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MathGroup Archive 2007

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Re: Beta function, Integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg79885] Re: [mg79858] Beta function, Integral
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Wed, 8 Aug 2007 04:45:51 -0400 (EDT)
  • Reply-to: hanlonr at cox.net

It is correct.

int1 = Integrate[t^(p - 1)*(1 - t)^(q - 1), {t, 0, 1}, 
    Assumptions -> {p > 0, q > 0}]

(Pi*Csc[Pi*q]*Gamma[p])/
   (Gamma[1 - q]*Gamma[p + q])

To force an alternate form add the assumption that p and q are integers.

int2 = Integrate[t^(p - 1)*(1 - t)^(q - 1), {t, 0, 1}, 
    Assumptions -> {Element[{p, q}, Integers], 
        p > 0, q > 0}]

(Gamma[p]*Gamma[q])/Gamma[p + q]

Verifying that the forma are equivalent and equal to Beta function

int1 == int2 == Beta[p, q] // FullSimplify

True

Hence

Solve[int1 == int2, Gamma[1 - q]][[1, 1]]

Gamma[1 - q] -> (Pi*Csc[Pi*q])/
     Gamma[q]

FullSimplify[Equal @@ %]

True


Bob Hanlon

---- Asim <maa48 at columbia.edu> wrote: 
> 
> Hi
> 
> The following integral does not seem to give the correct answer. The
> answer should be the Euler Beta function,  Beta[p,q]. Can anybody let
> me know what I am doing wrong? Or is this a bug?
> 
> 
> In[12]:= Integrate[t^{p - 1}*(1 - t)^(q - 1), {t, 0, 1},  Assumptions -
> > {p > 0, q > 0}]
> 
> Out[12]= {(\[Pi] Csc[\[Pi] q] Gamma[p])/(Gamma[1 - q] Gamma[p + q])}
> 
> 
> Thanks
> 
> Asim Ansari
> 
> 



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