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% David R. Wilkins
% School of Mathematics, Trinity College, Dublin 2, Ireland
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% Trinity College, 2000.
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\centerline{\Largebf ON AN EXPRESSION FOR THE NUMBERS OF}
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\centerline{\Largebf BERNOULLI, BY MEANS OF A DEFINITE}
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\centerline{\Largebf INTEGRAL, AND ON SOME CONNECTED}
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\centerline{\Largebf PROCESSES OF SUMMATION AND INTEGRATION}
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\centerline{\Largebf By}
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\centerline{\Largebf William Rowan Hamilton}
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\centerline{\largerm (Philosophical Magazine, 23 (1843), pp.\ 360--367.)}
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\centerline{\largerm Edited by David R. Wilkins}
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\centerline{\largerm 2000}
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\noindent
{\largeit On an Expression for the Numbers of\/}
{\largerm Bernoulli,} {\largeit by means of a Definite Integral;
and on some connected Processes of Summation and Integration.
By\/} {\largerm Sir}
{\largesc William Rowan Hamilton}, {\largeit LL.D., P.R.I.A.,
Member of several Scientific Societies at Home and Abroad,
Andrews' Professor of Astronomy in the University of Dublin, and
Royal Astronomer of Ireland.}
\bigbreak
\vskip 12pt
\centerline{[{\it The London, Edinburgh and Dublin Philosophical
Magazine and Journal of Science,}}
\centerline{3rd series, vol.~xxiii (1843), pp.\ 360--367.]}
\bigskip
The following analysis, extracted from a paper which has been in
part communicated to the Royal Irish Academy, but has not yet
been printed, may interest some readers of the Philosophical
Magazine.
\bigbreak
1.
Let us consider the function of two real variables, $m$ and $n$,
represented by the definite integral
$$y_{m,n}
= \int_0^\infty dx \, \left( {\sin x \over x} \right)^m
\cos nx;
\eqno (1.)$$
in which we shall suppose that $m$ is greater than zero; and
which gives evidently the general relation
$$y_{m,-n} = y_{m,n}.$$
By changing $m$ to $m + 1$; integrating first the factor
$x^{-m-1} \, dx$, and observing that
$$x^{-m} \sin x^{m+1} \cos nx$$
vanishes both when $x = 0$, and when $x = \infty$; and then
putting the differential coefficient
$\displaystyle {d \over dx} (\sin x^{m+1} \cos nx)$
under the form
$${\textstyle {1 \over 2}} \sin x^m
\{
(m + 1 + n) \cos (nx + x)
+ (m + 1 - n) \cos (nx - x)
\};$$
we are conducted to the following equation, in finite and partial
differences,
$$2 m y_{m+1, n}
= (m + 1 + n) y_{m, n + 1} + (m + 1 - n) y_{m, m - 1};
\eqno (2.)$$
and if we suppose that the difference between the two variables
on which $y$ depends is an even integer number, this equation
takes the form
$$m y_{m+1, m+1-2k}
= (m + 1 - k) y_{m, m+2-2k} + k y_{m, m - 2k}.
\eqno (3.)$$
The same equation in differences (2.) shows easily that
$$\eqalign{
& y_{m+1, n} = 0,
\hbox{ when } n = \hbox{ or } > m + 1,\cr
& \hbox{if } y_{m, n-1} = 0,
\hbox{ when } n - 1 > m;\cr}$$
but, by a well-known theorem, which in the present notation becomes
$$y_{1,0} = {\pi \over 2},
\eqno (4.)$$
it is easy to prove, not only that
$$y_{1,1} = y_{1,-1} = {\pi \over 4},
\eqno (5.)$$
but also that
$$y_{1,n} = 0, \hbox{ if } n^2 > 1;
\eqno (6.)$$
we have therefore, generally, for all whole values of $m > 1$,
and for all real values of $n$,
$$y_{m,n} = 0, \hbox{ unless } n^2 < m^2.
\eqno (7.)$$
\bigbreak
2.
If then we make
$${\rm T}_m = \Sigma y_{m, m - 2k} (-t)^k,
\eqno (8.)$$
the sign~$\Sigma$ indicating a summation which may be extended
from as large a negative to as large a positive whole value of
$k$ as we think fit, but which extends at least from $k = 0$ to
$k = m$, $m$ being here a positive whole number; this sum will
in general (namely when $m > 1$) include only $m - 1$ terms
different from $0$, namely those which correspond to
$k = 1, 2,\ldots \, m - 1$;
but in the particular case $m = 1$, the sum will have two such
terms, instead of none, namely those answering to $k = 0$ and
$k = 1$, so that we shall have
$${\rm T}_1
= y_{1,1} - y_{1,-1} t
= {\pi \over 4} (1 - t).
\eqno (9.)$$
Multiplying the first member of the equation in differences~(3.)
by $(-t)^k$, and summing with respect to $k$, we obtain
$m {\rm T}_{m+1}$, $m$ being here any whole number $> 0$.
Multiplying and summing in like manner the second member of the
same equation~(3.), the term $m y_{m, m + 2 - 2k}$ of that member
gives $-m t {\rm T}_m$, because we may change $k$ to $k + 1$
before effecting the indefinite summation; $k y_{m, m - 2k}$
gives
$\displaystyle t {d \over dt} {\rm T}_m$;
and $(1 - k) y_{m, m + 2 - 2k}$ gives
$\displaystyle t^2 {d \over dt} {\rm T}_m$; but
$$-mt {\rm T}_m + (t + t^2) {d \over dt} {\rm T}_m
= (1 + t)^{m+1} {td \over dt} (1 + t)^{-m} {\rm T}_m;$$
therefore
$$m (1 + t)^{-m-1} {\rm T}_{m+1}
= {d \over d \log t} (1 + t)^{-m} {\rm T}_m.
\eqno (10.)$$
This equation in mixed differences gives, by (9.),
$${\rm T}_m
= {\pi \over 4}
{(1 + t)^m \over 1 \mathbin{.} 2 \mathbin{.} 3
\, \ldots \, (m - 1)}
\left( {d \over d \log t} \right)^{m-1}
{1 - t \over 1 + t};
\eqno (11.)$$
the factorial denominator being considered as $= 1$, when $m =
1$, as well as when $m = 2$. If $m > 1$, we may change
$\displaystyle {1 - t \over 1 + t}$ to
$\displaystyle {2 \over 1 + t}$,
from which it only differs by a constant; and then by changing
also $t$ to $e^h$, and multiplying by
$\displaystyle {2 \over \pi}$, we obtain the formula:
$$\eqalignno{
{(e^h + 1)^m
\over 1 \mathbin{.} 2 \mathbin{.} 3 \, \ldots \, (m - 1)}
\left( {d \over dh} \right)^{m - 1} (e^h + 1)^{-1}
\hskip -12em \cr
&= {2 \over \pi} \Sigma_{(k) \,}{}_1^{m-1} \int_0^\infty dx \,
\left( {\sin x \over x} \right)^m (-e^h)^k
\cos (mx - 2kx);
&(12.)\cr}$$
which conducts to many interesting consequences. A few of them
shall be here mentioned.
\bigbreak
3.
The summation indicated in the second member of this formula can
easily be effected in general; but we shall here consider only
the two cases in which $m$ is an odd or an even whole number
greater than unity, while $h$ becomes $= 0$ after the $m - 1$
differentiations of $(e^h + 1)^{-1}$, which are directed in the
first member.
When $m$ is odd (and greater than one), each power, such as
$(- e^h)^k$ in the second member, is accompanied by another,
namely $(- e^h)^{m-k}$, which is multiplied by the cosine of the
same multiple of $x$; and these two powers destroy each other,
when added, if $h = 0$: we arrive therefore in this manner at the
known result, that
$$\left( {d \over dh} \right)^{2p} (e^h + 1)^{-1} = 0,
\quad\hbox{when}\quad h = 0, \quad\hbox{if}\quad p > 0.
\eqno (13.)$$
On the contrary, when $m$ is even, and $h = 0$, the powers
$(- e^h)^k$ and $(- e^h)^{m - k}$ are equal, and their sum is
double of either; and because
$$(-1)^p \{ 1 - 2 \cos 2x + 2 \cos 4x - \cdots
+ (-1)^{p-1} 2 \cos (2px - 2x) \}
= - {\cos (2px - x) \over \cos x},$$
by making $m = 2p$ we arrive at this other result, which perhaps
is new, that (if $p > 0$ and $h = 0$)
$$\left( {d \over dh} \right)^{2p - 1} (e^h + 1)^{-1}
= {-1 \mathbin{.} 2 \mathbin{.} 3 \, \ldots \, (2p - 1)
\over 2^{2p - 1} \pi}
\int_0^\infty dx \left( {\sin x \over x} \right)^{2p}
{\cos (2px - x) \over \cos x}.
\eqno (14.)$$
Developing therefore $(e^h + 1)^{-1}$ according to ascending
powers of $h$; subtracting the development from ${1 \over 2}$,
multiplying by $h$, and changing $h$ to $2h$; we obtain
$$h {e^h - e^{-h} \over e^h + e^{-h}}
= {2 \over \pi} \int_0^\infty {dx \over \cos x}
\Sigma_{(p) \,}{}_1^\infty
\left( {h \sin x \over x} \right)^{2p} \cos (2px - x);
\eqno (15.)$$
that is, effecting the summation, and dividing by $h^2$,
$${1 \over h} {e^h - e^{-h} \over e^h + e^{-h}}
= {2 \over \pi}
\int_0^\infty
{dx \, x^{-2} \sin x^2 (1 - h^2 x^{-2} \sin x^2)
\over 1 - 2 h^2 x^{-2} \sin x^2 \cos 2x
+ h^4 x^{-4} \sin x^4};
\eqno (16.)$$
or, integrating both members with respect to $h$,
$$\int_0^h {dh \over h} {e^h - e^{-h} \over e^h + e^{-h}}
= {1 \over \pi} \int_0^\infty {dx \over x} \tan x
\log \sqrt{\vphantom{\biggl\{}}
{1 + h x^{-1} \sin 2x + h^2 x^{-2} \sin x^2
\over 1 - h x^{-1} \sin 2x + h^2 x^{-2} \sin x^2}.
\eqno (17.)$$
It might seem, at first sight, from this equation, that the
integral in the first member ought to vanish, when taken from
$h = 0$ to $h = \infty$; because, if we set about to develope the
second member, according to the descending powers of $h$, we see
that the coefficient of $h^0$ vanishes; but when we find that, on
the same plan, the coefficient of $h^{-1}$ is infinite, being
$\displaystyle = {2 \over \pi} \int_0^\infty dx$,
we perceive that this mode of development is here inappropriate:
and in fact, it is clear that the first member of the formula
(17.) increases continually with $h$, while $h$ increases from
$0$.
\bigbreak
4.
Again, since
$${-h \over e^h + 1}
= \psi(2h) - \psi(h),
\quad\hbox{if}\quad
\psi(h) = {h \over e^h - 1},
\eqno (18.)$$
we shall have (for $p > 0$) the expression
$${\rm A}_{2p}
= {2^{1 - 2p} \pi^{-1} \over 2^{2p} - 1}
\int_0^\infty dx \left( {\sin x \over x} \right)^{2p}
{\cos (2px - x) \over \cos x},
\eqno (19.)$$
if, according to a known form of development, which the foregoing
reasonings would suffice to justify, we write
$${h \over e^h - 1} + {h \over 2}
= 1 + {\rm A}_2 h^2 + {\rm A}_4 h^4 + {\rm A}_6 h^6
+ \hbox{\&c.}
\eqno (20.)$$
If $p$ be a large number, the rapid and repeated changes of sign
of the numerator of the fraction
$\displaystyle {\cos (2px - x) \over \cos x}$
produce nearly a mutual destruction of the successive elements of
the integral (19.), except in the neighbourhood of those values
of $x$ which cause the denominator of the same fraction to
vanish; namely those values which are odd positive multiples of
${\pi \over 2}$ (the integral itself being not extended so as to
include any negative values of $x$). Making therefore
$$x = (2i - 1) {\pi \over 2} + \omega,
\eqno (21.)$$
in which $i$ is a whole number $> 0$, and $\omega$ is positive or
negative, but nearly equal to $0$; we shall have
$$\cos (2px - x)
= (-1)^{p+i-1} \sin (2p \omega - \omega),$$
exactly, and $\cos x = (-1)^i \omega$, nearly; changing also
$\displaystyle \left( {\sin x \over x} \right)^{2p}$
to the value which it has when $\omega = 0$, namely
$\displaystyle \left( {2 \over \pi} \right)^{2p} (2i - 1)^{-2p}$;
and observing that
$$\int_{-\omega}^\omega d \omega \,
{\sin ( 2p \omega - \omega ) \over \omega}
= \pi,
\quad\hbox{nearly,}
\eqno (22.)$$
even though the extreme values of $\omega$ may be small, if $p$
be very large; we find that the part of ${\rm A}_{2p}$,
corresponding to any one value of the number~$i$, is, at least
nearly, represented by the expression
$${(-1)^{p-1} 2 (2i - 1)^{-2p} \over (2^{2p} - 1) \pi^{2p}};
\eqno (23.)$$
which is now to be summed, with reference to $i$, from $i = 1$ to
$i = \infty$. But this summation gives rigorously the relation
$$\Sigma_{(i) \,}{}_1^\infty (2i - 1)^{-2p}
= (1 - 2^{-2p}) \Sigma_{(i) \,}{}_1^\infty i^{-2p};
\eqno (24.)$$
we are conducted, therefore, to the expression
$${\rm A}_{2p}
= (-1)^{p-1} 2 (2\pi)^{-2p}
\Sigma_{(i) \,}{}_1^\infty i^{-2p},
\eqno (25.)$$
as {\it at least\/} approximately true, when the number~$p$ is
large. But in fact the expression (25.) is {\it rigorous\/} for
all whole values of $p$ greater than $0$; as we shall see by
deducing from it an analogous expression for a Bernoullian
number, and comparing this with known results.
\bigbreak
5.
The development
$${1 \over e^h - 1} + {1 \over 2}
= h^{-1} + {\rm B}_1 {h \over 1 \mathbin{.} 2}
- {\rm B}_3 {h^3
\over 1 \mathbin{.} 2 \mathbin{.} 3 \mathbin{.} 4}
+ \hbox{\&c.},
\eqno (26.)$$
being compared with that marked (20.), gives, for the $p$th
Bernoullian number, the known expression
$${\rm B}_{2p-1}
= (-1)^{p-1} 1 \mathbin{.} 2 \mathbin{.} 3 \mathbin{.} 4
\, \ldots \, 2p {\rm A}_{2p};
\eqno (27.)$$
and therefore, rigorously, by the equation (19.) of the present
paper,
$${\rm B}_{2p-1}
= {(-1)^{p-1} 1 \mathbin{.} 2 \mathbin{.} \, \ldots \, 2p
\over 2^{2p-1} (2^{2p} - 1) \pi}
\int_0^\infty dx \left( {\sin x \over x} \right)^{2p}
{\cos (2px - x) \over \cos x};
\eqno (28.)$$
a formula which is believed to be new. Treating the definite
integral which it involves by the process just now used, we
necessarily obtain the same result as if we combine at once the
equations (25.) and (27.). We find, therefore, in this manner,
that the equation
$${2^{2p-1} \pi^{2p} {\rm B}_{2p-1}
\over 1 \mathbin{.} 2 \mathbin{.} 3 \mathbin{.} 4
\, \ldots \, 2p}
= \Sigma_{(i) \,}{}_1^\infty i^{-2p},
\eqno (29.)$$
(in which, by the notation here employed,
$$\Sigma_{(i) \,}{}_1^\infty i^{-2p}
= 1^{-2p} + 2^{-2p} + 3^{-2p} + \hbox{\&c.})$$
is {\it at least nearly true}, when $p$ is a large number, but
Euler has shown, in his {\it Institutiones Calculi
Differentialis\/} (vol.~i. cap.~v. p.~339. ed.~1787), that this
equation (29.) is {\it rigorous}, each member being the
coefficient of $u^{2p}$ in the development of
$\displaystyle {1 \over 2} (1 - \pi i \cot \pi u)$.
[See also Professor De Morgan's Treatise on the Diff.\ and
Int.\ Calc., `Library of Useful Knowledge,' part~xix. p.~581.]
The analysis of the present paper is therefore not only verified
generally, but also the modifications which were made in the form
of that definite integral which entered into our rigorous
expressions (19.) and (28.) for ${\rm A}_{2p}$ and
${\rm B}_{2p-1}$, by the process of the last article, (on the
ground that the parts omitted or introduced thereby must at least
nearly destroy each other, through what may be called the
``principle of fluctuation,'') are now seen to have produced no
ultimate error at all, their mutual compensation being perfect; a
result which may tend to give increased confidence in applying a
similar process of approximation, or transformation, to the
treatment of other similar integrals; although the logic of this
process may deserve to be more closely scrutinized. Some
assistance towards such a scrutiny may be derived from the essay
on ``Fluctuating Functions,'' which has been published by the
present writer in the second part of the nineteenth volume of the
Transactions of the Royal Irish Academy.
\bigbreak
6.
It may be worth while to notice, in conclusion, that the property
marked (7.) of the definite integral (1.), enables us to change
$\displaystyle {\cos (2px - x) \over \cos x}$
to $\sin 2px \, \tan x$, in the equations (14.), (15.), (19.),
(28.); so that the $p$th Bernoullian number may rigorously be
expressed as follows:---
$${\rm B}_{2p-1}
= {(-1)^{p-1} \mathbin{.} 1 \mathbin{.} 2 \, \ldots \, 2p
\over 2^{2p-1} (2^{2p} - 1) \pi}
\int_0^\infty dx \,
\left( {\sin x \over x} \right)^{2p}
\sin 2px \, \tan x;
\eqno (30.)$$
under which form the preceding deduction of its transformation
(29.) admits of being slightly simplified. The same modification
of the foregoing expressions conducts easily to the equation
$$\log {e^h + e^{-h} \over 2}
= {1 \over \pi} \int_0^\infty dx \, \tan x
\tan^{-1}
{h^2 \sin x^2 \sin 2x
\over x^2 - h^2 \sin x^2 \cos 2x};
\eqno (31.)$$
in which $\tan^{-1}$ is a characteristic equivalent to arc tang.,
and which may be made an expression for $\log \sec x$, by merely
changing the sign of $h^2$ in the last denominator; and from this
equation~(31.) it would be easy to return to an expression for
the coefficients in the development of
$\displaystyle {e^h - e^{-h} \over e^h + e^{-h}}$,
or in that of $\tan h$, and therefore to the numbers of
Bernoulli. Those numbers might thus be deduced from the
following very simple equation:
$$\pi \log \sec h = \int_0^\infty dx \, y \tan x;
\eqno (32.)$$
in which $y$ is connected with $x$ and $h$ by the relation
$${\sin y \over \sin (2x - y)}
= \left( {h \sin x \over x} \right)^2.
\eqno (33.)$$
\bigbreak
Observatory of Trinity College, Dublin,\par
\qquad October~6, 1843.
\bye