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MathGroup Archive 2006

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Hadamard Finite Part

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71427] Hadamard Finite Part
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Sat, 18 Nov 2006 04:40:58 -0500 (EST)

$VersionNumber
5.2

Consider the following function

f=x^4/(1+Exp[-x]);

The integral of f over {0,Infinity} is divergent

Block[{Message}, Integrate[f, {x, 0, Infinity}]]
Infinity

Here are some attempts to get the finite part of the integral in the
Hadamard sense.

First with the setting GenerateConditions->False

Integrate[f, {x, 0, Infinity}, GenerateConditions -> False]
0

Then using the following setting

List @@ Integrate[f, {x, 0, e}]
(Limit[#1, e -> Infinity] & ) /@ %
DeleteCases[%, _DirectedInfinity][[1]]
N[%]
{e^4*Log[1 + E^e], 4*e^3*PolyLog[2, -E^e], -12*e^2*PolyLog[3, -E^e],
24*e*PolyLog[4, -E^e], -24*PolyLog[5, -E^e], -((45*Zeta[5])/2)}
{Infinity, -Infinity, Infinity, -Infinity, Infinity, -((45*Zeta[5])/2)}
-((45*Zeta[5])/2)
-23.33087449072582

Directing removing the divergent term

Integrate[f - x^4, {x, 0, Infinity}]
N[%]
NIntegrate[f - x^4, {x, 0, Infinity}]
-((45*Zeta[5])/2)
-23.33087449072582
-23.330874489932825

Using the zeta function regularization technique

Integrate[fx, 0, Infinity}, GenerateConditions -> False]
% /. s -> 4
N[%]
(-(-2)^(-s))*(-1 + 2^s)*Gamma[1 + s]*Zeta[1 + s]
-((45*Zeta[5])/2)
-23.33087449072582

So the finite part of the integral is -((45*Zeta[5])/2) and not 0 as
GenerateConditions->True
setting might cheat us.

However using a convergence implying factor Exp[-e x] I got the
following

Integrate[f*Exp[(-e)*x], {x, 0, Infinity}, Assumptions -> e > 0]
(Limit[#1, e -> 0] & ) /@ List @@ Expand[FunctionExpand[%]]
DeleteCases[%, _DirectedInfinity][[1]]
N[%]
(3/4)*(Zeta[5, e/2] - Zeta[5, (1 + e)/2])
{(1/32)*PolyGamma[4, 1/2], Infinity}
(1/32)*PolyGamma[4, 1/2]
-24.108570307083355

Integrate[f*Exp[(-e)*x], {x, 0, Infinity}, Assumptions -> e > 0]
(Limit[#1, e -> 0] & ) /@ List @@ %
DeleteCases[%, _DirectedInfinity][[1]]
N[%]
(3/4)*(Zeta[5, e/2] - Zeta[5, (1 + e)/2])
{3/4, Infinity}
3/4
0.75

What I miss here?

Thanks a lot
Dimitris


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