proof of Euler-Maclaurin summation formula

Let a and b be integers such that a<b, and let f:[a,b] be continuousMathworldPlanetmathPlanetmath. We will prove by inductionMathworldPlanetmath that for all integers k0, if f is a Ck+1 function,

a<nbf(n)=abf(t)dt+r=0k(-1)r+1Br+1(r+1)!(f(r)(b)-f(r)(a))+(-1)k(k+1)!abBk+1(t)f(k+1)(t)dt (1)

where Br is the rth Bernoulli numberMathworldPlanetmathPlanetmath and Br(t) is the rth Bernoulli periodic function.

To prove the formulaMathworldPlanetmathPlanetmath for k=0, we first rewrite n-1nf(t)dt, where n is an integer, using integration by parts:

n-1nf(t)dt = n-1nddt(t-n+12)f(t)dt
= (t-n+12)f(t)|n-1n-n-1n(t-n+12)f(t)dt
= 12(f(n)+f(n-1))-n-1n(t-n+12)f(t)dt.

Because t-n+12=B1(t) on the interval (n-1,n), this is equal to


From this, we get


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:


which is the Euler-Maclaurin formula for k=0, since B1=-12.

Suppose that k>0 and the formula is correct for k-1, that is

a<nbf(n)=abf(t)dt+r=0k-1(-1)r+1Br+1(r+1)!(f(r)(b)-f(r)(a))+(-1)k-1k!abBk(t)f(k)(t)dt. (2)

We rewrite the last integral using integration by parts and the facts that Bk is continuous for k2 and Bk+1(t)=(k+1)Bk(t) for k0:

abBk(t)f(k)(t)dt = abBk+1(t)k+1f(k)(t)dt
= 1k+1Bk+1(t)f(k)(t)|ab-1k+1abBk+1(t)f(k+1)(t)dt.

Using the fact that Bk(n)=Bk for every integer n if k2, we see that the last term in Eq. 2 is equal to


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