# Euler–Maclaurin

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##### 2: 25.2 Definition and Expansions
###### §25.2(iii) Representations by the Euler–Maclaurin Formula
25.2.8 $\zeta\left(s\right)=\sum_{k=1}^{N}\frac{1}{k^{s}}+\frac{N^{1-s}}{s-1}-s\int_{N% }^{\infty}\frac{x-\left\lfloor x\right\rfloor}{x^{s+1}}\mathrm{d}x,$ $\Re s>0$, $N=1,2,3,\dots$.
25.2.9 $\zeta\left(s\right)=\sum_{k=1}^{N}\frac{1}{k^{s}}+\frac{N^{1-s}}{s-1}-\frac{1}% {2}N^{-s}+\sum_{k=1}^{n}\genfrac{(}{)}{0.0pt}{}{s+2k-2}{2k-1}\frac{B_{2k}}{2k}% N^{1-s-2k}-\genfrac{(}{)}{0.0pt}{}{s+2n}{2n+1}\int_{N}^{\infty}\frac{% \widetilde{B}_{2n+1}\left(x\right)}{x^{s+2n+1}}\mathrm{d}x,$ $\Re s>-2n$; $n,N=1,2,3,\dots$.
##### 3: 2.10 Sums and Sequences
###### §2.10(i) Euler–Maclaurin Formula
This is the EulerMaclaurin formula. … In both expansions the remainder term is bounded in absolute value by the first neglected term in the sum, and has the same sign, provided that in the case of (2.10.7), truncation takes place at $s=2m-1$, where $m$ is any positive integer satisfying $m\geq\frac{1}{2}(\alpha+1)$. For extensions of the EulerMaclaurin formula to functions $f(x)$ with singularities at $x=a$ or $x=n$ (or both) see Sidi (2004, 2012b, 2012a). …
##### 4: Bibliography E
• D. Elliott (1998) The Euler-Maclaurin formula revisited. J. Austral. Math. Soc. Ser. B 40 (E), pp. E27–E76 (electronic).
• ##### 5: 25.11 Hurwitz Zeta Function
###### §25.11(iii) Representations by the Euler–Maclaurin Formula
25.11.5 $\zeta\left(s,a\right)=\sum_{n=0}^{N}\frac{1}{(n+a)^{s}}+\frac{(N+a)^{1-s}}{s-1% }-s\int_{N}^{\infty}\frac{x-\left\lfloor x\right\rfloor}{(x+a)^{s+1}}\mathrm{d% }x,$ $s\neq 1$, $\Re s>0$, $a>0$, $N=0,1,2,3,\dots$.
25.11.6 $\zeta\left(s,a\right)=\frac{1}{a^{s}}\left(\frac{1}{2}+\frac{a}{s-1}\right)-% \frac{s(s+1)}{2}\int_{0}^{\infty}\frac{\widetilde{B}_{2}\left(x\right)-B_{2}}{% (x+a)^{s+2}}\mathrm{d}x,$ $s\neq 1$, $\Re s>-1$, $a>0$.
25.11.7 $\zeta\left(s,a\right)=\frac{1}{a^{s}}+\frac{1}{(1+a)^{s}}\left(\frac{1}{2}+% \frac{1+a}{s-1}\right)+\sum_{k=1}^{n}\genfrac{(}{)}{0.0pt}{}{s+2k-2}{2k-1}% \frac{B_{2k}}{2k}\frac{1}{(1+a)^{s+2k-1}}-\genfrac{(}{)}{0.0pt}{}{s+2n}{2n+1}% \int_{1}^{\infty}\frac{\widetilde{B}_{2n+1}\left(x\right)}{(x+a)^{s+2n+1}}% \mathrm{d}x,$ $s\neq 1$, $a>0$, $n=1,2,3,\dots$, $\Re s>-2n$.
##### 6: Bibliography S
• A. Sidi (2004) Euler-Maclaurin expansions for integrals with endpoint singularities: A new perspective. Numer. Math. 98 (2), pp. 371–387.
• A. Sidi (2012a) Euler-Maclaurin expansions for integrals with arbitrary algebraic endpoint singularities. Math. Comp. 81 (280), pp. 2159–2173.
• A. Sidi (2012b) Euler-Maclaurin expansions for integrals with arbitrary algebraic-logarithmic endpoint singularities. Constr. Approx. 36 (3), pp. 331–352.
• ##### 7: Bibliography H
• M. Hauss (1997) An Euler-Maclaurin-type formula involving conjugate Bernoulli polynomials and an application to $\zeta(2m+1)$ . Commun. Appl. Anal. 1 (1), pp. 15–32.
• ##### 8: Bibliography B
• B. C. Berndt (1975a) Character analogues of the Poisson and Euler-MacLaurin summation formulas with applications. J. Number Theory 7 (4), pp. 413–445.
• ##### 9: 5.7 Series Expansions
For 20D numerical values of the coefficients of the Maclaurin series for $\Gamma\left(z+3\right)$ see Luke (1969b, p. 299). …
If $k$ in (3.5.4) is not arbitrarily large, and if odd-order derivatives of $f$ are known at the end points $a$ and $b$, then the composite trapezoidal rule can be improved by means of the EulerMaclaurin formula (§2.10(i)). …