Re: Calculation of a not so simple integral

• To: mathgroup at smc.vnet.net
• Subject: [mg131225] Re: Calculation of a not so simple integral
• From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>
• Date: Wed, 19 Jun 2013 01:26:29 -0400 (EDT)
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• References: <20130617103132.997F66AF4@smc.vnet.net>

```Hi, Roberto,

You are right.
Now, I tried your integral on my machine, and it gave a definite analytic result:

Integrate[
Sin[x/2]^2/x^2/(x^2 - 4*Pi^2)^2/(x^2 + a^2)^2, {x, -Infinity,
Infinity}, Assumptions -> a > 0]

(3 E + 28 E \[Pi]^2 - 16 (-8 + E) \[Pi]^4 -
64 (-4 + E) \[Pi]^6)/(64 E (\[Pi] + 4 \[Pi]^3)^3)

I do not like it, since it is independent of a.
Also this:

Integrate[
Sin[x/2]^2/x^2/(x^2 - 4*Pi^2)^2/(x^2 + a^2)^2, {x, -Infinity,
Infinity}, Assumptions -> a > 0, PrincipalValue -> True]

-(1/(64 E (\[Pi] +
4 \[Pi]^3)^3))(8 I E^2 \[Pi]^3 (1 +
4 \[Pi]^2) ExpIntegralEi[-1] -
8 \[Pi]^3 (15 \[Pi] + 28 \[Pi]^3 + I ExpIntegralEi[1] +
4 I \[Pi]^2 ExpIntegralEi[1]) +
E (-3 - 28 \[Pi]^2 + 16 \[Pi]^4 + 64 \[Pi]^6 +
16 I \[Pi]^(5/2)
MeijerG[{{0, 0, 1/2}, {}}, {{0}, {}}, -2 I, 1/2] +
64 I \[Pi]^(9/2)
MeijerG[{{0, 0, 1/2}, {}}, {{0}, {}}, -2 I, 1/2]))

Shows no dependence upon a.

Further, this:

tbl = Table[{a,
NIntegrate[
Sin[x/2]^2/x^2/(x^2 - 4*Pi^2)^2/(x^2 + a^2)^2, {x, -Infinity,
Infinity}]}, {a, 0.01, 3, 0.01}];

works and produces a table of values {a, int} that can be evaluated:

ListPlot[tbl]

and exhibits a dependence upon a.

Alexei

Alexei BOULBITCH, Dr., habil.
IEE S.A.
ZAE Weiergewan,
11, rue Edmond Reuter,
L-5326 Contern, LUXEMBOURG

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e-mail: alexei.boulbitch at iee.lu

-----Original Message-----
From: Brambilla Roberto Luigi (RSE) [mailto:Roberto.Brambilla at rse-web.it]
Sent: Monday, June 17, 2013 12:53 PM
To: Alexei Boulbitch; mathgroup at smc.vnet.net
Subject: Re: Calculation of a not so simple integral

The integrand has NO poles on the real axis : in x=+/-(2*Pi) the integrand assume zero value :

Limit[Sin[x/2]^2/(x^2 - 4 Pi^2), x -> 2 Pi]
0

Rob.

-----Messaggio originale-----
Da: Alexei Boulbitch [mailto:Alexei.Boulbitch at iee.lu]
Inviato: luned=EC 17 giugno 2013 12.32
A: mathgroup at smc.vnet.net
Oggetto: Re: Calculation of a not so simple integral

(*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

Hi, Roland,

I checked your expression, it has a pole on the x axis. Check this

Sin[x/2]^2 / x^2 / (x^2-4*Pi^2 )^2 / (x^2 + a^2)^2//StandardForm

Could it be the case, that you have just written it down in a wrong way?

Alexei BOULBITCH, Dr., habil.
IEE S.A.
ZAE Weiergewan,
11, rue Edmond Reuter,
L-5326 Contern, LUXEMBOURG

Office phone :  +352-2454-2566
Office fax:       +352-2454-3566
mobile phone:  +49 151 52 40 66 44

e-mail: alexei.boulbitch at iee.lu

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