610
B Review of Traditional Calculus
which we may rewrite as
h
2
[f (x + h) + f (x)] =
x+h
x
f (t) dt
+
∞
k=1
B 2k
(2k)!
h
2k
f
(2k−1) (x + h) − f
(2k−1) (x)
.
Upon setting x = a + ih, and summing over i from 0 to n − 1, we obtain
1
2 h
n−1
i=0
[f (a + (i + 1)h) + f (a + ih)] =
a+nh
a
f (t) dt
+
∞
k=1
B 2k
(2k)!
h
2k
f
(2k−1) (a + nh) − f
(2k−1) (a)
.
(B.1.21)
Now, as the left-hand side of this expression can be reduced to h
n
i=0
f (a + ih)
−
1
2 h[f (a) + f (a + nh), Eq. (B.1.21) becomes
h
n
i=0
f (a + ih) =
a+nh
a
f (t) dt +
1
2 h[f (a) + f (a + nh)]
+
∞
k=1
B 2k
(2k)!
h
2k
f
(2k−1) (a + nh) − f
(2k−1) (a)
.
(B.1.22)
Substitution of h = 1 into Eq. (B.1.22) gives the summation of a function of integers
in the form
n
i=0
f (a + i) =
a+n
a
f (t) dt +
1
2 [f (a) + f (a + n)]
+
∞
k=1
B 2k
(2k)!
f
(2k−1) (a + n) − f
(2k−1) (a)
.
Upon setting a to 0 in the above result, we obtain an expression for the sum over
integers of a function that is defined on the real line in terms of the corresponding
definite integral over that function plus correction terms that depend upon the
derivatives of the function: explicitly, we find that 5
5 Note that the Bernoulli numbers B 2k , k ≥ 2 alternate in sign: this sign alternation is often taken
into account in versions of the Euler–Maclaurin summation formula by introducing a factor (−1) k