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
-------------------------------------
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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