Content-Length: 52930 | pFad | https://dlmf.nist.gov/./././././././././.././././../././../././../././../././././8.15#E2.info

DLMF: §8.15 Sums ‣ Incomplete Gamma Functions ‣ Chapter 8 Incomplete Gamma and Related Functions

About the Project

8 Incomplete Gamma and Related Functions Incomplete Gamma Functions 8.14 Integrals 8.16 Generalizations

§8.15 Sums

ⓘ

Keywords:: incomplete gamma functions, sums
Notes:: See Tricomi (1950b).
Permalink:: http://dlmf.nist.gov/8.15
Addition (effective with 1.1.3):: Equation (8.15.2) was added.
See also:: Annotations for Ch.8

8.15.1		$\gamma\left(a,\lambda x\right)=\lambda^{a}\sum_{k=0}^{\infty}\gamma\left(a+k,x% \right)\frac{(1-\lambda)^{k}}{k!}.$
ⓘ Symbols: $!$ : factorial (as in $n!$ ), $\gamma\left(\NVar{a},\NVar{z}\right)$ : incomplete gamma function, $x$ : real variable, $a$ : parameter and $k$ : nonnegative integer Permalink: http://dlmf.nist.gov/8.15.E1 Encodings: TeX, pMML, png See also: Annotations for §8.15 and Ch.8

8.15.2		$a\sum_{k=1}^{\infty}\left(\frac{{\mathrm{e}}^{2\pi\mathrm{i}k(z+h)}}{\left(2% \pi\mathrm{i}k\right)^{a+1}}\Gamma\left(a,2\pi\mathrm{i}kz\right)+\frac{{% \mathrm{e}}^{-2\pi\mathrm{i}k(z+h)}}{\left(-2\pi\mathrm{i}k\right)^{a+1}}% \Gamma\left(a,-2\pi\mathrm{i}kz\right)\right)=\zeta\left(-a,z+h\right)+\frac{z% ^{a+1}}{a+1}+\left(h-\tfrac{1}{2}\right)z^{a},$
$h\in[0,1]$ .
ⓘ Symbols: $\zeta\left(\NVar{s},\NVar{a}\right)$ : Hurwitz zeta function, $\pi$ : the ratio of the circumference of a circle to its diameter, $[\NVar{a},\NVar{b}]$ : closed interval, $\in$ : element of, $\mathrm{e}$ : base of natural logarithm, $\mathrm{i}$ : imaginary unit, $\Gamma\left(\NVar{a},\NVar{z}\right)$ : incomplete gamma function, $z$ : complex variable, $a$ : parameter and $k$ : nonnegative integer Proof sketch: Start with $a<-1$ and $z>0$ . For $\Gamma\left(a,\pm 2\pi\mathrm{i}kz\right)$ take integral representation (8.2.2) and use the substitution $t=\pm 2\pi\mathrm{i}kz(1+\tau)$ . The sum and the integral can be interchanged, and the sum can be evaluated via (4.6.1). Use integration by parts. This will result in $\left(h-\tfrac{1}{2}\right)z^{a}$ plus two integrals with infinitely many poles. The residue theorem (§1.10(iv)) will give us an infinite series which can be identified via (25.11.1). For other values of $a$ and $z$ use analytic continuation. Referenced by: §25.11(x), §25.11(x), §8.15, Erratum (V1.1.3) for Additions Permalink: http://dlmf.nist.gov/8.15.E2 Encodings: TeX, pMML, png Addition (effective with 1.1.3): This equation was added. See also: Annotations for §8.15 and Ch.8

For the Hurwitz zeta function $\zeta\left(s,a\right)$ see §25.11(i). For other infinite series whose terms include incomplete gamma functions, see Nemes (2017a), Reynolds and Stauffer (2021), and Prudnikov et al. (1986b, §5.2).

8.14 Integrals 8.16 Generalizations

© 2010–2024 NIST / Disclaimer / Feedback; Version 1.2.3; Release date 2024-12-15.

NIST

Site Privacy Accessibility Privacy Program Copyrights Vulnerability Disclosure No Fear Act Policy FOIA Environmental Policy Scientific Integrity Information Quality Standards Commerce.gov Science.gov USA.gov