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