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Re: integrate log*sinc

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  • Subject: [mg109205] Re: integrate log*sinc
  • From: "David Park" <djmpark at>
  • Date: Sat, 17 Apr 2010 06:03:04 -0400 (EDT)

Well, I'm treading on the edge of my knowledge, but what about doing a
series expansion?

First, evaluate the indefinite integral and then evaluate the result at the

intg0[x_] = Integrate[Log[x] Sin[x]/x, x]; 
Limit[%, x -> \[Infinity]] - Limit[%, x -> 0] 

Then do a series expansion of the integral that Mathematica returned.

series1 = 
 Series[intg0[x], {x, 0, 7}, Assumptions -> x > 0] // Normal // Expand 

Then do a series expansion of the initial integrand and integrate term by

Series[Log[x] Sin[x]/x, {x, 0, 7}] // Normal 
series2 = Integrate[#, x] & /@ % // Expand 

The two series match term for term (at least as far as we have gone) so they
must represent the same function.

David Park
djmpark at  

From: pimeja [mailto:sed.nivo at] 

Hi All,

For Integrate[Log[x] Sin[x]/x, {x, 0, \[Infinity]}] Mathematica
returns -EulerGamma \[Pi].
How to proof this analytical?

I've tried to use residue theory but it seems not suitable since
integrand has pool of second order in zero (for Jordan lema should be
first order only). Substitution x=Exp[y] returns even more strange

Thanks in advance.

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