# proof of Euler-Maclaurin summation formula

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

 $\sum_{a (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 $\int_{n-1}^{n}f(t)\mathrm{d}t$, where $n$ is an integer, using integration 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 is equal 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 $n=a+1,a+2,\ldots,b$, so that the middle term on the right telescopes away for the most part:

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

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 (2)

We rewrite the last integral using integration by parts and the facts that $B_{k}$ 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 $k\geq 2$, we see that the last term in Eq. 2 is equal to

 $\frac{(-1)^{k+1}B_{k+1}}{(k+1)!}(f^{(k)}(b)-f^{(k)}(a))+\frac{(-1)^{k}}{(k+1)!% }\int_{a}^{b}B_{k+1}(t)f^{(k+1)}(t)\mathrm{d}t.$

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

Title proof of Euler-Maclaurin summation formula ProofOfEulerMaclaurinSummationFormula 2013-03-22 13:28:41 2013-03-22 13:28:41 pbruin (1001) pbruin (1001) 5 pbruin (1001) Proof msc 65B15