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MathGroup Archive 2007

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Re: [Mathematica 6] Integrate strange result

  • To: mathgroup at smc.vnet.net
  • Subject: [mg78366] Re: [Mathematica 6] Integrate strange result
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Fri, 29 Jun 2007 05:46:11 -0400 (EDT)
  • References: <f5vs8m$kob$1@smc.vnet.net>

            Nasser Abbasi       :
> In[13]:= $Version
> Out[13]= 6.0 for Microsoft Windows (32-bit) (April 27, 2007)
>
> The problem is this: When I ask Mathematica to integrate something
> involving some constant parameter, it gives an answer for some range
> of the constant involved, say range A, and it says there is no answer
> for the other range, say range B
>
> But next, when I specify, using assumptions, that the constant is now
> in range B, then Mathematica does given an answer to the integral !
>
> Why does it in one case say there is no answer if the constant is in
> range B, and in the second case it gives an answer when the constant
> is in range B?
>
> case 1
> ========
>
> In[2]:= f = (1 + k*Sin[a]^2)^(1/2);
> In[12]:= Assuming[Element[k, Reals], Integrate[f, {a, 0, 2*Pi}]]
>
> Out[12]= If[k >= -1, 4*EllipticE[-k],   Integrate[Sqrt[1 +
> k*Sin[a]^2],    {a, 0, 2*Pi}, Assumptions ->     k < -1]]
>
> notice in the above, it says for k<-1 there is NO answer
>
> case 2
> =======
> In[2]:= f = (1 + k*Sin[a]^2)^(1/2);
> In[10]:= Assuming[Element[k, Reals] && k < -1,   Integrate[f, {a, 0,
> 2*Pi}]]
>
> Out[10]= 0
>
> Notice in the above, for k<-1  it gives an answer, which is zero.
>
> What is it I am missing here?
>
> thanks
> Nasser

When Assumptions return a of the form structure
If[o==True,o1,Integrate[...]]
it does not mean necessarily that if o!=True then the integral
cannot be evaluated.
Consider for example the following integral

Integrate[E^(-x^2 - x*y)/Sqrt[x], {x, 0, Infinity}]

We have

In[12]:=
Integrate[E^(-x^2 - x*y)/Sqrt[x], {x, 0, Infinity}]

Out[12]=
If[Re[y] < 0, (E^(y^2/8)*Pi*Sqrt[-y]*(BesselI[-(1/4), y^2/8] +
BesselI[1/4, y^2/8]))/(2*Sqrt[2]),
  Integrate[1/(E^(x*(x + y))*Sqrt[x]), {x, 0, Infinity}, Assumptions -
> Re[y] >= 0]]

That is for Re[y]<0 the integral is (E^(y^2/8)*Pi*Sqrt[-y]*(BesselI[-
(1/4), y^2/8] + BesselI[1/4, y^2/8]))/(2*Sqrt[2]);
a check can confirm its validity.
How about Re[y]>0? Is it divergent? There is not a closed form
expression?
Try it out!

In[14]:=
({#1, Integrate[E^(-x^2 - x*y)/Sqrt[x], {x, 0, Infinity}, Assumptions -
> #1[Re[y], 0]]} & ) /@ {Less, Greater}

Out[14]=
{{Less, (E^(y^2/8)*Pi*Sqrt[-y]*(BesselI[-(1/4), y^2/8] + BesselI[1/4,
y^2/8]))/(2*Sqrt[2])},
  {Greater, (1/2)*E^(y^2/8)*Sqrt[y]*BesselK[1/4, y^2/8]}}

That is, we get a different form of answer.

For y=0?

In[26]:=
With[{y = 0}, Integrate[E^(-x^2 - x*y)/Sqrt[x], {x, 0, Infinity}]]
FullSimplify[% == (1/2)*Gamma[1/4]]

Out[26]=
2*Gamma[5/4]
Out[27]=
True

Note also,

In[32]:=
{Limit[(E^(y^2/8)*Pi*Sqrt[-y]*(BesselI[-(1/4), y^2/8] + BesselI[1/4,
y^2/8]))/(2*Sqrt[2]), y -> 0, Direction -> 1],
  Limit[(1/2)*E^(y^2/8)*Sqrt[y]*BesselK[1/4, y^2/8], y -> 0, Direction
-> -1]}
(Plus[#1 - #1] & ) @@ %

Out[32]=
{Pi/(Sqrt[2]*Gamma[3/4]), (1/2)*Gamma[1/4]}
Out[33]=
0

That is the integral is a continuous function with respect to the
parameter y.

In[38]:=
Plot[NIntegrate[Exp[-x^2 - x*y]/Sqrt[x], {x, 0, Infinity}], {y, -2,
2}]

Dimitris



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