proof of Euler-Maclaurin summation formula

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

\\sum_{a<n\\leq b}f(n)=\\int_{a}^{b}f(t)\\mathrm{d}t+\\sum_{r=0}^{k}\\frac{(-1)^{r+1%}B_{r+1}}{(r+1)!}(f^{(r)}(b)-f^{(r)}(a))+\\frac{(-1)^{k}}{(k+1)!}\\int_{a}^{b}B_%{k+1}(t)f^{(k+1)}(t)\\mathrm{d}t

(1)

where $B_{r}$ is the $r$ th Bernoulli number and $B_{r}(t)$ is the $r$ thBernoulli periodic function.

To prove the formula for $k=0$ , we first rewrite $\\int_{n-1}^{n}f(t)\\mathrm{d}t$ , where $n$ is an integer, usingintegration by parts:

$\\displaystyle\\int_{n-1}^{n}f(t)\\mathrm{d}t$	$\\displaystyle=$	$\\displaystyle\\int_{n-1}^{n}\\frac{\\mathrm{d}}{\\mathrm{d}t}(t-n+\\frac{1}{2})f(t)%\\mathrm{d}t$
	$\\displaystyle=$	$\\displaystyle(t-n+\\frac{1}{2})f(t)\\big{\|}_{n-1}^{n}-\\int_{n-1}^{n}(t-n+\\frac{1%}{2})f^{\\prime}(t)\\mathrm{d}t$
	$\\displaystyle=$	$\\displaystyle\\frac{1}{2}(f(n)+f(n-1))-\\int_{n-1}^{n}(t-n+\\frac{1}{2})f^{\\prime%}(t)\\mathrm{d}t.$

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

\\int_{n-1}^{n}f(t)\\mathrm{d}t=\\frac{1}{2}(f(n)+f(n-1))-\\int_{n-1}^{n}B_{1}(t)f%^{\\prime}(t)\\mathrm{d}t.

From this, we get

f(n)=\\int_{n-1}^{n}f(t)\\mathrm{d}t+\\frac{1}{2}(f(n)-f(n-1))+\\int_{n-1}^{n}B_{1%}(t)f^{\\prime}(t)\\mathrm{d}t.

Now we take the sum of this expression for $n=a+1,a+2,\\ldots,b$ , sothat the middle term on the right telescopes away for the most part:

\\sum_{n=a+1}^{b}f(n)=\\int_{a}^{b}f(t)\\mathrm{d}t+\\frac{1}{2}(f(b)-f(a))+\\int_{%a}^{b}B_{1}(t)f^{\\prime}(t)\\mathrm{d}t

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

\\sum_{a<n\\leq b}f(n)=\\int_{a}^{b}f(t)\\mathrm{d}t+\\sum_{r=0}^{k-1}\\frac{(-1)^{r%+1}B_{r+1}}{(r+1)!}(f^{(r)}(b)-f^{(r)}(a))+\\frac{(-1)^{k-1}}{k!}\\int_{a}^{b}B_%{k}(t)f^{(k)}(t)\\mathrm{d}t.

(2)

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

	$\\displaystyle\\int_{a}^{b}B_{k}(t)f^{(k)}(t)\\mathrm{d}t$	$\\displaystyle=$	$\\displaystyle\\int_{a}^{b}\\frac{B_{k+1}^{\\prime}(t)}{k+1}f^{(k)}(t)\\mathrm{d}t$
		$\\displaystyle=$	$\\displaystyle\\frac{1}{k+1}B_{k+1}(t)f^{(k)}(t)\\big{\|}_{a}^{b}-\\frac{1}{k+1}%\\int_{a}^{b}B_{k+1}(t)f^{(k+1)}(t)\\mathrm{d}t.$

Using the fact that $B_{k}(n)=B_{k}$ for every integer $n$ if $k\\geq 2$ , wesee 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)\\mathrm{d}t.

Substituting this and absorbing the left term into the summationyields Eq. 1, as required.