+2)(z+3)\dots (z+k)}{k}^z,$

where $k$ is naturally restricted to be a positive integer, and with this restriction, $\mathrm{\Pi }(k,z)$ represents a function determined solely by the two quantities $k$ and $z$. Then it is easy to understand that the theorem proposed at the end of the preceding article can be expressed as follows:

[39]	$F(\alpha ,\beta ,\gamma ,1)=\frac{\mathrm{\Pi }(k,\gamma -1)\cdot \mathrm{\Pi }(k,\gamma -\alpha -\beta -1)}{\mathrm{\Pi }(k,\gamma -\alpha -1)\cdot \mathrm{\Pi }(k,\gamma -\beta -1)}\cdot F(\alpha ,\beta ,\gamma +k,1)$

It will be worthwhile to examine the nature of the function $\mathrm{\Pi }(k,z)$ in more detail. Whenever $z$ is a negative integer, the function evidently has an infinitely large value, as long as a sufficiently large value is assigned to $k$. For non-negative integer values of $z$, we have: Template:C etc., and generally:

[40]	$\mathrm{\Pi }(k,z)=\frac{1.2.3\dots \dots z}{(1+\frac{1}{k})(1+\frac{2}{k})(1+\frac{3}{k})\dots (1+\frac{z}{k})}$

For arbitrary values of $z$, we have:

[41]	$\mathrm{\Pi }(k,z+1)=\mathrm{\Pi }(k,z)\cdot \frac{1+z}{1+\frac{1+z}{k}},$
[42]	$\mathrm{\Pi }(k+1,z)=\mathrm{\Pi }(k,z)\cdot \left\{\frac{{(1+\frac{1}{k})}^{z+1}}{1+\frac{1+z}{k}}\right\},$

and therefore, since $\mathrm{\Pi }(1,z)=\frac{1}{z+1},$

[43]	$\mathrm{\Pi }(k,z)=\frac{1}{z+1}\cdot \frac{2^{z+1}}{1^z\cdot (2+z)}\cdot \frac{3^{z+1}}{2^z(3+z)}\cdot \frac{4^{z+1}}{3^z(4+z)}\cdots \frac{k^{z+1}}{(k-1{)}^z(k+z)}$

It is worth giving special attention to the limit toward which the function \Pi(k, z) continually converges, as k increases to infinity. First, let $h$ be a finite value of $k$ which is greater than $z$. Then it is clear that, as $k$ increases from $h$ to $h+1,$ the logarithm of $\mathrm{\Pi }(k,z)$ receives an increment which can be expressed by the following convergent series: Template:C Therefore, as $k$ increases from $h$ to $h+n,$ the logarithm of $\mathrm{\Pi }(k,z)$ will receive an increment Template:C which will always remain finite, even when $n$ tends to infinity, as can be easily demonstrated. Therefore, unless an infinite factor is already present in $\mathrm{\Pi }(h,z),$ i.e., unless $z$ is a negative integer, the limit of $\mathrm{\Pi }(k,z)$ as $k$ tends to infinity will certainly be a finite quantity. Hence, it is evident that $\mathrm{\Pi }(\mathrm{\infty },z)$ depends solely on $z,$ or in other words, it is a function of $z$ alone, which we will simply denote by $\mathrm{\Pi }z.$ We therefore define the function $\mathrm{\Pi }z$ as the value of the product: Template:C for $k=\mathrm{\infty },$ or, if one prefers, as the limit of the infinite product Template:C

Immediately following from equation 41, we have the fundamental equation:

[44]	$\mathrm{\Pi }(z+1)=(z+1)\mathrm{\Pi }z$

Hence, in general, for any positive integer $n,$

[45]	$\mathrm{\Pi }(z+n)=(z+1)(z+2)(z+3)\dots \dots (z+n)\mathrm{\Pi }z$

For a negative integer value of $z,$ the value of the function $\mathrm{\Pi }z$ will be infinitely large; for non-negative integer values, we have Template:C and, in general

[46]

$\mathrm{\Pi }z=1.2.3\dots z$

However, this property of our function should not be mistaken as its definition, as it is inherently limited to integer values and there exist countless other functions (e.g., $\cos 2\pi z.\mathrm{\Pi }z,$ $\cos \pi z^{2n}\mathrm{\Pi }z$, etc., where $\pi$ denotes the circumference of a circle of radius $=1$) that share the same property.

Although the function $\mathrm{\Pi }(k,z)$ may appear to be more general than $\mathrm{\Pi }z,$ it will henceforth be redundant for us, as it can easily be reduced to the latter. Indeed, it follows from the combination of equations 38, 45, and 46 that

[47]	$\mathrm{\Pi }(k,z)=\frac{k^z\mathrm{\Pi }k.\mathrm{\Pi }z}{\mathrm{\Pi }(k+z)}$

Moreover, the connection of these functions with that which Template:Sc has called "facultates numericae" is evident. Specifically, the facultates numericae, which this author denotes by $a^{bIc}$, can be expressed in our notation as: Template:C However, it seems more advisable to introduce a function of one variable into the analysis, rather than a function of three variables, especially when the latter can be reduced to the former.

The continuity of the function $\mathrm{\Pi }z$ is interrupted whenever its value becomes infinitely large, i.e., for negative integer values of $z.$ Therefore, it will be positive from $z=-1$ to $z=\mathrm{\infty },$ and since for each limit $\mathrm{\Pi }z$ obtains an infinitely large value, there will be a minimum value between them, which we found to be $=0.8856024,$ corresponding to the value $z=0.4616321.$ Between the limits $z=-1$ and $z=-2,$ the value of the function $\mathrm{\Pi }z$ is negative, between $z=-2$ and $z=-3$ it is positive again, and so forth, as follows from equation 44. Furthermore, it is clear that if one knows all the values of the function $\mathrm{\Pi }z$ between two arbitrary limits that differ by unity, e.g. from $z=0$ to $z=1,$ then the value of the function for any other real value of $z$ can be easily deduced from equation 45. To this end, we constructed a table, appended to this section, which gives the Briggsian logarithms of the function $\mathrm{\Pi }z$ to twenty figures, from $z=0$ to $z=1.$ However, it should be noted that the final twentieth figure may be subject to an error of one or two units.

Since the limit of the function $F(\alpha ,\beta ,\gamma +k,1)$ as $k$ increases to infinity is clearly unity, equation 39 transforms into the following:

[48]	$F(\alpha ,\beta ,\gamma ,1)=\frac{\mathrm{\Pi }(\gamma -1).\mathrm{\Pi }(\gamma -\alpha -\beta -1)}{\mathrm{\Pi }(\gamma -\alpha -1).\mathrm{\Pi }(\gamma -\beta -1)}$

This formula provides the complete solution to the question posed in this section. The following elegant equations follow automatically:

[49]	$F(\alpha ,\beta ,\gamma ,1)=F(-\alpha ,-\beta ,\gamma -\alpha -\beta ,1),$
[50]	$F(\alpha ,\beta ,\gamma ,1).F(-\alpha ,\beta ,\gamma -\alpha ,1)=1,$
[51]	$F(\alpha ,\beta ,\gamma ,1).F(\alpha ,-\beta ,\gamma -\beta ,1)=1.$

In the first equation, $\gamma$ must be a positive quantity, and so must be $\gamma -\beta$ in the second and $\gamma -\alpha$ in the third.

Let us apply formula 48 to some of the equations from art. 5. By setting $t=90^{\circ }=\frac{1}{2}\pi$ in Formula XIII, we find that $\frac{1}{2}\pi =F(\frac{1}{2},\frac{1}{2},\frac{3}{2},1)$, which is equivalent to the well-known equation Template:C Therefore, since formula 48 gives us $F(\frac{1}{2},\frac{1}{2},\frac{3}{2},1)=\frac{\mathrm{\Pi }(\frac{1}{2})\cdot \mathrm{\Pi }(-\frac{1}{2})}{\mathrm{\Pi }(0)\cdot \mathrm{\Pi }(0)}$, and since $\mathrm{\Pi }(0)=1$, $\mathrm{\Pi }(\frac{1}{2})=\frac{1}{2}\mathrm{\Pi }(-\frac{1}{2})$, we have $\pi =(\mathrm{\Pi }(-\frac{1}{2}){)}^2$, or

[52]	$\mathrm{\Pi }(-\frac{1}{2})=\sqrt{\pi }$
[53]	$\mathrm{\Pi }(\frac{1}{2})=\frac{1}{2}\sqrt{\pi }$

Formula XVI of art. 5, which is equivalent to the well-known equation Template:C holds generally for any value of $n$, as long as $t$ remains between the limits $-90^{\circ }$ and $+90^{\circ }$. For $t=\frac{1}{2}\pi$, we have Template:C and from this we derive the elegant formula Template:C or setting $n=2z$,

[54]	$\mathrm{\Pi }(-z)\cdot \mathrm{\Pi }(+z)=\frac{z\pi }{\sin z\pi }$
[55]	$\mathrm{\Pi }(-z)\cdot \mathrm{\Pi }(z-1)=\frac{\pi }{\sin z\pi }$

and writing $z+\frac{1}{2}$ for $z,$

[56]	$\mathrm{\Pi }(-\frac{1}{2}+z)\cdot \mathrm{\Pi }(-\frac{1}{2}-z)=\frac{\pi }{\cos z\pi }$

From formula 54, combined with the definition of the function $\mathrm{\Pi }$, it follows that $\frac{z\pi }{\sin z\pi }$ is the limit of the product Template:C as $k$ tends to infinity, and therefore Template:C Similarly, from 56, we deduce Template:C These are well-known formulas, that have been derived by analysts using entirely different methods.

Let $n$ be an integer. Then the value of expression Template:C is found to be Template:C Thus it is independent of $z$, or remains the same regardless of the value assigned to $z$. Therefore, since $\mathrm{\Pi }(k,0)=\mathrm{\Pi }(nk,0)=1$, it is given by the product Template:C As $k$ increases to infinity, we obtain Template:C According to formula 55, the product on the right, when multiplied by itself with order of the factors reversed, yields Template:C Hence, we have the elegant theorem

[57]	$\frac{n^{nz}\mathrm{\Pi }z\cdot \mathrm{\Pi }(z-\frac{1}{n})\cdot \mathrm{\Pi }(z-\frac{2}{n})\dots \dots \mathrm{\Pi }(z-\frac{n-1}{n})}{\mathrm{\Pi }nz}=\frac{(2\pi {)}^{\frac{1}{2}(n-1)}}{\sqrt{n}}.$

The integral $\int x^{\lambda -1}(1-x^{\mu }{)}^{\nu }\mathrm{d}x,$ taken in such a way that it vanishes for $x=0,$ can be expressed by the following series, provided that $\lambda ,$ $\mu$ are positive quantities: Template:C Hence its value for $x=1$ will be Template:C From this theorem, all the relations that the illustrious Template:Sc once painstakingly developed emerge naturally. Thus, by setting e.g. Template:C we have $A=\frac{\mathrm{\Pi }\frac{1}{4}.\mathrm{\Pi }(-\frac{1}{2})}{\mathrm{\Pi }(-\frac{1}{4})},$ $B=\frac{\mathrm{\Pi }\frac{3}{4}.\mathrm{\Pi }(-\frac{1}{2})}{3\mathrm{\Pi }\frac{1}{4}}=\frac{\mathrm{\Pi }(-\frac{1}{4}).\mathrm{\Pi }(-\frac{1}{2})}{4\mathrm{\Pi }\frac{1}{4}},$ and thus $AB=\frac{1}{4}\pi .$ Furthermore, it follows from this, since $\mathrm{\Pi }\frac{1}{4}.\mathrm{\Pi }(-\frac{1}{4})=\frac{\frac{1}{4}\pi }{\sin \frac{1}{4}\pi }=\frac{\pi }{\sqrt{8}},$ that Template:C The numerical value of $A$ computed by Template:Sc is $=1,31102877714605987,$ and the value of $B,$ according to the same author, is $=0,59907011736779611,$ while from our own calculation, employing a particular method, it is Template:C

In general, it can be easily shown that the value of the function $\mathrm{\Pi }z$, if $z$ is a rational quantity $=\frac{m}{\mu },$ where $m,$ $\mu$ are integers, can be deduced from $\mu -1$ values of such integrals evaluated at $x=1$, and indeed in many different ways. Indeed, taking an integer value for $\lambda$ and a fraction with denominator $=\mu$ for $\nu ,$ the value of that integral is always reduced to three $\mathrm{\Pi }z$, where $z$ is a fraction with a denominator $=\mu ;$ any such $\mathrm{\Pi }z$ can be reduced to $\mathrm{\Pi }(-\frac{1}{\mu }),$ or to $\mathrm{\Pi }(-\frac{2}{\mu }),$ or to $\mathrm{\Pi }(-\frac{3}{\mu })$ etc., or to $\mathrm{\Pi }(-\frac{\mu -1}{\mu })$ by formula 45, if $z$ is a fraction; for indeed, if $z$ is an integer, then $\mathrm{\Pi }z$ itself is known. From those values of the integrals, generally speaking, any $\mathrm{\Pi }(-\frac{m}{\mu })$ can be obtained by elimination, provided that $m<\mu .$ Indeed, it suffices to take half as many integrals if we also invoke formula 54. Thus, setting e.g. Template:C we will have Template:C Therefore, since $\mathrm{\Pi }\frac{1}{5}=\frac{1}{5}\mathrm{\Pi }(-\frac{4}{5}),$ we have Template:C Formulas 54, 55 still yield Template:C so that two integrals $D$ and $E,$ or $E$ and $F,$ suffice to compute all values of $\mathrm{\Pi }(-\frac{1}{5}),$ $\mathrm{\Pi }(-\frac{2}{5}),$ etc.

Setting $y=\nu x,$ and $\mu =1,$ the value of the integral $\int \frac{y^{2-1}(1-\frac{y}{\nu }{)}^{\nu }\mathrm{d}y}{{\nu }^2}$ from $y=0$ to $y=\nu ,$ or the value of the integral $\int y^{\lambda -1}(1-\frac{y}{\nu }{)}^{\nu }\mathrm{d}y$ between the same limits, is $=\frac{{\nu }^{\lambda }\mathrm{\Pi }\lambda .\mathrm{\Pi }\nu }{\lambda \mathrm{\Pi }(\lambda +\nu )}$$=\frac{\mathrm{\Pi }(\nu ,\lambda )}{\lambda }$ (form. 47), provided that $\nu$ is an integer. Now, the limit of $\mathrm{\Pi }(\nu ,\lambda )$ as $\nu$ increases to infinity will be $=\mathrm{\Pi }\lambda ,$ and the limit of $(1-\frac{y}{\nu }{)}^{\nu }$ will be $e^{-y},$ where $e$ denotes the base of the hyperbolic logarithm. Therefore, if $\lambda$ is positive, the value of the integral $\int y^{\lambda -1}{e}^{-y}\mathrm{d}y$ from $y=0$ to $y=\mathrm{\infty }$ will be $\frac{\mathrm{\Pi }\lambda }{\lambda }$ or $\mathrm{\Pi }(\lambda -1),$ or by writing $\lambda$ for $\lambda -1,$ the value of the integral $\int y^{\lambda }{e}^{-y}\mathrm{d}y$ from $y=0$ to $y=\mathrm{\infty }$ will be $\mathrm{\Pi }\lambda ,$ provided that $\lambda +1$ is a positive quantity.

More generally, by setting $y=z^{\alpha },$ $\alpha \lambda +\alpha -1=\beta ,$ the integral $\int y^{\lambda }{e}^{-y}\mathrm{d}y$ becomes $\int \alpha z^{\beta }{e}^{-z^{\alpha }}\mathrm{d}z,$ which, when taken between the limits $z=0$ and $z=\mathrm{\infty },$ is expressed by $\mathrm{\Pi }(\frac{\beta +1}{\alpha }-1),$ or in other words,

The value of the integral $\int z^{\beta }{e}^{-z^{\alpha }}\mathrm{d}z,$ from $z=0$ to $z=\mathrm{\infty }$ is $=\frac{\mathrm{\Pi }(\frac{\beta +1}{\alpha }-1)}{\alpha }=\frac{\mathrm{\Pi }\frac{\beta +1}{\alpha }}{\beta +1}$ provided that both $\alpha$ and $\beta +1$ are positive quantities (if either is negative, the value of the integral is $-\frac{\mathrm{\Pi }\frac{\beta +1}{\alpha }}{\beta +1}$). Thus for e.g. $\beta =0,$ $\alpha =2,$ the value of the integral $\int e^{-zz}\mathrm{d}z$ is found to be $=\mathrm{\Pi }\frac{1}{2}=\frac{1}{2}\sqrt{\pi }.$

The illustrious Template:Sc derived, for the sum of logarithms Template:C, the series Template:C where $𝔄=\frac{1}{6},$ $𝔅=\frac{1}{30},$ $ℭ=\frac{1}{42}$, etc. are the Template:Sc numbers. Thus, this series evaluates to $\log \mathrm{\Pi }z;$ although at first glance this conclusion may seem restricted to integer values, upon closer inspection it will be found that the method employed by Template:Sc (Instit. Calc. Diff. Cap. vi. 159) can be applied, at least for positive fractional values, with the same justification as for integers: he assumes only that the function of $z$ to be developed in a series, can be expressed in such a way that its decrement, if $z$ changes to $z-1,$ can be found using Taylor's theorem, and simultaneously that the same decrement is $=\log z.$ The former condition relies on the continuity of the function, and therefore does not apply to negative values of $z,$ to which it is not permitted to extend the series; however, the latter condition applies generally to the function $\log \mathrm{\Pi }z$ without restriction to integer values of $z.$ Therefore, we set

[58]	$\log \mathrm{\Pi }z=(z+\frac{1}{2})\log z-z+\frac{1}{2}\log 2\pi +\frac{𝔄}{1.2z}-\frac{𝔅}{3.4z^3}+\frac{ℭ}{5.6z^5}-\frac{𝔇}{7.8z^7}+\dots$

From this, since we have Template:C and by setting $n=2$ in formula 57, Template:C we get

[59]	$\log \mathrm{\Pi }(z-\frac{1}{2})=z\log z-z+\frac{1}{2}\log 2\pi -\frac{𝔄}{1.2.2z}+\frac{7𝔅}{3.4.8z^3}-\frac{31ℭ}{5.6.32z^5}+\frac{127𝔇}{7.8.128z^7}-\dots$

For large values of $z,$ these two series converge sufficiently rapidly from the beginning that an approximate and sufficiently exact sum can be conveniently obtained. However, it should be noted that for any given value of $z$, no matter how large, only limited precision can be achieved, since the Bernoulli numbers constitute a hypergeometric series, and therefore these series, if extended sufficiently far, will certainly turn from convergent to divergent. It cannot be denied that the theory of such divergent series is still fraught with difficulties, which we may discuss in more detail on another occasion.

From formula 38, it follows that Template:C Taking logarithms and expanding into infinite series yields

[60]

$\begin{array}{cccc}\log \mathrm{\Pi }(k,z+\omega )& =& & \log \mathrm{\Pi }(k,z)\\ & & & +\omega (\log k-\frac{1}{z+1}-\frac{1}{z+2}-\frac{1}{z+3}-\dots -\frac{1}{z+k})\\ & & & +\frac{1}{2}\omega \omega (\frac{1}{(z+1{)}^2}+\frac{1}{(z+2{)}^2}+\frac{1}{(z+3{)}^3}+\dots +\frac{1}{(z+k{)}^2})\\ & & & -\frac{1}{3}{\omega }^3(\frac{1}{(z+1{)}^3}+\frac{1}{(z+2{)}^3}+\frac{1}{(z+3{)}^3}+\dots +\frac{1}{(z+k{)}^3})\\ & & & +\dots \end{array}$

The series multiplied by $\omega$, which, if one prefers, can also be expressed as Template:C consists of a finite number of terms, but as $k$ tends to infinity, it converges to a certain limit, which presents to us a new species of transcendental functions, to be denoted from now on by $\mathrm{\Psi }z$.

Furthermore, if we denote the sums of the infinitely extended series Template:C by $P,$ $Q,$ $R$ etc. respectively (for which it seems less than necessary to introduce functional symbols), we will have

[61]	$\log \mathrm{\Pi }(z+\omega )=\log \mathrm{\Pi }z+\omega \mathrm{\Psi }z+\frac{1}{2}\omega \omega P-\frac{1}{3}{\omega }^{3}Q+\frac{1}{4}{\omega }^{4}R-\dots$

The function $\mathrm{\Psi }z$ will clearly be the first derivative of the function $\log \mathrm{\Pi }z,$ and therefore

[62]	$\frac{\mathrm{d}\mathrm{\Pi }z}{\mathrm{d}z}=\mathrm{\Pi }z.\mathrm{\Psi }z$

Similarly, we have $P=\frac{\mathrm{d}\mathrm{\Psi }z}{\mathrm{d}z},Q=-\frac{\mathrm{d}\mathrm{d}{\mathrm{\Psi }}_z}{2\mathrm{d}{z}^2},R=+\frac{{\mathrm{d}}^3\mathrm{\Psi }z}{2.3\mathrm{d}{z}^3},$ etc.

The function $\mathrm{\Psi }z$ is almost as remarkable as the function $\mathrm{\Pi }z;$ we will gather here some significant relations pertaining to it. Differentiating equation 44 yields

[63]	$\mathrm{\Psi }(z+1)=\mathrm{\Psi }z+\frac{1}{z+1}$

from which we have

[64]	$\mathrm{\Psi }(z+n)=\mathrm{\Psi }z+\frac{1}{z+1}+\frac{1}{z+2}+\frac{1}{z+3}+\dots +\frac{1}{z+n}$

This formula allows us to progress from smaller values of $z$ to larger ones, or to regress from larger values to smaller ones. For larger positive values of $z,$ the numerical values of the function are quite conveniently computed by the following formulas, obtained by differentiating equations 58, 59, to which the same considerations apply as in art. 29 regarding formulas 58 and 59.

[65]	$\mathrm{\Psi }z=\log z+\frac{1}{2z}-\frac{𝔄}{2zz}+\frac{𝔅}{4z^4}-\frac{ℭ}{6z^{\mathrm{6}}}+\dots$
[66]	$\mathrm{\Psi }(z-\frac{1}{2})=\log z+\frac{𝔄}{2.2zz}-\frac{7𝔅}{4.8z^4}+\frac{31ℭ}{6.32z^6}-\dots$

Thus for $z=10,$ we have calculated Template:C from which we regress to Template:C For positive integer values of $z,$ we generally have

[67]	Template:Nowrap

For negative integer values, however, it is clear that $\mathrm{\Psi }z$ becomes infinitely large.

Formula 55 provides us with $\log \mathrm{\Pi }(-z)+\log \mathrm{\Pi }(z-1)=\log \pi -\log \sin z\pi ,$ and differentiating this yields

[68]	$\mathrm{\Psi }(-z)-\mathrm{\Psi }(z-1)=\pi \cot z\pi$

Moreover, from the definition of the function $\mathrm{\Psi },$ we generally have

[69]	Template:Nowrap

This yields the well-known series Template:C

Similarly, by differentiation formula 57, we obtain

[70]	$\mathrm{\Psi }z+\mathrm{\Psi }(z-\frac{1}{n})+\mathrm{\Psi }(z-\frac{2}{n})+\dots +\mathrm{\Psi }(z-\frac{n-1}{n})=n\mathrm{\Psi }nz-n\log n$

and therefore, by setting $z=0$,

[71]	$\mathrm{\Psi }(-\frac{1}{n})+\mathrm{\Psi }(-\frac{2}{n})+\mathrm{\Psi }(-\frac{3}{n})+\dots +\mathrm{\Psi }(-\frac{n-1}{n})=(n-1)\mathrm{\Psi }0-n\log n$}}

For example, we have Template:C and thus Template:C

Just as in the previous article we reduced $\mathrm{\Psi }(-\frac{1}{2})$ to $\mathrm{\Psi }0$ and logarithms, so in generally can we reduce $\mathrm{\Psi }(-\frac{m}{n}),$ where $m,$ $n$ are integers with $m<n,$ to $\mathrm{\Psi }0$ and logarithms. Let us set $\frac{2\pi }{n}=\omega ,$ and let $\phi$ be any one of the angles $\omega ,$ $2\omega ,$ $3\omega \dots (n-1)\omega$; then $1=\cos n\phi =\cos 2n\phi =\cos 3n\phi$ etc., $\cos \phi =\cos (n+1)\phi =\cos (n+2)\phi$ etc., $\cos 2\phi =\cos (n+2)\phi$ etc., and also Template:Nowrap Thus, we have Template:C and by summation, Template:C But we generally have, for values of $x$ not greater than unity, Template:C which easily follows from the expansion of $\log (1-rx)+\log (1-\frac{x}{r}),$ where $r$ denotes the quantity $\cos \phi +\sqrt{-1}.\sin \phi .$ Hence, the preceding equation becomes

[72]	$\cos \phi .\mathrm{\Psi }\frac{1-n}{n}+\cos 2\phi .\mathrm{\Psi }\frac{2-n}{n}+\cos 3\phi .\mathrm{\Psi }\frac{3-n}{n}+\dots +\cos (n-1)\phi .\mathrm{\Psi }(-\frac{1}{n})$
	$=-\mathrm{\Psi }0+\frac{1}{2}n\log (2-2\cos \phi )$

Now, let us set $\phi =\omega ,$ $\phi =2\omega ,$ $\phi =3\omega$, etc. up to $\phi =(n-1)\omega ,$ and multiply these equations in their respective order by $\cos m\omega ,$ $\cos 2m\omega ,$ $\cos 3m\omega ,$ etc. up to $\cos (n-1)m\omega ,$ and add the sum of these products to equation 71: Template:C If we now consider that Template:C where $k$ denotes any one of the numbers $1,$ $2,$ $3\dots (n-1)$ except for $m$ and $n-m,$ for which the sum is $=\frac{1}{2}n,$ then it will is apparent, after adding these equations and dividing by $\frac{n}{2},$ that:

[73]	$\mathrm{\Psi }(-\frac{m}{n})+\mathrm{\Psi }(-\frac{n-m}{m})=$ $\begin{array}{cc}& 2\mathrm{\Psi }0-2\log n+\cos m\omega .\log (2-2\cos \omega )+\cos 2m\omega .\log (2-2\cos 2\omega )\\ & +\cos 3m\omega .\log (2-2\cos 3\omega )+\dots +\cos (n-1)m\omega .\log (2-2\cos (n-1)\omega )\end{array}$

The last term of this equation is clearly $=\cos m\omega .\log (2-2\cos \omega ),$ the penultimate one $=\cos 2m\omega .\log (2-2\cos 2\omega )$, etc., so that the terms are always equal in pairs, except when $n$ is even, in which case there is a singular term $\cos \frac{n}{2}.m\omega \log (2-2\cos \frac{n}{2}\omega ),$ which is either $=+2\log 2$ for even $m$ or $=-2\log 2$ for odd $m$. Combining the equation Template:C with equation 73, we obtain, for odd values of $n,$

[74]

$\begin{array}{cc}\mathrm{\Psi }(-\frac{m}{n})=& \mathrm{\Psi }0+\frac{1}{2}\pi \mathrm{cotang}\frac{m\pi }{n}-\log n+\cos \frac{2m\pi }{n}.\log (2-2\cos \frac{2\pi }{n})\\ & +\cos \frac{4m\pi }{n}.\log (2-2\cos \frac{4\pi }{n})+\cos \frac{6m\pi }{n}.\log (2-2\cos \frac{6\pi }{n})+\dots \\ & +\cos \frac{(n-1)m\pi }{n}.\log (2-2\cos \frac{(n-1)\pi }{n}),\end{array}$

provided that $m$ is a positive integer less than $n.$ For even values of $n,$ we have

[75]

$\begin{array}{cc}\mathrm{\Psi }(-\frac{m}{n})=& \mathrm{\Psi }0+\frac{1}{2}\pi \mathrm{cotang}\frac{m\pi }{n}-\log n+\cos \frac{2m\pi }{n}\log (2-2\cos \frac{2\pi }{n})\\ & +\cos \frac{4m\pi }{n}\log (2-2\cos \frac{4\pi }{n})+\dots +\cos \frac{(n-2)m\pi }{n}\log (2-2\cos \frac{(n-2)\pi }{n})\\ & \pm \log 2\end{array}$

where the upper sign holds for even $m$ and the lower sign holds for odd $m$. For example, we find: Template:C

In fact, upon combining these equations with equation 64, it is evident that $\mathrm{\Psi }z$ can generally be determined for "arbitrary rational values" of $z,$ whether positive or negative, in terms of $\mathrm{\Psi }0$ and logarithms, which is a most remarkable theorem.

According to art. 28, $\mathrm{\Pi }\lambda$ represents the value of the integral $\int y^{\lambda }{e}^{-y}\mathrm{d}y$ from $y=0$ to $y=\mathrm{\infty },$ provided that $\lambda +1$ is a positive quantity. Therefore, differentiating with respect to $\lambda ,$ we obtain Template:C or

[76]	Template:Nowrap

More generally, by setting $y=z^{\alpha }$ and $\alpha \lambda +\alpha -1=\beta ,$ the value of the integral $\int z^{\epsilon }{e}^{-z^a}\log z\mathrm{d}z,$ from $z=0$ to $z=\mathrm{\infty },$ becomes Template:C provided that both $\beta +1$ and $\alpha$ are positive quantities, or equal to the same quantity with the opposite sign, in the case where one of $\beta +1$ and $\alpha$ is negative.

Not only the product $\mathrm{\Pi }\lambda .\mathrm{\Psi }\lambda ,$ but also the function $\mathrm{\Psi }\lambda$ itself can be expressed through a definite integral. Letting $k$ denote a positive integer, it is clear that the value of the integral $\int \frac{x^{\lambda }-x^{\lambda +k}}{1-x}.\mathrm{d}x,$ from $x=0$ to $x=1$ is Template:C Moreover, since the value of the integral $\int (\frac{1}{1-x}-\frac{k{x}^{k-1}}{1-x^k})\mathrm{d}x$ is generally $=\text{Const.}+\log \frac{1-x^k}{1-x},$ it will be $=\log k$ between the limits $x=0$ and $x=1.$ Hence, it is clear that the value of the integral $S=\int (\frac{1-x^2+x^{2+k}}{1-x}-\frac{k{x}^{k-1}}{1-x^k})\mathrm{d}x$ between the same limits is Template:C which we denote by $\mathrm{\Omega }.$ Let us break down the integral $S$ into two parts Template:C After the substitution $x=y^k,$ the first part $\int \frac{1-x^{\lambda }}{1-x}.\mathrm{d}x$ becomes Template:C from which it is clear that its value from $x=0$ to $x=1$ is equal to the value of the integral Template:C between the same limits, since it is clear that the letter $y$ can be substituted by $x$ under this restriction. Hence the integral $S$, between the same limits, becomes Template:C Now, by setting $x^k=z,$ this integral becomes Template:C which therefore, between the limits $z=0$ and $z=1$, must be equal to $\mathrm{\Omega }.$ However, when $k$ increases to infinity, the limit of $\mathrm{\Omega }$ is $\mathrm{\Psi }\lambda ,$ the limit of $\frac{\lambda +1}{k}$ is $0,$ and the limit of $k(1-z{)}^{\frac{1}{k}}$ is $\log \frac{1}{z}$ or $-\log z.$ Therefore, we have

[77]	$\mathrm{\Psi }\lambda =\int (\frac{1}{\log \frac{1}{z}}-\frac{z^{\lambda }}{1-z})\mathrm{d}z=\int (-\frac{1}{\log z}-\frac{z^{\lambda }}{1-z})\mathrm{d}z$

from $z=0$ to $z=1.$

The definite integrals by which the functions $\mathrm{\Pi }\lambda ,$ $\mathrm{\Pi }\lambda .\mathrm{\Psi }\lambda ,$ have been expressed above must be restricted to values of $\lambda$ such that $\lambda +1$ becomes a positive quantity: this restriction arose naturally from the derivation itself, and indeed it is easily understood that for other values of $\lambda$ those integrals always become infinite, even if the functions $\mathrm{\Pi }\lambda ,$ $\mathrm{\Pi }\lambda .\mathrm{\Psi }\lambda$ might remain finite. The truth of formula 77 surely requires the same condition, that $\lambda +1$ be a positive quantity (otherwise, the integral would certainly become infinite, even if the function $\mathrm{\Psi }\lambda$ remains finite): but at first glance, the deduction of the formula seems to be general and not subject to any restriction. However, upon closer inspection, it is easy to see that this restriction is already inherent in the analysis itself by which the formula was derived. Namely, we tacitly assumed that the integral $\int \frac{1-x^{\lambda }}{1-x}\mathrm{d}x,$ which is equal to $\int \frac{k{x}^{k-1}-k{x}^{\lambda k+k-1}}{1-x^k}\mathrm{d}x,$ has a finite value, a condition that requires $\lambda +1$ to be a positive quantity. From our analysis, it indeed follows that these two integrals are always equal if the latter is extended from $x=0$ to $x=1-\omega ,$ and the former from $x=0$ to $x=(1-\omega {)}^k,$ however small the quantity $\omega$ may be, as long as it is not equal to zero. But notwithstanding this, in the case where $\lambda +1$ is not a positive quantity, the two integrals, extended from $x=0$ to the same limit $x=1-\omega ,$ do not converge to equality, but rather their difference grows infinitely as $\omega$ becomes infinitely small. This example shows how much circumspection is needed in dealing with infinite quantities, which in our analytic reasoning are only to be admitted insofar as they can be reduced to the theory of limits.

By setting $z=e^{-u},$ it is clear that formula 77 can also be expressed as

	Template:Nowrap
[78]	$\mathrm{\Psi }\lambda =\int (\frac{e^{-u}}{u}-\frac{e^{-2u}}{e^u-1})\mathrm{d}u,$ from $u=0$ to $u=\mathrm{\infty }$

(Thus by setting $e^{-y}=0,$ the value of $\mathrm{\Pi }\lambda$ in art. 28 becomes Template:C

Furthermore, it is clear from formula 77 that

[79]	$\mathrm{\Psi }\lambda -\mathrm{\Psi }\mu =\int \frac{z^{\mu }-z^{\lambda }}{1-z}\mathrm{d}z,$ from $z=0$ to $z=1$

where not only $\lambda +1$ but also $\mu +1$ must be a positive quantity.

In the same formula 77, if we set $z=u^{\alpha },$ where $\alpha$ denotes a positive quantity, we get Template:C and since, for a positive value of $\beta ,$ we likewise have Template:C it is clear that Template:C or equivalently, Template:C with all of these integrals being extended from $u=0$ to $u=1.$ However, by setting $\lambda =0,$ the latter integral can be evaluated "indefinitely"; namely, it is $=\log \frac{1-u^{\alpha }}{1-u^{\beta }},$ if it is to vanish for $u=0;$ therefore, since for $u=1$ we must set $\frac{1-u^{\alpha }}{1-u^{\beta }}=\frac{\alpha }{\beta },$ the integral becomes $\log \frac{\alpha }{\beta }=\int \frac{u^{\alpha -1}-u^{\beta -1}}{\log u}\mathrm{d}u,$ from $u=0$ to $u=1,$ a theorem which was formerly deduced by the illustrious Template:Sc using other methods.