162
Infinite Series
So B0 = 1 and B1 =
1/2. The remaining odd Bernoulli numbers vanish
B2n+1 = 0
for n > 0
(4.105)
and the remaining even ones are given by Euler’s zeta function (4.92) formula
B2n =
( 1)n 1 2(2n)!
⇣(2n)
(2⇡)2n
for n > 0.
(4.106)
The Bernoulli numbers occur in the power series for many transcendental
functions, for instance
1
1 X 22k B2k 2k
coth x = +
x
x
(2k)!
1
for x2 < ⇡ 2 .
(4.107)
k=1
Bernoulli’s polynomials Bn (y) are defined by the series
1
X
xexy
xn
=
Bn (y)
x
e
1
n!
(4.108)
n=0
xexy /(ex
for the generating function
1).
Some authors (Whittaker and Watson, 1927, p. 125–127) define Bernoulli’s
numbers instead by
Z 1 2n 1
2(2n)!
t
dt
Bn =
⇣(2n) = 4n
(4.109)
2⇡t
(2⇡)2n
e
1
0
a result due to Carda.
4.12 Asymptotic Series
A series
sn (x) =
n
X
ak
k=0
xk
(4.110)
is an asymptotic expansion for a real function f (x) if the remainder Rn
Rn (x) = f (x)
sn (x)
(4.111)
satisfies the condition
lim xn Rn (x) = 0
x!1
(4.112)
for fixed n. In this case, one writes
f (x) ⇡
1
X
ak
k=0
xk
(4.113)
4.12 Asymptotic Series
163
where the wavy equal sign indicates equality in the sense of (4.112). Some
authors add the condition:
lim xn Rn (x) = 1
(4.114)
n!1
for fixed x.
Example 4.14 (The Asymptotic Series for E1 )
totic expansion for the function
Z 1
dy
E1 (x) =
e y
y
x
Let’s develop an asymp-
(4.115)
which is related to the exponential-integral function
Z x
dy
Ei(x) =
ey
y
1
(4.116)
by the tricky formula E1 (x) =
Ei( x). Since
✓ y◆
e y
d
e
e y
=
y
dy
y
y2
(4.117)
we may integrate by parts, getting
E1 (x) =
Z
x
e
x
1
y
e
x
dy
.
y2
(4.118)
Integrating by parts again, we find
E1 (x) =
x
e
x
e x
+2
x2
Z
1
Rn (x) = ( 1)n n!
Z
dy
.
y3
y
x
Eventually, we develop the series
✓
0!
1!
2!
E1 (x) = e x
+
x
x2 x3
with remainder
e
3!
4!
+
x4 x5
1
e
y
x
(4.119)
...
dy
.
y n+1
◆
(4.120)
(4.121)
Setting y = u + x, we have
n! e x
Rn (x) = ( 1)
xn+1
n
Z
1
0
e
u
du
1+
u n+1
x
(4.122)