Re: question (2nd)

*To*: mathgroup at smc.vnet.net*Subject*: [mg75263] Re: question (2nd)*From*: dimitris <dimmechan at yahoo.com>*Date*: Mon, 23 Apr 2007 05:34:54 -0400 (EDT)*References*: <f0f8ud$k47$1@smc.vnet.net>

I think I have just encountered the quickest verification... In[2]:= Timing[Developer`ZeroQ[Gamma[1/4]^2/(2*Sqrt[2*Pi]) - 2*EllipticF[Pi/4, 2]]] Out[2]= {0.07800000000000001*Second, True} 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

**Follow-Ups**:**Re: Re: question (2nd)***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>