proof of Euler-Maclaurin summation formula

Let $a$ and $b$ be integers such that , and let $f:[a,b]\to ℝ$ be continuous. We will prove by induction that for all integers $k\ge 0$, if $f$ is a $C^{k+1}$ function,

(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 ${\int }_{n-1}^{n}f(t)dt$, where $n$ is an integer, using integration by parts:

${\displaystyle {\int }_{n-1}^n}f(t)dt$ $=$ ${\displaystyle {\int }_{n-1}^n}{\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}}(t-n+{\displaystyle \frac{1}{2}})f(t)dt$

$=$ ${(t-n+{\displaystyle \frac{1}{2}})f(t)|}_{n-1}^n-{\displaystyle {\int }_{n-1}^n}(t-n+{\displaystyle \frac{1}{2}})f^{\prime }(t)dt$

$=$ ${\displaystyle \frac{1}{2}}(f(n)+f(n-1))-{\displaystyle {\int }_{n-1}^n}(t-n+{\displaystyle \frac{1}{2}})f^{\prime }(t)dt.$

Because $t-n+\frac{1}{2}=B_1(t)$ on the interval $(n-1,n)$, this is equal to

$${\int }_{n-1}^{n}f(t)dt=\frac{1}{2}(f(n)+f(n-1))-{\int }_{n-1}^n{B}_1(t)f^{\prime }(t)dt.$$

From this, we get

$$f(n)={\int }_{n-1}^{n}f(t)dt+\frac{1}{2}(f(n)-f(n-1))+{\int }_{n-1}^n{B}_1(t)f^{\prime }(t)dt.$$

Now we take the sum of this expression for $n=a+1,a+2,\mathrm{\dots },b$, so that the middle term on the right telescopes away for the most part:

$$\sum _{n=a+1}^{b}f(n)={\int }_a^{b}f(t)dt+\frac{1}{2}(f(b)-f(a))+{\int }_a^b{B}_1(t)f^{\prime }(t)dt$$

which is the Euler-Maclaurin formula for $k=0$, since $B_1=-\frac{1}{2}$.

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

(2)

We rewrite the last integral using integration by parts and the facts that $B_k$ is continuous for $k\ge 2$ and $B_{k+1}^{\prime }(t)=(k+1)B_k(t)$ for $k\ge 0$:

${\displaystyle {\int }_a^b}{B}_k(t)f^{(k)}(t)dt$ $=$ ${\displaystyle {\int }_a^b}{\displaystyle \frac{B_{k+1}^{\prime }(t)}{k+1}}{f}^{(k)}(t)dt$

$=$ ${{\displaystyle \frac{1}{k+1}}{B}_{k+1}(t)f^{(k)}(t)|}_a^b-{\displaystyle \frac{1}{k+1}}{\displaystyle {\int }_a^b}{B}_{k+1}(t)f^{(k+1)}(t)dt.$

Using the fact that $B_k(n)=B_k$ for every integer $n$ if $k\ge 2$, we see that the last term in Eq. 2 is equal to

$$\frac{{(-1)}^{k+1}{B}_{k+1}}{(k+1)!}(f^{(k)}(b)-f^{(k)}(a))+\frac{{(-1)}^k}{(k+1)!}{\int }_a^b{B}_{k+1}(t)f^{(k+1)}(t)dt.$$

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