[Maxima] About computing integral integrate(exp(-x^k),x,0,1)

Richard Hennessy rich.hennessy at verizon.net
Sat Jan 7 15:16:02 CST 2012


“The situation is confusing as to what form Maxima really uses.”  6.5.3 makes sense to me.  As for my email I think I misunderstood what you did.  Sorry for your confusion, your answer is correct.

Rich


From: James Nesta 
Sent: Friday, January 06, 2012 11:02 PM
To: Richard Hennessy 
Cc: maxima at math.utexas.edu 
Subject: Re: [Maxima] About computing integral integrate(exp(-x^k),x,0,1)

Hi everybody, 

Thanks for checking my math on this Rich. In looking back at what I wrote I noticed that I had omitted a factor of (1/k) in the transform of the differential. The transform should look like dx = (1/k)*z^(1/k - 1)*dz. I'm sorry if this might have caused confusion; I should monitor my typing more carefully!

The version of the incomplete gamma function that I used is from Abramowitz & Stegun (6.5.3); this is the version that I was familiar with from other books. Using this definition I still get the answer that I originally got. In the Maxima reference on this function the integral form of this function follows A&S (6.5.3) but it is identified as A&S (6.5.2), an alternative version. The situation is confusing as to what form Maxima really uses. I think that your solution Rich corresponds to the integral form of incomplete gamma from zero to x, vs my form from x to infinity. In doing a numerical test on Maxima's gamma_incomplete(5,0) I get 24 = 4!, which seems to correspond to the integral form that I used. So it seems to me that the formula reference in Maxima's manual has a typo that is highly misleading. I again apologize to the readers for any confusion that might have arisen from any of my statements and extend my thanks to those, including you, who attempted to clarify the problem.

Jim

On Wed, Jan 4, 2012 at 9:45 PM, Richard Hennessy <rich.hennessy at verizon.net> wrote:

  I meant the integral becomes (1/k)*[gamma(1/k)-incomplete_gamma(1/k, z)]. Typo.

  Rich

  -----Original Message----- From: Richard Hennessy
  Sent: Thursday, January 05, 2012 12:43 AM
  To: James Nesta ; maxima at math.utexas.edu 

  Subject: Re: [Maxima] About computing integral integrate(exp(-x^k),x,0,1)

  "integral becomes(1/k)*[gamma(1/k)-incomplete_gamma(1/k, 1)]"

  I think you made a mistake, it should be

  the integral becomes (1/k)*[gamma(1/k)-incomplete_gamma(1/k, k)]

  Rich

  -----Original Message----- From: James Nesta
  Sent: Wednesday, January 04, 2012 10:48 PM
  To: maxima at math.utexas.edu
  Subject: Re: [Maxima] About computing integral integrate(exp(-x^k),x,0,1)

  In fooling with this integral I was able to get what I think is a manageable
  solution. As follows:

  Transform the variable x to z by the transformation x = z^(1/k), dx =
  z^(1/k-1)dz,

  the integral becomes(1/k)*[gamma(1/k)-incomplete_gamma(1/k, 1)]. I think
  that Maxima should be able to handle each of these functions. I hope that
  this helps.

  Jim
  _______________________________________________
  Maxima mailing list
  Maxima at math.utexas.edu
  http://www.math.utexas.edu/mailman/listinfo/maxima 

-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://www.math.utexas.edu/pipermail/maxima/attachments/20120107/1a2f13f8/attachment-0001.html>


More information about the Maxima mailing list