       Re: Beta function, Integral

• To: mathgroup at smc.vnet.net
• Subject: [mg79945] Re: Beta function, Integral
• From: Asim <maa48 at columbia.edu>
• Date: Thu, 9 Aug 2007 05:23:08 -0400 (EDT)
• References: <f990ds\$btq\$1@smc.vnet.net><f9c1q6\$75a\$1@smc.vnet.net>

```On Aug 8, 2:19 pm, dimitris <dimmec... at yahoo.com> wrote:
> On 7    , 08:37, Asim <ma... 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:= Integrate[t^{p - 1}*(1 - t)^(q - 1), {t, 0, 1},  Assumptions -
>
> > > {p > 0, q > 0}]
>
> > Out= {(\[Pi] Csc[\[Pi] q] Gamma[p])/(Gamma[1 - q] Gamma[p + q])}
>
> > Thanks
>
> > Asim Ansari
>
> First note that you used List where you should have used parentheses!
> A common mistake.
> It must be t^(p-1); not {p-1}.
>
> In the Mathematica I work, I took:
>
> In:=
> \$Version
>
> Out=
> "5.2 for Microsoft Windows (June 20, 2005)"
>
> In:=
> Integrate[t^(p - 1)*(1 - t)^(q - 1), {t, 0, 1}, Assumptions -> {p > 0,
> q > 0}]
> FunctionExpand[Beta[p, q] - %]
>
> Out=
> (Gamma[p]*Gamma[q])/Gamma[p + q]
>
> Out=
> 0
>
> Ommiting { } from your output, we have also
>
> In:=
> FullSimplify[Beta[p, q] - (Pi*Csc[Pi*q]*Gamma[p])/(Gamma[1 -
> q]*Gamma[p + q])]
>
> Out=
> 0
>
> as it must be.
>
> Note also that
>
> In:= Beta[p, q] // FunctionExpand
> Out= (Gamma[p] Gamma[q])/Gamma[p + q]
>
> and
>
> In:=
> FullSimplify[(Gamma[p]*Gamma[q])/Gamma[p + q] == Beta[p, q]]
>
> Out=
> True
>
> but
>
> In:=
> FullSimplify[c
>
> Out=
> c
>
> Regards
> Dimitris

Thanks for all who responded to by Beta function integral post.

I used Mathematica 6.0, and by mistake used t^{p-1} in the integral,
I was expecting   (Gamma[p] Gamma[q])/Gamma[p + q] which can be
obtained from Mathematica 5.2 as the answer, but Mathematica 6.0
returns (\[Pi] Csc[\[Pi] q] Gamma[p])/(Gamma[1 - q] Gamma[p + q]). The
two are indeed the same, but the former is the more "standard" answer
given the definition of the Beta function.

Regards

Asim Ansari

```

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