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