Re: question
- From: dimitris <dimmechan@xxxxxxxxx>
- Date: 23 Apr 2007 04:36:20 -0700
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])
Ο/Η dimitris έγραψε:
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 (by hand or a CAS) 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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