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21: 1.13 Differential Equations

… ►The substitution $\xi =1/z$ in (1.13.1) gives … ►The substitution … ►In (1.13.1) substitute …

22: 18.23 Hahn Class: Generating Functions

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23: 20.11 Generalizations and Analogs

… ►With the substitutions $a=q{\mathrm{e}}^{2\mathrm{i}z}$, $b=q{\mathrm{e}}^{-2\mathrm{i}z}$, with $q={\mathrm{e}}^{\mathrm{i}\pi \tau }$, we have …

24: 28.33 Physical Applications

… ►Substituting $z=\omega t$, $a=b/{\omega }^2$, and $2q=f/{\omega }^2$, we obtain Mathieu’s standard form (28.2.1). …

25: 14.5 Special Values

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26: 10.41 Asymptotic Expansions for Large Order

… ►To establish (10.41.12) we substitute into (10.34.3), with $m=0$ and $z$ replaced by $\nu z$, by means of (10.41.13) observing that when $|z|$ is large the effect of replacing $z$ by $z{\mathrm{e}}^{\pm \pi \mathrm{i}}$ is to replace $\eta$, ${(1+z^2)}^{\frac{1}{4}}$, and $p$ by $-\eta$, $\pm \mathrm{i}{(1+z^2)}^{\frac{1}{4}}$, and $-p$, respectively. … ►Lastly, we substitute into (10.4.4), again with $z$ replaced by $\nu z$. …

27: 18.9 Recurrence Relations and Derivatives

… ►Further $n$-th derivative formulas relating two different Jacobi polynomials can be obtained from §15.5(i) by substitution of (18.5.7). … ►Further $n$-th derivative formulas relating two different Laguerre polynomials can be obtained from §13.3(ii) by substitution of (13.6.19). …

28: 18.17 Integrals

… ►Formula (18.17.9), after substitution of (18.5.7), is a special case of (15.6.8). Formulas (18.17.9), (18.17.10) and (18.17.11) are fractional generalizations of $n$-th derivative formulas which are, after substitution of (18.5.7), special cases of (15.5.4), (15.5.5) and (15.5.3), respectively. … ►Formulas (18.17.14) and (18.17.15) are fractional generalizations of $n$-th derivative formulas which are, after substitution of (13.6.19), special cases of (13.3.18) and (13.3.20), respectively. … ►

18.17.21_1 $$\frac{1}{\sqrt{2\pi c}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{1}{2}{x}^2/c}{H}_n\left(x\right){\mathrm{e}}^{\mathrm{i}xy}dx={\left(\mathrm{i}\sqrt{2c-1}\right)}^n{\mathrm{e}}^{-\frac{1}{2}c{y}^2}{H}_n\left(\frac{cy}{\sqrt{2c-1}}\right),$$ $\mathrm{\Re }\left(c\right)>0$, $c\ne \frac{1}{2}$,

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Symbols:

$H_n\left(x\right)$: Hermite polynomial, $\pi$: the ratio of the circumference of a circle to its diameter, $dx$: differential of $x$ , $\mathrm{e}$: base of natural logarithm, $\mathrm{i}$: imaginary unit, $\int$: integral, $\mathrm{\Re }$: real part, $y$: real variable, $n$: nonnegative integer and $x$: real variable

Sources:

Erdélyi et al. (1954a, §1.10(8), §2.10(10)); Erdélyi et al. (1953b, §10.13(30))

Proof sketch:

In the integrand of the left-hand side substitute (18.10.8) with data for $H_n\left(x\right)$ given by Table 18.10.1. Interchange integrals. Then the inner integral can be evaluated. The resulting integral can again be evaluated by (18.10.8) for Hermite.

Referenced by:

§18.17(v), §18.17(v), Erratum (V1.2.0) Section 18.17

Permalink:

http://dlmf.nist.gov/18.17.E21_1

Encodings:

pMML, png, TeX

See also:

Annotations for §18.17(v), §18.17(v), §18.17 and Ch.18

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18.17.34_5 $${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-xz}{L}_m^{(\alpha )}\left(x\right)L_n^{(\alpha )}\left(x\right){\mathrm{e}}^{-x}{x}^{\alpha }dx=\frac{\mathrm{\Gamma }\left(\alpha +m+1\right)\mathrm{\Gamma }\left(\alpha +n+1\right)}{\mathrm{\Gamma }\left(\alpha +1\right)m!n!}\frac{z^{m+n}}{{(z+1)}^{\alpha +m+n+1}}{}_2{}^{}F_1^{}(\genfrac{}{}{0pt}{}{-m,-n}{\alpha +1};z^{-2}),$$ $\mathrm{\Re }z>-1$.

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Symbols:

$\mathrm{\Gamma }\left(z\right)$: gamma function, ${}_2{}^{}F_1^{}(a,b;c;z)$: $=F(a,b;c;z)$ notation for Gauss’ hypergeometric function, $L_n^{(\alpha )}\left(x\right)$: Laguerre (or generalized Laguerre) polynomial, $dx$: differential of $x$ , $\mathrm{e}$: base of natural logarithm, $!$: factorial (as in $n!$), $\int$: integral, $\mathrm{\Re }$: real part, $z$: complex variable, $m$: nonnegative integer, $n$: nonnegative integer and $x$: real variable

Source:

Erdélyi et al. (1954a, 4.11(35))

Proof sketch:

First assume $z>-1$. Make change of integration variable $x\to x/(z+1)$. Twice substitute (18.18.12) with $\lambda ={(z+1)}^{-1}$. Then use the orthogonality relation for the Laguerre polynomials, see Table 18.3.1. Finally perform analytic continuation with respect to $z$.

Referenced by:

Erratum (V1.2.0) Section 18.17

Permalink:

http://dlmf.nist.gov/18.17.E34_5

Encodings:

pMML, png, TeX

See also:

Annotations for §18.17(vi), §18.17(vi), §18.17 and Ch.18

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29: 4.13 Lambert $W$-Function

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4.13.9_1 $$W_0\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}{d}_n{\left(\mathrm{e}z+1\right)}^{n/2},$$ , ,

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Symbols:

$W\left(z\right)$: Lambert $W$-function, $\pi$: the ratio of the circumference of a circle to its diameter, $\frac{df}{dx}$: derivative of $f$ with respect to $x$ , $\mathrm{e}$: base of natural logarithm, $\mathrm{ph}$: phase, $n$: integer and $z$: complex variable

Proof sketch:

Substitute (4.13.9_1) into $z(1+W)\frac{dW}{dz}=W$.

Referenced by:

(4.13.9_1), §4.13, §4.13, Erratum (V1.1.7) for Expansion

Permalink:

http://dlmf.nist.gov/4.13.E9_1

Encodings:

pMML, png, TeX

Addition (effective with 1.1.7):

This equation was added.

See also:

Annotations for §4.13 and Ch.4

… ►For integrals of $W\left(z\right)$ use the substitution $w=W\left(z\right)$, $z=w{\mathrm{e}}^w$ and $dz=(w+1){\mathrm{e}}^{w}dw$. …

30: 10.18 Modulus and Phase Functions

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