Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2001
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2001

[Date Index] [Thread Index] [Author Index]

Search the Archive

RE: Transformation of Gamma Function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg31293] RE: [mg31266] Transformation of Gamma Function
  • From: "Higinio Ramos" <higra at usal.es>
  • Date: Sat, 27 Oct 2001 01:08:02 -0400 (EDT)
  • References: <200110260828.EAA05608@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

----- Original Message -----
From: Andre Kronimus <kronimus at whu.edu>
To: mathgroup at smc.vnet.net
Subject: [mg31293] [mg31266] Transformation of Gamma Function


> Hi,
>
> when evaluating the integral (Exp[-Exp[-k t]-t], {t,0,infinity},{k>0})
> in Mathematica, it returns a transformed version including Gamma
> functions, namely Gamma[1+1/k]-Gamma[1/k,1]/k. The two expressions are
> identical for some numerical values that I checked. However, I have been
> unable to figure out which transformation rules Mthematica used for
> transforming the original integral. Does anybody have an idea which
> rules have been used or how to force Mathematica to output every
> intermediate step of the transformation?
>
> Andre
>
Doing the transformation E^(k t)=1/z, the original integral becomes in
In[211]:=
Integrate[(1/k)Exp[-z]z^(1/k - 1) , {z, 0, 1}, Assumptions -> k > 0], whose
solution is:
Out[211]=
\!\(\(Gamma[1\/k] - Gamma[1\/k, 1]\)\/k\),
and using the identity: Gamma[a+1]=a*Gamma[a],
results in the solution that Mathematica returns.
Higinio



  • Prev by Date: Re: Reading 16-bit TIFF Images into Mathematica
  • Next by Date: RE: how to run Mathematica with Lap-top
  • Previous by thread: Transformation of Gamma Function
  • Next by thread: 2D to 3D graphics?