Re: Troubles with Integrate

• To: mathgroup at smc.vnet.net
• Subject: [mg70279] Re: Troubles with Integrate
• From: Peter Pein <petsie at dordos.net>
• Date: Wed, 11 Oct 2006 01:53:43 -0400 (EDT)
• References: <eg84hk\$nir\$1@smc.vnet.net>

```Roman V. Kiseliov schrieb:
> If I try to do following calculation Mathematica 5.2.0.0 can't evaluate
> integrals.
>
>  psi:= (-Pi^(-1))*((Sin[Pi*((t - 5/4)/3)]/(t - 5/4))*Cos[Pi*(t - 5/4)] +
>       (Sin[Pi*(t + 1/4)]/(t + 1/4))*Cos[Pi*(t + 1/4)]) +
>     (1/Pi)*((Sin[2*Pi*((t - 7/8)/3)]/(t - 7/8))*Sin[2*Pi*(t - 7/8)] -
>       (Sin[2*Pi*((t - 1/8)/3)]/(t - 1/8))*Sin[2*Pi*(t - 1/8)]);
>
>  Integrate[psi*psi, t]
>  Integrate[psi*psi, {t, -inf, +inf}]
>
>  If I try to interrupt calculation, Mathematica sometimes says
>  'Mathematica has detected a possible internal error. If possible, report
>  the error to
>  support at wolfram.com, quoting "Assertion 'interruptMenuType != 0' failed at
>  KernelPackets.c:726", and describe in as much detail as possible what
>  you were
>  doing when the error occurred.'
>
>  Note also that Integrate[psi, {t, -inf, +inf}] in Mathematica 4.2.0.0
>  evaluates as 0
>  but Mathematica 5.2.0.0 give value -1/2
>
> Roman V. Kiseliov
> Kursk State University
> Theor. Phys. Dept.
>
Hi Roman,

is it correct to have a factor 1/3 resp. 2/3 in the first sine of the
trig. products in term 1,3 and 4 but not in term2? I ask, because the
outcome would be significantly simpler:

In[10]:=
\[Psi] := -(((Sin[(1/3)*Pi*(t - 5/4)]*Cos[Pi*(t - 5/4)])/(t - 5/4) +
(Sin[(1/3)*Pi*(t + 1/4)]*Cos[Pi*(t + 1/4)])/(t + 1/4))/Pi) +
((Sin[(2/3)*Pi*(t - 7/8)]*Sin[2*Pi*(t - 7/8)])/(t - 7/8) -
(Sin[(2/3)*Pi*(t - 1/8)]*Sin[2*Pi*(t - 1/8)])/(t - 1/8))/Pi
In[11]:=
Timing[(Integrate[#1 /. t -> Cases[Factor[Denominator[#1]],
(c_.)*t + (a_) /; c =!= 0 :> x - a/c, Infinity, 1],
{x, -Infinity, Infinity}] & ) /@ Expand[\[Psi]^2]]
Out[11]=
{28.061999999999998*Second, {1}}

Greetings,
Peter

```

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