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Revision History for A163931

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Showing entries 1-10 | older changes
Decimal expansion of the higher-order exponential integral E(x, m=2, n=1) at x=1.
(history; published version)
#38 by R. J. Mathar at Wed Jan 25 06:06:45 EST 2023
STATUS

editing

approved

#37 by R. J. Mathar at Wed Jan 25 06:06:35 EST 2023
COMMENTS

The series expansions of the higher-order exponential integrals are dominated by the constants alpha(k,n), see A163927, and gamma(k,n) = G(k,n), see A163930A090998.

CROSSREFS

Cf. A163927 (alpha(k,n)), A163930 A090998 (gamma(k,n) = G(k,n)), A163932.

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approved

editing

#36 by Susanna Cuyler at Fri Dec 06 21:43:01 EST 2019
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proposed

approved

#35 by Jon E. Schoenfield at Fri Dec 06 19:51:53 EST 2019
STATUS

editing

proposed

#34 by Jon E. Schoenfield at Fri Dec 06 19:51:49 EST 2019
COMMENTS

We define the higher-order exponential integrals by E(x,m,n) = x^(n-1)*int(E(t,m-1,n)/t^n, Integral_{t=x..infinity} E(t,m-1,n) /t^n for m >= 1 and n >= 1 with E(x,m=0,n) = exp(-x), see Meijer and Baken.

FORMULA

E(x=1,m=2,n=1) = gamma^2/2 + Pi^2/12 + Sum_{k>=1..infinity} ((-1)^k/(k^2*k!)).

E(x=0,n,m) = (1/(n-1))^m for n >= 2.

int(Integral_{t=0..x} E(t,m,n), t=0..x) = 1/n^m - E(x,n,n+1).

E(x,m,n+1) = (1/n)*(E(x,m-1,n+1) - x*E(x,m,n)).

E(x,m,n) = (-1)^m * ((-x)^(n-1)/(n-1)!) * Sum_{kz=0..floor(m/2)}(alpha (kz, n)*G(m-2*kz, n)) + (-1) ^m * ((-x)^(n-1)/(n-1)!) * Sum_{kz=0..floor(m/2)}(Sum_{i=1..m-2*kz}(alpha (kz, n) *G(m-2*kz-i, n)*lnlog(x)^i/i!)) + (-1)^m * Sum_{ kx=0..n-2}((-x)^kx/((kx-n+1)^m*kx!) + (-1)^m * Sum_{ky>=n..infinity}((-x)^ky /(( ky-n+1)^m*ky!)).

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approved

editing

#33 by N. J. A. Sloane at Fri Sep 28 11:08:09 EDT 2018
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proposed

approved

#32 by Michel Marcus at Fri Sep 28 10:40:13 EDT 2018
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editing

proposed

#31 by Michel Marcus at Fri Sep 28 10:39:46 EDT 2018
LINKS

M. S. Milgram, <a href="http://dx.doi.org/10.1090/S0025-5718-1985-0777276-4">The generalized integro-exponential function</a>, Math. of Computation, Vol. 44, pp. 443-458, 1985.

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proposed

editing

#30 by Johannes W. Meijer at Fri Sep 28 10:08:55 EDT 2018
STATUS

editing

proposed

#29 by Johannes W. Meijer at Fri Sep 28 10:08:16 EDT 2018
COMMENTS

We define the higher-order exponential integrals by E(x,m,n) = x^(n-1)*int(E(t,m-1,n)/t^n, t=x..infinity) for m => 0 =1 and n => =1 with E(x,m=0,n) = exp(-x), see Meijer and Baken.

FORMULA

E(x=1,m=2,n=1) = gamma^2/2 + Pi^2/12 + Sum_{k=1..infinity} ((-1)^k/(k^2*k!)).

E(x=0,n,m) = (1/(n-1))^m for n=>=2.

E(x,m,n) = (-1)^m * ((-x)^(n-1)/(n-1)!) *sum Sum_{kz=0..floor(m/2)}(alpha (kz, n)*(G(m-2*kz, n) ) + (-1) ^m * ((-x)^(n-1)/(n-1)!) * Sum_{kz=0..floor(m/2)} (Sum_{i=1..m-2*kz}(alpha (kz, n) *G(m-2*kz-i, n)*logln(x)^i/i!,i=)) + (-1..)^m-2 *kz))) + Sum_{ kx=0..(n-2)} ((-x)^kx/((kx-n+1)^m*kx!)) + (-1)^m * Sum_{ky=n..infinity}((-x)^ky /(( ky-n+1)^m*ky!))).

AUTHOR

Johannes W. Meijer & and _Nico Baken (n.h.g.baken(AT)tudelft.nl), _, Aug 13 2009, Aug 17 2009

STATUS

approved

editing

Discussion
Fri Sep 28
10:08
Johannes W. Meijer: Formula repaired, see style sheet, plus some minor edits.