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MathGroup Archive 2007

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Re: question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75306] Re: question
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Tue, 24 Apr 2007 03:30:45 -0400 (EDT)
  • References: <f0f8ud$k47$1@smc.vnet.net>

Note also that

In[18]:=
Integrate[1/Sin[x]^m, {x, 0, Pi/2}, Assumptions -> m < 1]
% /. m -> 1/2

Out[18]=
(Sqrt[Pi]*Gamma[1/2 - m/2])/(2*Gamma[1 - m/2])

Out[19]=
(Sqrt[Pi]*Gamma[1/4])/(2*Gamma[3/4])

Dimitris

=CF/=C7 dimitris =DD=E3=F1=E1=F8=E5:
> Hello.
>
> In Mathematica 5.2 I got
>
> In[73]:=
> Integrate[1/Sqrt[Sin[u]], {u, 0, Pi/2}]
>
> Out[73]=
> 2*EllipticF[Pi/4, 2]
>
> The result ic correct as a numerical integration indicates.
>
> In[74]:=
> {Chop[N[%]], NIntegrate[1/Sqrt[Sin[u]], {u, 0, Pi/2}]}
>
> Out[74]=
> {2.6220575244897986, 2.622057554312378}
>
> It can be proved that the definite integral under discussion is
> given by Gamma[1/4]^2/(2*Sqrt[2*Pi]).
>
> In[75]:=
> N[Gamma[1/4]]^2/(2*Sqrt[2*Pi])
>
> Out[75]=
> 2.6220575542921196
>
> Can somebody point me out a way to show that the
> Gamma[1/4]^2/(2*Sqrt[2*Pi]) is equal to 2*EllipticF[Pi/4, 2] or/and a
> series
> of steps from taking from the latter to the former and vice versa?
>
> Thanks
> Dimitris



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