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21: 15.6 Integral Representations

… ►However, this reverses the direction of the integration contour, and in consequence (15.6.5) would need to be multiplied by $-1$. …

22: Errata

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This subsection has been significantly updated. In particular, the following formulae have been corrected. Equation (19.25.35) has been replaced by

19.25.35 $$z+2\omega =\pm R_F(\mathrm{\wp }\left(z\right)-e_1,\mathrm{\wp }\left(z\right)-e_2,\mathrm{\wp }\left(z\right)-e_3),$$

in which the left-hand side $z$ has been replaced by $z+2\omega$ for some $2\omega \in 𝕃$, and the right-hand side has been multiplied by $\pm 1$. Equation (19.25.37) has been replaced by

19.25.37 $$\zeta \left(z+2\omega \right)+(z+2\omega )\mathrm{\wp }\left(z\right)=\pm 2R_G(\mathrm{\wp }\left(z\right)-e_1,\mathrm{\wp }\left(z\right)-e_2,\mathrm{\wp }\left(z\right)-e_3),$$

in which the left-hand side $\zeta \left(z\right)+z\mathrm{\wp }\left(z\right)$ has been replaced by $\zeta \left(z+2\omega \right)+(z+2\omega )\mathrm{\wp }\left(z\right)$ and the right-hand side has been multiplied by $\pm 1$. Equation (19.25.39) has been replaced by

19.25.39 $$\zeta \left({\omega }_j\right)+{\omega }_j{e}_j=2R_G(0,e_j-e_k,e_j-e_{\mathrm{\ell }}),$$

in which the left-hand side ${\eta }_j$ was replaced by $\zeta \left({\omega }_j\right)$, for some $2{\omega }_j\in 𝕃$ and $\mathrm{\wp }\left({\omega }_j\right)=e_j$. Equation (19.25.40) has been replaced by

19.25.40 $$z+2\omega =\pm \sigma \left(z\right)R_F({\sigma }_1^2(z),{\sigma }_2^2(z),{\sigma }_3^2(z)),$$

in which the left-hand side $z$ has been replaced by $z+2\omega$, and the right-hand side was multiplied by $\pm 1$. For more details see §19.25(vi).

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Bounds have been sharpened. The second paragraph now reads, “The $n$th error term is bounded in magnitude by the first neglected term multiplied by $\chi (n+\sigma )+1$ where $\sigma =\frac{1}{6}$ for (9.7.7) and $\sigma =0$ for (9.7.8), provided that $n\ge 0$ in the first case and $n\ge 1$ in the second case.” Previously it read, “In (9.7.7) and (9.7.8) the $n$th error term is bounded in magnitude by the first neglected term multiplied by $2\chi (n)\exp \left(\sigma \pi /(72\zeta )\right)$ where $\sigma =5$ for (9.7.7) and $\sigma =7$ for (9.7.8), provided that $n\ge 1$ in both cases.” In Equation (9.7.16)

9.7.16

$$\mathrm{Bi}\left(x\right)\le \frac{{\mathrm{e}}^{\xi }}{\sqrt{\pi }{x}^{1/4}}\left(1+\left(\chi (\frac{7}{6})+1\right)\frac{5}{72\xi }\right),$$

$${\mathrm{Bi}}^{\prime }\left(x\right)\le \frac{x^{1/4}{\mathrm{e}}^{\xi }}{\sqrt{\pi }}\left(1+\left(\frac{\pi }{2}+1\right)\frac{7}{72\xi }\right),$$

the bounds on the right-hand sides have been sharpened. The factors $\left(\chi (\frac{7}{6})+1\right)\frac{5}{72\xi }$, $\left(\frac{\pi }{2}+1\right)\frac{7}{72\xi }$, were originally given by $\frac{5\pi }{72\xi }\exp \left(\frac{5\pi }{72\xi }\right)$, $\frac{7\pi }{72\xi }\exp \left(\frac{7\pi }{72\xi }\right)$, respectively.

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Bounds have been sharpened. The first paragraph now reads, “The $n$th error term in (9.7.5) and (9.7.6) is bounded in magnitude by the first neglected term multiplied by

9.7.17

provided that $n\ge 0$, $\sigma =\frac{1}{6}$ for (9.7.5) and $n\ge 1$, $\sigma =0$ for (9.7.6).” Previously it read, “When $n\ge 1$ the $n$th error term in (9.7.5) and (9.7.6) is bounded in magnitude by the first neglected term multiplied by

9.7.17

Here $\sigma =5$ for (9.7.5) and $\sigma =7$ for (9.7.6).”

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14.19.2 $$P_{\nu -\frac{1}{2}}^{\mu }\left(\cosh \xi \right)=\frac{\mathrm{\Gamma }\left(\frac{1}{2}-\mu \right)}{{\pi }^{1/2}{\left(1-{\mathrm{e}}^{-2\xi }\right)}^{\mu }{\mathrm{e}}^{(\nu +(1/2))\xi }}𝐅(\frac{1}{2}-\mu ,\frac{1}{2}+\nu -\mu ;1-2\mu ;1-{\mathrm{e}}^{-2\xi }),$$ $\mu \ne \frac{1}{2},\frac{3}{2},\frac{5}{2},\mathrm{\dots }$

Originally the argument to $𝐅$ in this equation was incorrect (${\mathrm{e}}^{-2\xi }$, rather than $1-{\mathrm{e}}^{-2\xi }$), and the condition on $\mu$ was too weak ($\mu \ne \frac{1}{2}$, rather than $\mu \ne \frac{1}{2},\frac{3}{2},\frac{5}{2},\mathrm{\dots }$). Also, the factor multiplying $𝐅$ was rewritten to clarify the poles; originally it was $\frac{\mathrm{\Gamma }\left(1-2\mu \right)2^{2\mu }}{\mathrm{\Gamma }\left(1-\mu \right){\left(1-{\mathrm{e}}^{-2\xi }\right)}^{\mu }{\mathrm{e}}^{(\nu +(1/2))\xi }}$.

Reported 2010-11-02 by Alvaro Valenzuela.

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23: 18.30 Associated OP’s

… ►For other cases there may also be, in addition to a possible integral as in (18.30.10), a finite sum of discrete weights on the negative real $x$-axis each multiplied by the polynomial product evaluated at the corresponding values of $x$, as in (18.2.3). …

24: 19.29 Reduction of General Elliptic Integrals

… ►If both square roots in (19.29.22) are 0, then the indeterminacy in the two preceding equations can be removed by using (19.27.8) to evaluate the integral as $R_G(a_1{b}_2,a_2{b}_1,0)$ multiplied either by $-2/(b_1{b}_2)$ or by $-2/(a_1{a}_2)$ in the cases of (19.29.20) or (19.29.21), respectively. …

25: 25.11 Hurwitz Zeta Function

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25.11.24 $$\sum _{r=1}^{k-1}{\zeta }^{\prime }(s,\frac{r}{k})=(k^s-1){\zeta }^{\prime }\left(s\right)+k^s\zeta \left(s\right)\ln k,$$ $s\ne 1$, $k=1,2,3,\mathrm{\dots }$.

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Symbols:

$\zeta (s,a)$: Hurwitz zeta function, $\zeta \left(s\right)$: Riemann zeta function, $\ln z$: principal branch of logarithm function, $a$: real or complex parameter, $s$: complex variable and $k$: integer

Keywords:

derivative

Proof sketch:

Derivable from (25.11.15) with $a=1/k$, multiplying by $k^s$, differentiating with respect to $s$, and using (25.11.2).

Referenced by:

(25.11.21), §25.11(vi)

Permalink:

http://dlmf.nist.gov/25.11.E24

Encodings:

pMML, png, TeX

See also:

Annotations for §25.11(vi), §25.11(vi), §25.11 and Ch.25

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26: Bibliography M

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Two-point quasi-fractional approximations to the Airy function $\mathrm{Ai}(x)$ . J. Comput. Phys. 99 (2), pp. 337–340.

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Notes:

There is an error in Eq. (5): the second term in parentheses needs to be multiplied by $x$.

External Links:

ISSN 0021-9991, Document, MathReview Entry, ZentralBlatt

Referenced by:

1st item.

Encodings:

BibTeX

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