Re: Troubles with Integrate

• To: mathgroup at smc.vnet.net
• Subject: [mg70253] Re: Troubles with Integrate
• From: ab_def at prontomail.com
• Date: Tue, 10 Oct 2006 06:12:00 -0400 (EDT)
• References: <eg84hk\$nir\$1@smc.vnet.net>

```Roman V. Kiseliov wrote:
> 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.

Integrate term by term and simplify the integrands:

In[2]:= # /. s : a_*(k_.*t + t0_)^p_ :>
Integrate[s /. t :> t - t0/k, {t, -Infinity, Infinity}]& /@
Expand[psi]

Out[2]= -1/2

In[3]:= (ad = # /. s : a_*(k_.*t + t0_)^p_ :>
Integrate[s /. t :> t - t0/k, t]& /@
Expand[psi^2]) //
FreeQ[#, Integrate]&

Out[3]= True

In[4]:= Limit[ad, t -> Infinity] - Limit[ad, t -> -Infinity] // Expand

Out[4]= 4/3 - 8*Sqrt[2]/(9*Pi)

Another way is to use the Fourier transform:

In[5]:= Psi = FourierTransform[Expand[psi], t, p]

(Without Expand we get some unwelcome Sign'[...] terms in the result.)

In[6]:= Sqrt[2*Pi]*Psi /. p -> 0

Out[6]= -1/2

By the Parseval's identity, the integral of Abs[psi]^2 is the same as
the integral of Abs[Psi]^2 (because the scalar product is conserved):

In[7]:= Integrate[Abs[Psi]^2, {p, -Infinity, Infinity}] // Expand

Out[7]= 4/3 - 8*Sqrt[2]/(9*Pi)

Integrating the image is easier because Psi has a finite support.

Maxim Rytin
m.r at inbox.ru

```

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