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