# proof of Euler-Maclaurin summation formula

Let $a$ and $b$ be integers such that $$, and let
$f:[a,b]\to \mathbb{R}$ 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:

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

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

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

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 $k\ge 0$:

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

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

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 |