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. Would you please suggest something? > > 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 inte= gral > along the relevant half of the imaginary axis for each exponent separatel= y > you should be able to get it) > > In[8]:= Integrate[x^(eps -1)*Sin[x],{x,0,Infinity}, Assumptions-> > {0<=eps<1}] > > 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 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. > > > Thanks in advance.

**Follow-Ups**:**Re: integrate log*sinc***From:*Leonid Shifrin <lshifr@gmail.com>