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MathGroup Archive 2007

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some questions (unecessary-?-complicated results by Integrate)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74847] some questions (unecessary-?-complicated results by Integrate)
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Sun, 8 Apr 2007 05:03:45 -0400 (EDT)

Hello to all.

I have a pair of queries. Any help will be greatly appreciate.

Consider the integral

In[2]:=
f = HoldForm[Integrate[1/(x^4 + 2*a*x^2 + 1)^2, {x, 0, Infinity}]]

>From the following I think that the integral converges for a>-1.

In[7]:=
1/(x^4 + 2*a*x^2 + 1)^2 + O[x, 0]^6
1/(x^4 + 2*a*x^2 + 1)^2 + O[x, Infinity]^16
ToRadicals[Reduce[(x^4 + 2*a*x^2 + 1)^2 == 0 && x >= 0 && a > -1, x,
Reals]]
ToRadicals[Reduce[(x^4 + 2*a*x^2 + 1)^2 == 0 && x >= 0 && a <= -1, x,
Reals]]

Out[7]=
SeriesData[x, 0, {1, 0, -4*a, 0, -2 + 12*a^2}, 0, 6, 1]

Out[8]=
SeriesData[x, Infinity, {1, 0, -4*a, 0, (-8 + 48*a^2)/4, 0, (40*a -
4*a*(-8 + 48*a^2))/6}, 8, 16, 1]

Out[9]=
False

Out[10]=
(a < -1 && (x == Sqrt[-a + Sqrt[-1 + a^2]] || x == -Sqrt[-a + Sqrt[-1
+ a^2]])) || (a == -1 && x == 1)


Mathematica however does not see able to understand it.
And moreover it returns an unecessary complicated symbolic result if
we add the assumption a>-1.

In[12]:=
ReleaseHold[f /. Integrate[x___] :> Integrate[x, Assumptions -> a >
-1]]

Out[12]=
If[Im[Sqrt[-a - Sqrt[-1 + a^2]]] > 0 && Im[Sqrt[-a + Sqrt[-1 + a^2]]]
> 0,
  (I*(-3*Sqrt[-a - Sqrt[-1 + a^2]] + 3*Sqrt[-a + Sqrt[-1 + a^2]] +
     a^2*(Sqrt[-a - Sqrt[-1 + a^2]] - Sqrt[-a + Sqrt[-1 + a^2]]) +
     a*Sqrt[-1 + a^2]*(Sqrt[-a - Sqrt[-1 + a^2]] + Sqrt[-a + Sqrt[-1 +
a^2]]))*Pi)/
   (16*(-1 + a^2)^(3/2)*Sqrt[-a - Sqrt[-1 + a^2]]*Sqrt[-a + Sqrt[-1 +
a^2]]),
  Integrate[1/(1 + 2*a*x^2 + x^4)^2, {x, 0, Infinity},
   Assumptions -> a > -1 && (Im[Sqrt[-a - Sqrt[-1 + a^2]]] <= 0 ||
Im[Sqrt[-a + Sqrt[-1 + a^2]]] <= 0)]]

However the integral is simply

In[13]:=
mine = ((2*a + 3)/(2^(7/2)*(a + 1)^(3/2)))*Pi;

Indeed

In[14]:=
ReleaseHold[(f /. a -> #1 & ) /@ Range[-4/5, 2, 1/5]]

Out[14]=
{(7/8)*Sqrt[5/2]*Pi, (9*Sqrt[5]*Pi)/32, (11/24)*Sqrt[5/6]*Pi,
(13/64)*Sqrt[5/2]*Pi, (3*Pi)/(8*Sqrt[2]), (17/96)*Sqrt[5/3]*Pi,
(19/56)*Sqrt[5/14]*Pi, (21*Sqrt[5]*Pi)/256, (23/216)*Sqrt[5/2]*Pi,
(5*Pi)/32, (27/88)*Sqrt[5/22]*Pi, (29/192)*Sqrt[5/6]*Pi,
(31/104)*Sqrt[5/26]*Pi, (33/224)*Sqrt[5/7]*Pi, (7*Pi)/(24*Sqrt[6])}

In[15]:=
(mine /. a -> #1 & ) /@ Range[-4/5, 2, 1/5]

Out[15]=
{(7/8)*Sqrt[5/2]*Pi, (9*Sqrt[5]*Pi)/32, (11/24)*Sqrt[5/6]*Pi,
(13/64)*Sqrt[5/2]*Pi, (3*Pi)/(8*Sqrt[2]), (17/96)*Sqrt[5/3]*Pi,
(19/56)*Sqrt[5/14]*Pi, (21*Sqrt[5]*Pi)/256, (23/216)*Sqrt[5/2]*Pi,
(5*Pi)/32, (27/88)*Sqrt[5/22]*Pi, (29/192)*Sqrt[5/6]*Pi,
(31/104)*Sqrt[5/26]*Pi, (33/224)*Sqrt[5/7]*Pi, (7*Pi)/(24*Sqrt[6])}

Of course Mathematica's symbolic result math is correct.
Let set the GenerateConditions to False. In this way we get a simpler
result than previously obtained.

In[17]:=
math = ReleaseHold[f /. Integrate[x___] :> Integrate[x,
GenerateConditions -> False]]

Out[17]=
((Sqrt[1/(a + Sqrt[-1 + a^2])]*(3 + a*(-a + Sqrt[-1 + a^2])) + Sqrt[a
+ Sqrt[-1 + a^2]]*(-3 + a*(a + Sqrt[-1 + a^2])))*Pi)/(16*(-1 +
a^2)^(3/2))

In[18]:=
FullSimplify[Block[{Message}, (math /. a -> #1 & ) /@ Range[-4/5, 2,
1/5]]]

Out[18]=
{(7/8)*Sqrt[5/2]*Pi, (9*Sqrt[5]*Pi)/32, (11/24)*Sqrt[5/6]*Pi,
(13/64)*Sqrt[5/2]*Pi, (3*Pi)/(8*Sqrt[2]), (17/96)*Sqrt[5/3]*Pi,
(19/56)*Sqrt[5/14]*Pi, (21*Sqrt[5]*Pi)/256, (23/216)*Sqrt[5/2]*Pi,
Indeterminate, (27/88)*Sqrt[5/22]*Pi,
  (29/192)*Sqrt[5/6]*Pi, (31/104)*Sqrt[5/26]*Pi,
(33/224)*Sqrt[5/7]*Pi, (7*Pi)/(24*Sqrt[6])}

For the value a=1 special care is needed for the symbolic result math.
(Note that my formula does not need special care for a=1. Indeed this
feuture is due to the
complcated structure of Mathematica's result).

In[20]:=
Limit[math, a -> 1]

Out[20]=
(5*Pi)/32


The question still remains! What is the reason of the complicated
structure of Mathematica's analytic result?

And here comes my second question.

For a>-1 Mathematica's complicated result and mine simpler one are
equal.

Plot[Chop[math - mine], {a, -0.99, 3}, Axes -> False, Frame -> True]

However I was able to show this only by the following code

In[26]:=
math == mine;
(#1^2 & ) /@ %;
FullSimplify[(Expand[#1] & ) /@ %];
PowerExpand[%]

Out[29]=
True

Why can't Mathematica simplify math-mine to zero with just specifying
a>-1?
(I have already tried it but I failed!).

For everyone wonder about mine result, try

In[33]:=
(1*Dt[x])/(x^4 + 2*a*x^2 + 1)^2 /. x -> Sqrt[y] /. Dt[y] -> 1
Integrate[%, {y, 0, Infinity}]

Out[33]=
1/(2*Sqrt[y]*(1 + 2*a*y + y^2)^2)

Out[34]=
(1/2)*If[a > 0, ((3 + 2*a)*Pi)/(4*Sqrt[2]*(1 + a)^(3/2)), Integrate[1/
(Sqrt[y]*(1 + 2*a*y + y^2)^2), {y, 0, Infinity},
    Assumptions -> a <= 0]]


But not that even now, Mathematica can't see that the integral
converges for a>-1 and it
erroneoussly returns the convergence region a>0.

E.g.

In[37]:=
(1*Dt[x])/(x^4 + 2*(-6/7)*x^2 + 1)^2 /. x -> Sqrt[y] /. Dt[y] -> 1
Integrate[%, {y, 0, Infinity}]

Out[37]=
1/(2*Sqrt[y]*(1 - (12*y)/7 + y^2)^2)

Out[38]=
(9/8)*Sqrt[7/2]*Pi


Dimitris



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