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MathGroup Archive 2007

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Re: question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75262] Re: question
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Mon, 23 Apr 2007 05:34:23 -0400 (EDT)
  • References: <f0f8ud$k47$1@smc.vnet.net>

After response from Peter Pein I realise something new (for me!).
You can do the well known substitution x=tan(u/2) and for non-rational
functions of sin(u) and cos(u).

So even though it was not exactly what I ask...

In[248]:=
Dt[u]/Sqrt[Sin[u]]
% /. Sin[u] -> Cos[u/2]^2 - Sin[u/2]^2
(Simplify[#1, 0 < x < 1] & )[% /. u -> 2*ArcTan[x]]
Integrate[% /. Dt[x] -> 1, {x, 0, 1}]

Out[248]=
Dt[u]/Sqrt[Sin[u]]

Out[249]=
Dt[u]/Sqrt[Cos[u/2]^2 - Sin[u/2]^2]

Out[250]=
(2*Dt[x])/Sqrt[1 - x^4]

Out[251]=
(2*Sqrt[Pi]*Gamma[5/4])/Gamma[3/4]

In[252]:=
FullSimplify[(2*Sqrt[Pi]*Gamma[5/4])/Gamma[3/4] == Gamma[1/4]^2/
(2*Sqrt[2*Pi])]

N::meprec: Internal precision limit $MaxExtraPrecision....
Out[252]=
True


=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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