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Re: integrate log*sinc

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  • Subject: [mg109206] Re: integrate log*sinc
  • From: Leonid Shifrin <lshifr at>
  • Date: Sat, 17 Apr 2010 06:03:15 -0400 (EDT)


I don't have a complete rigorous solution, just a couple of observations.

First, you can represent Log[x]/x as a coefficient of the linear term in
<eps> of expansion x^(eps-1) around eps = 0:

In[7]:= Series[x^(eps-1),{eps,0,1}]

Out[7]= 1/x+(Log[x] eps)/x+O[eps]^2

Now, this integral I did not check in all details manually, but it should be
not very hard to get
(by writing Sin[x] = 1/(2I)(Exp[I x] - Exp[-Ix]) and performing an integral
along the relevant half of the imaginary axis for each exponent separately
you should be able to get it)

In[8]:= Integrate[x^(eps -1)*Sin[x],{x,0,Infinity}, Assumptions->

Out[8]= Gamma[eps] Sin[(eps \[Pi])/2]

Expanding this around <eps> = 0 and picking the first - order term, we get
the result:

In[10]:= Series[Gamma[eps] Sin[(eps \[Pi])/2],{eps,0,1}]

Out[10]= \[Pi]/2-1/2 (EulerGamma \[Pi]) eps+O[eps]^2

which is, by the way, twice smaller than you quoted - and direct computation
in Mathematica
does confirm it.

Hope this helps.


On Fri, Apr 16, 2010 at 1:52 PM, pimeja <sed.nivo at> wrote:

> 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
> result.
> Thanks in advance.

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