Re: integrate log*sinc

*To*: mathgroup at smc.vnet.net*Subject*: [mg109206] Re: integrate log*sinc*From*: Leonid Shifrin <lshifr at gmail.com>*Date*: Sat, 17 Apr 2010 06:03:15 -0400 (EDT)

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: 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-> {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 computation in Mathematica does confirm it. Hope this helps. Regards, Leonid On Fri, Apr 16, 2010 at 1:52 PM, pimeja <sed.nivo 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. > >