Re: question

Hi P^2.
Very nice your function ChangeVar.
Keep on the good job!

I didn't know that you can do the substitution x == Tan[u/2] even in
non-rational functions
of sin(x) and cos(x). Now I really learnt something!

Now I learnt it...let's use it!

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

But I think, it's interesting, that Mathematica gives different forms of
the same expression, when using slightly different substitutions in Dimitri's
integral, because I would have guessed, there is some kind of normalization
inside Mathematica's Integrate[].

My workout indicates you are again right!

Thanks again!
Dimitris


Ο/Η Peter Pein έγραψε:
David W. Cantrell schrieb:
...
With t -> Pi/4, your 2*EllipticF[Pi/4, 2] becomes 1/2 Beta[1, 1/2, 1/4] and
then version 5.2 gives

In[2]:=
1/2 Beta[1, 1/2, 1/4]

Out[2]=
(Sqrt[Pi]*Gamma[1/4])/(2*Gamma[3/4])

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

Out[3]=
0

David

Hi,

It is 2AM in germany and I'm far too tired to have a closer look, whether the
identity 1/2 Beta[1, 1/2, 1/4] == Beta[Tan[Pi/8]^2, 1/4, 1/2] is trivial or
not. But I think, it's interesting, that Mathematica gives different forms of
the same expression, when using slightly different substitutions in Dimitri's
integral, because I would have guessed, there is some kind of normalization
inside Mathematica's Integrate[].

ChangeVar[x == Tan[u/4], Integrate[1/Sqrt[Sin[u]], {u, 0, Pi/2}], x]
--> Beta[Tan[Pi/8]^2, 1/4, 1/2]

good night,
Peter

.