       Re: integrate log*sinc

• To: mathgroup at smc.vnet.net
• Subject: [mg109234] Re: integrate log*sinc
• From: pimeja <sed.nivo at gmail.com>
• Date: Sun, 18 Apr 2010 05:58:42 -0400 (EDT)
• References: <hqc0vk\$fte\$1@smc.vnet.net>

```Leonid,

On Apr 17, 1:02 pm, Leonid Shifrin <lsh... at gmail.com> wrote:
> Hi,
>
> 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:

Thanks, this something more advanced than integration by parts :)
Only issue I suffer with this is that I hoped to extend the approach
to more interesting case:
Integrate[
Log[Abs[x - a]]*Sin[x - b]/(x - b), {x, -Infinity, Infinity}]
This integral appears in solving boundary problems in semi-infinite
domain.

>
> In:= Series[x^(eps-1),{eps,0,1}]
>
> Out= 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 inte=
gral
> along the relevant half of the imaginary axis for each exponent separatel=
y
> you should be able to get it)
>
> In:= Integrate[x^(eps -1)*Sin[x],{x,0,Infinity}, Assumptions->
> {0<=eps<1}]
>
> Out= Gamma[eps] Sin[(eps \[Pi])/2]
>
> Expanding this around <eps> = 0 and picking the first - order term, we =
get
> the result:
>
> In:= Series[Gamma[eps] Sin[(eps \[Pi])/2],{eps,0,1}]
>
> Out= \[Pi]/2-1/2 (EulerGamma \[Pi]) eps+O[eps]^2
>
> which is, by the way, twice smaller than you quoted - and direct computat=
ion
> in Mathematica
> does confirm it.
>
> Hope this helps.
>
> Regards,
> Leonid
>
>
>
> On Fri, Apr 16, 2010 at 1:52 PM, pimeja <sed.n... at gmail.com> 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.
>