M^2 (definite integral)
- To: mathgroup at smc.vnet.net
- Subject: [mg78765] M^2 (definite integral)
- From: dimitris <dimmechan at yahoo.com>
- Date: Mon, 9 Jul 2007 01:32:42 -0400 (EDT)
In[1600]:=
Integrate[Sqrt[z]*(Sqrt[1 + z]/Sqrt[z^5 + 1]), {z, 0, Infinity}]
Out[1600]=
-(MeijerG[{{1/5, 2/5, 3/5, 4/5, 4/5, 1}, {}}, {{-(1/10), 1/10, 3/10,
3/10, 1/2, 7/10}, {}}, 1]/(160*Sqrt[5]*Pi^5))
After several hours Mathematica (5.2 and 6 as I was informed see below
http://groups.google.gr/group/sci.math.symbolic/browse_thread/thread/24aa28ecd6734521/1657342a0b03492c?hl=el#1657342a0b03492c
) fails to give a numerical estimation with N[] of this MeijerG
function.
Oleksandr Pavlyk of WRI mentiones that this problem is fixed in the
development
version and he also offers a workaround (see above mentioned link).
Let's see how we can avoid the MeijerG function until the new
version of Mathematica comes on our hands...
This is a numerical estimation of the integral
In[1611]:=
NIntegrate[Sqrt[z]*(Sqrt[1 + z]/Sqrt[z^5 + 1]), {z, 0, Infinity},
WorkingPrecision -> 50, PrecisionGoal -> 30,
MaxRecursion -> 16]
Out[1611]=
2.968824946844772905814687775851795916334391176498`30.22201746305465
First write the integrand as follows
In[1656]:=
ff=Factor //@ (Simplify[#1, z > 0] & )[Sqrt[z]*(Sqrt[1 + z]/Sqrt[z^5
+
1])]
Out[1656]=
Sqrt[z/(1 - z + z^2 - z^3 + z^4)]
Then take the antiderivative of ff.
In[1666]:=
F = (FullSimplify[#1, z > 0] & )[Integrate[ff, z]]
Out[1666]=
(1/(Sqrt[5 + Sqrt[5]]*(-1 + z^2)))*(I*Sqrt[z]*(Sqrt[3 + Sqrt[5]]*(-1
+
z)*Sqrt[2 + 1/z + z]*
EllipticF[ArcSin[Sqrt[(1/2)*(-1 + Sqrt[5]) + 1/z + z]/5^(1/4)],
-
(2/(-1 + Sqrt[5]))] +
Sqrt[((-5 + Sqrt[5])*(-1 + z)^2)/z]*(1 +
z)*EllipticF[ArcSin[Sqrt[(1/2)*(-1 + Sqrt[5]) + 1/z + z]/5^(1/4)],
(1/2)*(-5 + 3*Sqrt[5])]))
There is a jump discontinuity exactly at z=1
In[1668]:=
(Plot[#1[F], {z, 0, 10}] & ) /@ {Re, Im};
(*plots to be displayed*)
Unfortunately applying the NL formula does not
work since Mathematica seems not to be able to evaluate
the limits at zero and infinity in reasonable time
(for z->1 left and right there is not such a problem).
The problem is due to the expression ArcSin[Sqrt[(1/2)*(-1 + Sqrt[5])
+ 1/z + z]/5^(1/4)] and its behavior as z tends to zero and infinity
In[1671]:=
(Limit[ArcSin[Sqrt[(1/2)*(-1 + Sqrt[5]) + 1/z + z]/5^(1/4)], z -> #1]
& ) /@ {0, Infinity}
Out[1671]=
{DirectedInfinity[-I], DirectedInfinity[-I]}
But we can overcome this problem if we consult to the very definition
of the EllipticF integral
In[1672]:=
Integrate[1/Sqrt[1 - m*Sin[ ]^2], { , 0, u}, GenerateConditions ->
False]
Out[1672]=
EllipticF[u, m]
For the application of the Newton Leibniz forula one needs
the limits of F at z={0,1(+,-),Infinity}.
We have
In[1674]:=
f1 = ToRadicals[RootReduce[-Limit[F, z -> 1, Direction -> -1] +
Limit[F, z -> 1, Direction -> 1]]]
Out[1674]=
(-1 + Sqrt[5])*EllipticF[ArcSin[Sqrt[1/2 + 3/(2*Sqrt[5])]], (1/2)*(-5
+ 3*Sqrt[5])]
For the limits at zero and Infinity we will work as follows
In[1677]:=
f2 = FullSimplify[F /. EllipticF[u_, m_] -> Integrate[1/Sqrt[1 -
m*Sin[ ]^2], { , 0, (-I)*Infinity}, GenerateConditions -> False]]
Out[1677]=
(1/(Sqrt[5 + Sqrt[5]]*(-1 + z^2)))*(I*Sqrt[z]*(Sqrt[2]*(-1 +
Sqrt[5])*(2 + Sqrt[5])*(-1 + z)*Sqrt[2 + 1/z + z]*
EllipticF[ArcCot[Sqrt[1 + Sqrt[5]]], 9 + 4*Sqrt[5]] -
(4*Sqrt[10]*Sqrt[((-5 + 2*Sqrt[5])*(-1 + z)^2)/z]*(1 + z)*
EllipticF[(-(1/4))*I*(Log[5] - 4*Log[1 + Sqrt[1 + Sqrt[5]]]),
5])/(-5 + 3*Sqrt[5])))
In[1685]:=
f3 = FullSimplify[Limit[f2, z -> Infinity] - Limit[f2, z -> 0]]
Out[1685]=
4*EllipticF[(-(1/4))*I*(Log[5] - 4*Log[1 + Sqrt[1 + Sqrt[5]]]), 5]
And (at last!) the requeted definite integral (no MeijerG please!)
In[1696]:=
res = FullSimplify[f3 + f1]
Out[1696]=
(-1 + Sqrt[5])*EllipticF[ArcSin[Sqrt[(1/10)*(5 + 3*Sqrt[5])]],
(1/2)*(-5 + 3*Sqrt[5])] +
4*EllipticF[(-(1/4))*I*(Log[5] - 4*Log[1 + Sqrt[1 + Sqrt[5]]]), 5]
How about check it?
In[1697]:=
N[res, 30]
Out[1697]=
2.968824946844772905814687775851795916332930758021`30.15051499783199
+
0``29.67793040712738*I
Dimitris
PS
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