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Euler-Maclaurin Integration Formula

To start,

$\displaystyle \int^1_0 f(x)dx = \int^1_0 f(x)B_0(x)dx$ (2.10)

It is also obvious that $ B_1'(x)=B_0(x)=1$ so that,

$\displaystyle \int^1_0 f(x)dx= f(1)B_1(1)-f(0)B_1(0) - \int^1_0 f'(x)B_1(x)dx$ (2.11)

Carrying on and seeing that $ B_{2n}(1)=B_{2n}(0)=B_2n$ and $ B_{2n+1}(1,0)=0$, we would find,

$\displaystyle \int^1_0f(x)dx = \frac{1}{2}[f(1)+f(0)] - \sum^{q}_{p=1} \frac{1}...
...}[f^{(2p-1)}(1)-f^{(2p-1)}(0)] + \frac{1}{(2q)!}\int^1_0 f^{(2q)}(x)B_{2q}(x)dx$ (2.12)

The transform $ x\rightarrow x+1$, $ x+1 \rightarrow x+2$ ,etcetera, yields,
$\displaystyle \int^n_0f(x)dx$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left( f(0) + f(n)\right) + \sum_{k=1}^{n-1}f(k)$ (2.13)
    $\displaystyle -\sum^{q}_{p=1} \frac{1}{(2p)!}B_{2p}[f^{(2p-1)}(1)-f^{(2p-1)}(0)] + \frac{1}{(2q)!}\int^1_0 B_{2q}(x)\sum^{n-1}_{\chi=0}f^{(2q)}(x+\chi)dx$  

root 2006-09-15