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MathGroup Archive 1995

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Definite integral difficulty

  • To: mathgroup at christensen.cybernetics.net
  • Subject: [mg1707] Definite integral difficulty
  • From: sergio at shark.inst.bnl.gov (Sergio Rescia)
  • Date: Mon, 17 Jul 1995 04:44:33 -0400

math #:L1030-9293

I was trying to solve the integral:

In[1]:= Integrate[Sin[u]^2 (Sin[x u]/(x u))^4,{u,0,Infinity}] //Simplify

                         2
        Pi (-1 + 2 Sqrt[x ])
Out[1]= --------------------
                   4
                8 x

The result is WRONG, since the integrand is always positive, so the integral must be 
positive for any x.

A workaround is to split the integral:

In[2]:= int1=Integrate[(Sin[x u]/(x u))^4,{u,0,Infinity}]

                                2
                       Pi Sqrt[x ]
Out[2]= If[Im[x] == 0, -----------, ComplexInfinity]
                             2
                          3 x

In[3]:= int2=Integrate[Cos[u]^2 (Sin[x u]/(x u))^4,{u,0,Infinity}]

Out[3]= If[Im[2 - 4 x] == 0 && Im[2 - 2 x] == 0 && Im[x] == 0 && 
 
>     Im[2 + 2 x] == 0 && Im[2 + 4 x] == 0, 
 
                             2                   2
>    (Pi (6 - 4 Sqrt[(-1 + x) ] + 8 Sqrt[(-1 + x) ] x - 
 
                         2   2      2       2                  2
>         4 Sqrt[(-1 + x) ] x  + 8 x  Sqrt[x ] - 4 Sqrt[(1 + x) ] - 
 
                          2       2             2                   2
>         8 x Sqrt[(1 + x) ] - 4 x  Sqrt[(1 + x) ] + Sqrt[(-1 + 2 x) ] - 
 
                             2       2                2                  2
>         4 x Sqrt[(-1 + 2 x) ] + 4 x  Sqrt[(-1 + 2 x) ] + Sqrt[(1 + 2 x) ] + 
 
                            2       2               2           4
>         4 x Sqrt[(1 + 2 x) ] + 4 x  Sqrt[(1 + 2 x) ])) / (48 x ), 
 
>    ComplexInfinity]



Since my x is real positive I ca also:


In[4]:= p1=(Pi*(x^2)^(1/2))/(3*x^2) /. Sqrt[x_^2]->x //Simplify

        Pi
Out[4]= ---
        3 x

In[6]:= p2= (Pi*(6 - 4*((-1 + x)^2)^(1/2) + 8*((-1 + x)^2)^(1/2)*x - 
        4*((-1 + x)^2)^(1/2)*x^2 + 8*x^2*(x^2)^(1/2) - 
        4*((1 + x)^2)^(1/2) - 8*x*((1 + x)^2)^(1/2) - 
        4*x^2*((1 + x)^2)^(1/2) + ((-1 + 2*x)^2)^(1/2) - 
        4*x*((-1 + 2*x)^2)^(1/2) + 4*x^2*((-1 + 2*x)^2)^(1/2) + 
        ((1 + 2*x)^2)^(1/2) + 4*x*((1 + 2*x)^2)^(1/2) + 
        4*x^2*((1 + 2*x)^2)^(1/2)))/(48*x^4) /. 
         {((-1+x)^2)^(1/2)->Abs[-1+x],
          ((-1+2 x)^2)^(1/2)->Abs[-1+2 x],
          ((1+x)^2)^(1/2)->(1+x),
          ((1+2 x)^2)^(1/2)->(1+2 x),
          Sqrt[x^2]->x} //Simplify

In[7]:= res=p1 - p2 //Simplify

Out[7]= (Pi (-3 + 6 x + 4 x  + 4 Abs[-1 + x] - 8 x Abs[-1 + x] + 
 
            2
>        4 x  Abs[-1 + x] - Abs[-1 + 2 x] + 4 x Abs[-1 + 2 x] - 
 
            2                        4
>        4 x  Abs[-1 + 2 x])) / (48 x )

and by further manipulation, more easily done on paper (at least for me!):

In[8]:= res2=Pi/(48 x^2) ( (4*x^3 + 6 x -3) + 4*Abs[x - 1]^3 - Abs[ 2*x - 1]^3)

                          3                3                3
        Pi (-3 + 6 x + 4 x  + 4 Abs[-1 + x]  - Abs[-1 + 2 x] )
Out[8]= ------------------------------------------------------
                                    2
                                48 x


It is not the first time I am facing problems with definite integrals.
Can you shed some light?

Regards,

Sergio Rescia
-------------------------------------
Reply to :
Sergio Rescia
Brookhaven National Laboratory
Instrumentation Division - Bld 535B
UPTON, NY 11973-5000 - U.S.A.
E-mail: Rescia at bnl.gov (Internet)
	Rescia@bnl     (Bitnet)
	BNL::RESCIA    (DECNET) 
Fax: 516-282-5773
-------------------------------------


 


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