DOC.
14
EINSTEIN
AND
BESSO MANUSCRIPT
379
[p. 10]
(Einstein)
[45][Pp.
1011]
are on one
side
of
a
sheet double the
size
of
the other
sheets of
the
manuscript.
In
the
top
right
corner a
page
number
is
added,
in
Besso's hand, which
presumably
refers
to
both
pages.
The sheet
is
folded
in
such
a way
that
a
fourpage set
is
formed.
[P. 11]
is
the
first
page,
[p.
10]
the
last
one.
The inside
pages
contain
only one
formula related
to the
calculations
at
this
point
in
the
manuscript
and
some totally
unrelated
thermodynamic
formulas
and
diagrams,
all
in
Einstein's hand. These inside
pages
are
not
reproduced
here.
The
one
related formula reads:
"+ ^
+

B2c4(l

+ )A"
(the
factors
c04
and
c4
in
the second
term
are
canceled
against one
another).
[46]On
[pp.
1012]
some
contour integrations
are
performed to
obtain
a
general
formula for
integrals
of
the form
of the
integral
in
[eq.
68]
on [p.
9]
for
the
angle
between
perihelion
and
aphelion.
[47]The
calculation carried
out in
[eqs.
6973] proceeds
as
follows. The
starting point is
[eq.
70], the
lefthand
side
of which
is just
the lefthand side of
[eq. 69].
The
integral
on
the
lefthand
side of
[eq.
69]
has
essentially
the
same
form
as
the
one
in
[eq.
68]; only
the
1/rterm
in
the
numerator
and
the constant in
front of
the
whole
integral are
omitted
(these
features
are
taken
into
account
on [p.
11]).
The
polynomial
under
the
square root
sign
in
[eq. 68]
is
written
in its
factorized form. The
roots
are
called
r1, r2,
r',
and
r".
The
integral is
to
be taken
between
r1
and
r2,
the values of
r
at perihelion
and
aphelion, respectively.
The
other
two
roots,
r'
and
r",
are
very
small
compared to
r1
and
r2.
Hence, it is
convenient
to expand
the
integrand in
r'/r
and
r"/r,
neglecting
all terms
of second order in these
quantities and smaller.
The result of this
expansion
is
given
in
[eq.
71]
(r

r2
should
be
r2

r).
The
integral is
now
broken
up
into two
parts:
one
part
with 1/r
y/(r

r1)(r2

r)
in the
integrand
and another
with
1/r2
y/(r

r1)(r2

r).
These
two
parts are
dealt with
separately (for
a
discussion
see
notes
48 and
49).
The
results,
[eq. 72]
and
[eq. 73]
(both containing
an
erroneous
factor
2), are
inserted
into
[eq.
71], giving
the
righthand
side of
[eq.
69].
At
first,
the
factor
1/2(r'
+ r") in front
of
the
second
part
of the
integral
is
omitted.
On
the
next
line,
this
is
corrected.
Finally,
[eq.
77]
is used for
r'
+
r"
(see
note
51).
The
mistake with
the
factor
2
is
noted
at
the
bottom
of
[p.
11].
[48]A
contour integration is performed to
evaluate the
first
part
of
the integral in
[eq. 71] (see
note 47).
The
factor
(r

r2)
on
the lefthand
side
of
[eq. 72]
should
be
(r2

r)
as
in
[eq. 70].
The
result
is
given as
2n/Jr1r2
(see
[eq. 72]).
This
should
be
t/r1r2.
This
is the
mistake referred
to
in
note
47. The
integration is
done
as
follows.
Call
the integral
that
is
to be
evaluated I. A
way
to
find I,
or
rather 2I, is
to integrate along
a
contour
in the
complex
plane
going
from
r1
to
r2
just
above the real
axis
and back from
r2
to
r1
just
underneath
(see
the
figure next
to
[eq.
72];
note
that
the
values the
function in
the
integrand
takes
on
just
above and
just
underneath the
real
axis have
opposite signs,
since
one
is
dealing
with different sheets of
the function).
The result
will be
the
same
for
any
other
contour
that
can
be
obtained from the
contour
in
the
figure
through
continuous
deformation,
i.e., through
a
deformation that does
not require passing through
any
of the
poles
of the
integrand (the points 0,
r1,
and
r2 on
the real axis).
The
following
contour
can
be
obtained
in this
way.
Start
out
from
a
point,
say
A, at
minus
infinity just
underneath the
real
axis,
go
to
zero staying just
underneath the
real axis,
circle around
the
pole
at
r
=
0,
and
go
back,
staying just
above the real
axis,
to
a
point,
say
B, at
minus
infinity
just
above
the
real
axis.
Complete
the
contour
with
an
almost closed circle of infinite radius
going
from
B to
A.
The
only
contribution
to
the
integral along
this contour
comes
from the
small
circle around the
pole
at
r
=
0
(see the
little circle
on
the left
in
the
figure
that
is
called the "0circle"
["0Kreis"]
in
[eq.
73]). Hence,
on
account
of
Cauchy's
theorem,
2I
is
equal
to
2ni
times the residue
in
r
=
0,
which
is
1/i
/r1r2.
In
[eq.
72],
2n/Jr1r2
is
mistakenly
set
equal
to
I rather than
to
2I.
[49]Another
contour integration, very
similar
to
the
one
discussed
in
note 48, is performed
to
evaluate
the
second
part
of the
integral
in
[eq. 71] (but
for the
factor
1/2(r'
+ r")). The result