Hello, the gamma distribution must be equivalent to an exponential for alpha=1 so the pdf for x=0 must be equal to 1/beta not 0.

To the definition of the GAMMADIST function, GAMMADIST(0;alpha,beta,0) is always zero.

Please let core developers close bug. This is not invalid as being 0 will break several assumptions.

references in support of defining 0^0=1 rather than 'undefined' (note that no references suggest to _define_ 0^0=0, every time lim 0^x x->0 is used is to argue for undefined, not to argue for 0^0=0 http://www.mathforum.org/dr.math/faq/faq.0.to.0.power.html http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/ and of course, in these days and age google is the ultimate reference :-) http://www.google.com/search?q=0^0&btnG=Search bc agrees: $ echo "0^0" | bc 1 Furthermore the claim that "To the definition of the GAMMADIST function, GAMMADIST(0;alpha,beta,0) is always zero." need a bit more substantiation. based on the behavior of the function one could argue for undefined in the name of purity or 1 in the name of prolongation by continuity. Note: ODFF says: If Cumulative is FALSE(), GAMMADIST returns 0 if x < 0 and the value (1/(beta^alpha).gamma(alpha))).(x^(alpha-1)).exp(-(x/beta)) so if the above function is to be defined for x=0, extending by continuation for x -> 0 we have to have GAMMADIST(0,1,1,0) = 1

Sorry, i must accuse me, you are right. I'm wrong, In 3.5 I'll correct this

in interpre6.cxx we must pay attention to the case x=0. For alpha < 1 -> #div/0 for alpha = 1 -> 1/beta for alpha > 1 -> 0 I'll implement this for 3.5.0 @Luca, is this correct.

(In reply to comment #5) > in interpre6.cxx we must pay attention to the case x=0. > For alpha < 1 -> #div/0 > for alpha = 1 -> 1/beta > for alpha > 1 -> 0 > > I'll implement this for 3.5.0 > > @Luca, is this correct. That should be fine, thanks.

fixed in master with commit 9f397368cce0b94f76d4f1d74150ccc75c498e8b

closing