proof of Euler-Maclaurin summation formula

proof of Euler-Maclaurin summation formula

Let

a

and

b

be

integers

such that

a

<

b

, and let

f

:

[

a

,

b

]

→

ℝ

be

continuous

. We will prove by

induction

that
for all integers

k

≥

0

, if

f

is a

C

k

+

1

function

,

∑

a

<

n

≤

b

f

⁢

(

n

)

=

∫

a

b

f

⁢

(

t

)

⁢

d

t

+

∑

r

=

0

k

(

-

1

)

r

+

1

⁢

B

r

+

1

(

r

+

1

)

!

⁢

(

f

(

r

)

⁢

(

b

)

-

f

(

r

)

⁢

(

a

)

)

+

(

-

1

)

k

(

k

+

1

)

!

⁢

∫

a

b

B

k

+

1

⁢

(

t

)

⁢

f

(

k

+

1

)

⁢

(

t

)

⁢

d

t

(1)

where

B

r

is the

r

th

Bernoulli number

and

B

r

⁢

(

t

)

is the

r

th

Bernoulli periodic function

.

To prove the

formula

for

k

=

0

, we first rewrite

∫

n

-

1

n

f

⁢

(

t

)

⁢

d

t

, where

n

is an integer, using

integration by parts

:

∫

n

-

1

n

f

⁢

(

t

)

⁢

d

t

=

∫

n

-

1

n

d

d

⁢

t

⁢

(

t

-

n

+

1

2

)

⁢

f

⁢

(

t

)

⁢

d

t

=

(

t

-

n

+

1

2

)

⁢

f

⁢

(

t

)

|

n

-

1

n

-

∫

n

-

1

n

(

t

-

n

+

1

2

)

⁢

f

′

⁢

(

t

)

⁢

d

t

=

1

2

⁢

(

f

⁢

(

n

)

+

f

⁢

(

n

-

1

)

)

-

∫

n

-

1

n

(

t

-

n

+

1

2

)

⁢

f

′

⁢

(

t

)

⁢

d

t

.

Because

t

-

n

+

1

2

=

B

1

⁢

(

t

)

on the

interval

(

n

-

1

,

n

)

, this is
equal to

∫

n

-

1

n

f

⁢

(

t

)

⁢

d

t

=

1

2

⁢

(

f

⁢

(

n

)

+

f

⁢

(

n

-

1

)

)

-

∫

n

-

1

n

B

1

⁢

(

t

)

⁢

f

′

⁢

(

t

)

⁢

d

t

.

From this, we get

f

⁢

(

n

)

=

∫

n

-

1

n

f

⁢

(

t

)

⁢

d

t

+

1

2

⁢

(

f

⁢

(

n

)

-

f

⁢

(

n

-

1

)

)

+

∫

n

-

1

n

B

1

⁢

(

t

)

⁢

f

′

⁢

(

t

)

⁢

d

t

.

Now we take the sum of this

expression

for

n

=

a

+

1

,

a

+

2

,

…

,

b

, so
that the middle term on the right telescopes away for the most part:

∑

n

=

a

+

1

b

f

⁢

(

n

)

=

∫

a

b

f

⁢

(

t

)

⁢

d

t

+

1

2

⁢

(

f

⁢

(

b

)

-

f

⁢

(

a

)

)

+

∫

a

b

B

1

⁢

(

t

)

⁢

f

′

⁢

(

t

)

⁢

d

t

which is the Euler-Maclaurin formula for

k

=

0

, since

B

1

=

-

1

2

.

Suppose that

k

>

0

and the formula is correct for

k

-

1

, that is

∑

a

<

n

≤

b

f

⁢

(

n

)

=

∫

a

b

f

⁢

(

t

)

⁢

d

t

+

∑

r

=

0

k

-

1

(

-

1

)

r

+

1

⁢

B

r

+

1

(

r

+

1

)

!

⁢

(

f

(

r

)

⁢

(

b

)

-

f

(

r

)

⁢

(

a

)

)

+

(

-

1

)

k

-

1

k

!

⁢

∫

a

b

B

k

⁢

(

t

)

⁢

f

(

k

)

⁢

(

t

)

⁢

d

t

.

(2)

We rewrite the last

integral

using integration by parts and the facts
that

B

k

is continuous for

k

≥

2

and

B

k

+

1

′

⁢

(

t

)

=

(

k

+

1

)

⁢

B

k

⁢

(

t

)

for

k

≥

0

:

∫

a

b

B

k

⁢

(

t

)

⁢

f

(

k

)

⁢

(

t

)

⁢

d

t

=

∫

a

b

B

k

+

1

′

⁢

(

t

)

k

+

1

⁢

f

(

k

)

⁢

(

t

)

⁢

d

t

=

1

k

+

1

⁢

B

k

+

1

⁢

(

t

)

⁢

f

(

k

)

⁢

(

t

)

|

a

b

-

1

k

+

1

⁢

∫

a

b

B

k

+

1

⁢

(

t

)

⁢

f

(

k

+

1

)

⁢

(

t

)

⁢

d

t

.

Using the fact that

B

k

⁢

(

n

)

=

B

k

for every integer

n

if

k

≥

2

, we
see that the last term in Eq.

2

is equal to

(

-

1

)

k

+

1

⁢

B

k

+

1

(

k

+

1

)

!

⁢

(

f

(

k

)

⁢

(

b

)

-

f

(

k

)

⁢

(

a

)

)

+

(

-

1

)

k

(

k

+

1

)

!

⁢

∫

a

b

B

k

+

1

⁢

(

t

)

⁢

f

(

k

+

1

)

⁢

(

t

)

⁢

d

t

.

Substituting this and absorbing the left term into the

summation

yields Eq.

1

, as required.

Title

proof of Euler-Maclaurin summation formula

Canonical

name

ProofOfEulerMaclaurinSummationFormula

Date of creation

2013-03-22 13:28:41

Last modified on

2013-03-22 13:28:41

Owner

pbruin (1001)

Last modified by

pbruin (1001)

Numerical id

5

Author

pbruin (1001)

Entry type

Proof

Classification

msc 65B15

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