Re: Beta function, Integral

• To: mathgroup at smc.vnet.net
• Subject: [mg79893] Re: Beta function, Integral
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Wed, 8 Aug 2007 04:50:02 -0400 (EDT)
• Organization: The Open University, Milton Keynes, UK
• References: <f990ds\$btq\$1@smc.vnet.net>

```Asim wrote:
> 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])}

The /mathematical/ expression that is returned by Mathematica 6 is
correct, indeed. However, the answer has funny extra curly brackets and
if one try to use the expression to check its validity against Beta[p,q]
the expression returned unevaluated. On the other hand, if one type in
the expression by hand, the simplification occurs and the identity is
checked positively.

In[1]:= \$Version

Out[1]= "6.0 for Microsoft Windows (32-bit) (June 19, 2007)"

In[2]:= sol =
Integrate[t^{p - 1}*(1 - t)^(q - 1), {t, 0, 1},
Assumptions -> {p > 0, q > 0}]

Out[2]= {(\[Pi] Csc[\[Pi] q] Gamma[p])/(Gamma[1 - q] Gamma[p + q])}
--------^---------------------------------------------------------^
Note the spurious curly brackets.

In[3]:= FullSinplify[Beta[p, q] == sol[[1]]]

Out[3]= FullSinplify[
Beta[p, q] == (\[Pi] Csc[\[Pi] q] Gamma[p])/(
Gamma[1 - q] Gamma[p + q])]

Even though we took out the contains of the list, Mathematica returns
the FullSimplify unevaluated.

In[4]:= FullSimplify[
Beta[p, q] == (Pi Csc[Pi q] Gamma[p])/(Gamma[1 - q] Gamma[p + q])]

Out[4]= True

Now, having entered the expression by hand, Mathematica is able to check
the identity.

A similar behavior can be seen with Mathematica 5.2, though the
expression returned is different.

In[1]:=
\$Version

Out[1]=
"5.2 for Microsoft Windows (June 20, 2005)"

In[2]:=
sol = Integrate[t^{p - 1}*(1 - t)^(q - 1), {t, 0, 1},
Assumptions -> {p > 0, q > 0}]

Out[2]=
{(Gamma[p]*Gamma[q])/Gamma[p + q]}

In[3]:=
FullSinplify[Beta[p, q] == sol[[1]]]

Out[3]=
FullSinplify[Beta[p, q] == (Gamma[p]*Gamma[q])/
Gamma[p + q]]

In[4]:=
FullSimplify[Beta[p, q] == (Gamma[p]*Gamma[q])/
Gamma[p + q]]

Out[4]=
True

--
Jean-Marc

```

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