Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2007
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

more in Assumptions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74989] more in Assumptions
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Fri, 13 Apr 2007 02:11:38 -0400 (EDT)

On the same vein with some recent posts of mine, let me consider the
integral of f[x,a,c] in the range [0,infinity), where

In[93]:=
f[x_, a_, c_] := 1/(x^4 + 2*a*x^2 + 1 + c)

Let investegate ranges of a,c in which the integral converges.
(I am interested in real values.)

In the integration limits we have

In[95]:=
f[x, a, c] + O[x]^6
f[x, a, c] + O[x, Infinity]^6

Out[95]=
SeriesData[x, 0, {(1 + c)^(-1), 0, (-2*a)/(1 + c)^2, 0, (4*a^2)/(1 +
c)^3 - (1 + c)^(-2)}, 0, 6, 1]
Out[96]=
SeriesData[x, Infinity, {1}, 4, 6, 1]

So in Infinity there is not problem, whereas in zero, it must be c !=
-1.

With this in mind we examine the denominator of f[x,a,c] for possible
poles
in the integration range.

In[100]:=
ToRadicals[Reduce[Denominator[f[x, a, c]] == 0 && c < -1 && x > 0, x,
Reals]]
ToRadicals[Reduce[Denominator[f[x, a, c]] == 0 && c > -1 && x > 0, x,
Reals]]

Out[100]=
c < -1 && x == -Sqrt[-a - Sqrt[-1 + a^2 - c]]
Out[101]=
a < 0 && ((-1 < c < -1 + a^2 && (x == Sqrt[-a + Sqrt[-1 + a^2 - c]] ||
x == -Sqrt[-a + Sqrt[-1 + a^2 - c]])) ||
   (c == -1 + a^2 && x == Sqrt[-a + Sqrt[-1 + a^2 - c]]))

In[102]:=
ToRadicals[Reduce[Denominator[f[x, a, c]] == 0 && c > -1 && x > 0 && a
> 0, x, Reals]]
ToRadicals[Reduce[Denominator[f[x, a, c]] == 0 && c > -1 && x > 0 && a
< 0, x, Reals]]

Out[102]=
False
Out[103]=
a < 0 && ((-1 < c < -1 + a^2 && (x == Sqrt[-a + Sqrt[-1 + a^2 - c]] ||
x == -Sqrt[-a + Sqrt[-1 + a^2 - c]])) ||
   (c == -1 + a^2 && x == Sqrt[-a + Sqrt[-1 + a^2 - c]]))

In[104]:=
ToRadicals[Reduce[Denominator[f[x, a, c]] == 0 && c > -1 + a^2 && x >
0 && a < 0, x, Reals]]

Out[104]=
False

I think we have enough evidence that the integral converges for
{a<0&&c > -1 + a^2 and a>0&&c>-1}.

However Mathematica for each one of the following commands
return alsmost the same unecessary complicated If[...] structure
(*outputs are ommited*)

In[124]:=
Integrate[f[x, a, c], {x, 0, Infinity}, Assumptions -> Element[{a,
c} , Reals]]

In[125]:=
Integrate[f[x, a, c], {x, 0, Infinity}, Assumptions -> a < 0]

In[126]:=
Integrate[f[x, a, c], {x, 0, Infinity}, Assumptions -> a > 0 && c >
-1]

In[127]:=
Integrate[f[x, a, c], {x, 0, Infinity}, Assumptions ->a < 0 && c > -1
+ a^2]

Let see some examples with specific values for {a,c}

In[132]:=
Integrate[f[x, 1, -9/10], {x, 0, Infinity}](*a>0&&c>-1*)
{N[%], NIntegrate[f[x, 1, -9/10], {x, 0, Infinity}]}
Integrate[f[x, -2, 3 + 1/10], {x, 0, Infinity}](*a<0&&c>-1+a^2*)
{N[%], NIntegrate[f[x, -2, 3 + 1/10], {x, 0, Infinity}]}
Integrate[f[x, 1, -1], {x, 0, Infinity}](*a>0&&c<=-1*)
Integrate[f[x, -2, 3], {x, 0, Infinity}](*a<0&&c<=-1+a^2*)

Out[132]=
(1/6)*Sqrt[50 - 5*Sqrt[10]]*Pi
Out[133]=
{3.061535891296801, 3.0615358912965314}
Out[134]=
(5*Pi)/Sqrt[82*(-20 + Sqrt[410])]
Out[135]=
{3.48007,3.48007}
Integrate::idiv: Integral of 1/(2*x^2 + x^4) does not converge on \
{0,Infinity}.
Out[136]=
Integrate[1/(2*x^2 + x^4), {x, 0, Infinity}]
Integrate::idiv: Integral of 1/(-2 + x^2)^2 does not converge on
{0,Infinity}.
Out[137]=
Integrate[1/(4 - 4*x^2 + x^4), {x, 0, Infinity}]

and compare them with the result of the symbolic definite integration

In[145]:=
ff = Integrate[f[x, a, c], {x, 0, Infinity}]
FullSimplify[% /. {{a -> 1, c -> -9/10}, {a -> -2, c -> 3 + 1/10}}]
N[%]

Out[145]=
If[Im[Sqrt[-a - Sqrt[-1 + a^2 - c]]] > 0 && Im[Sqrt[-a + Sqrt[-1 + a^2
- c]]] > 0,
  -((I*Pi)/(2*(Sqrt[-a - Sqrt[-1 + a^2 - c]] + Sqrt[-a + Sqrt[-1 + a^2
- c]])*Sqrt[-a - Sqrt[-1 + a^2 - c]]*
     Sqrt[-a + Sqrt[-1 + a^2 - c]])), Integrate[1/(1 + c + 2*a*x^2 +
x^4), {x, 0, Infinity},
   Assumptions -> Im[Sqrt[-a - Sqrt[-1 + a^2 - c]]] <= 0 || Im[Sqrt[-a
+ Sqrt[-1 + a^2 - c]]] <= 0]]
Out[146]=
{(5*Pi)/Sqrt[2*(10 + Sqrt[10])], (5*Pi)/Sqrt[82*(-20 + Sqrt[410])]}
Out[147]=
{3.061535891296801, 3.480066694467329}


Indeed the symbolic definite integration result is correct, but the
question of its
unecessary (?) complicated form still remains.

Seeing carefully the following I think there is a connection with
Mathematica's antiderivative

In[191]:=
Integrate[f[x, a, c], x]

Out[191]=
(ArcTan[x/Sqrt[a - Sqrt[-1 + a^2 - c]]]/Sqrt[a - Sqrt[-1 + a^2 - c]] -
ArcTan[x/Sqrt[a + Sqrt[-1 + a^2 - c]]]/
    Sqrt[a + Sqrt[-1 + a^2 - c]])/(2*Sqrt[-1 + a^2 - c])

In[190]:=
Integrate[f[x, a, c], {x, 0, Infinity}, Assumptions -> Im[Sqrt[-a -
Sqrt[-1 + a^2 - c]]] > 0 &&
    Im[Sqrt[-a + Sqrt[-1 + a^2 - c]]] > 0]

Out[190]=
-((I*Pi)/(2*(Sqrt[-a - Sqrt[-1 + a^2 - c]] + Sqrt[-a + Sqrt[-1 + a^2 -
c]])*Sqrt[-a - Sqrt[-1 + a^2 - c]]*
    Sqrt[-a + Sqrt[-1 + a^2 - c]]))

but I don't see any mathematical reason that could support this fact.

Any insight will be greatly appreciate.

Dimitris



  • Prev by Date: Re: Variable containing code
  • Next by Date: Re: Re: Finding unknown parameters using Mathematica
  • Previous by thread: Re: Memory issue in SVD
  • Next by thread: SparseArray