Re: Calculation of a not so simple integral

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• Subject: [mg131253] Re: Calculation of a not so simple integral
• From: Roland Franzius <roland.franzius at uos.de>
• Date: Fri, 21 Jun 2013 06:00:50 -0400 (EDT)
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• References: <kph7rp\$2e4\$1@smc.vnet.net>

```Am 15.06.2013 10:15, schrieb Roland Franzius:
> (*Partial Fractions decomposition,  Fourier Integrals
>
> The problem diagnostics:
>
> Calculate
>
> Integrate[ Sin[x/2]^2 / x^2 / (x^2-4*Pi^2 )^2 / (x^2 + a^2)^2 ,
> {x,-oo,oo}, Assumptions-> a>0]
>
> The integrand is nonnegative, has no poles on the real line and decays
> rapidly  ~ x^-10 as x->+-oo
>
> Mathematica  version 6+8 are failing to give a result in a reasonable
> time of calculation.
> Time constraint does not seem to work
>
> The indefinite integral is calculated readily with an obscure complex
> result, the primitive function giving completely useless limits for
> x->0,+-oo
>
> The FourierTransform with Limit k->0 seems to give a correct result
> compared to numerical integration for values of
> a -> 1/20
>
> *)

Remains to be noted that suddenly today in the morning  my Mathematica 8
(local install with license server) seems to have been updated over the
internet with respect to this integral.

While yesterday the result was  MeiersG, when I came back after 2 hours,
suddenly we have amuch faster and slightly better result

In[14]:= ires=Timing[ Integrate[Evaluate[Sin[x/2]^2
f[x,2\[Pi],a]],{x,-\[Infinity],\[Infinity]},Assumptions->0<a<1]]

Out[14]= {64.865, (
3 a^7 + 28 a^5 \[Pi]^2 + 16 a^2 (-7 + 6 a) \[Pi]^4 +
64 (-3 + 2 a) \[Pi]^6)/(64 a^5 \[Pi]^3 (a^2 + 4 \[Pi]^2)^3)}

still wrong, the exponential terms are missing.

But at least for a>1 we get the algebraically correct result.

In[31]:= iresg=Timing[ Integrate[Evaluate[Sin[x/2]^2
f[x,2\[Pi],a]],{x,-\[Infinity],\[Infinity]},Assumptions->1<a<\[Infinity]]]

Out[31]= {63.477, (
E^-a (16 a^2 (7 + a) \[Pi]^4 + 64 (3 + a) \[Pi]^6 +
E^a (3 a^7 + 28 a^5 \[Pi]^2 + 16 a^2 (-7 + 6 a) \[Pi]^4 +
64 (-3 + 2 a) \[Pi]^6)))/(64 a^5 (a^2 \[Pi] + 4 \[Pi]^3)^3)}

With  Assumptions -> a>0, Integrate  produces both results as a
conditional expression. It Contains the difference of two imaginary Log's.

So I conclude there seem to exists problems with the partial fractions
decomposition engine and in the simplification of complex Log's and Atan's.

Of course this is a nontrivial matter because it involves complex
contour integrals which cannot be desribed by the endpoints only.

--

Roland Franzius

```

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