improper integral using complex analysis
            
            
          
             
     


             improper intergral using complex analysis 



            Hi,
            

            concerning second order poles on the real axis and my integral from 
            o to inf

            sin²(alnx)/(x-1)²?
            In a bib I consult a book on complex variables by F Bayen and C 
            Margaria.They
            give a solution for a second order pole on the real axis.

            lim r->0{int from 0 to 1-r x.expa/(x-1)²dx+int from 1+r to inf 
            xexp.a /(x-1)²dx -2/r}= -pi.a.cotg pi.a (I)

            We use the keyhole contour with the branch 0<phi<2.pi
            and with an identitation on the upper cut an lower cut.

            W integrate around the keyhole contour the function 
            f(z)=z.expa/(x-1)²

            and becomes:

            (1 - exp.2ia.pi)int from o to 1-r xexpa/(x-1)² +int from 1+r to inf 
            xexp.a/(x-1)²

            +int from pi to 0 f(1+r exp.i.phi)irexp i.phi d phi+int from 0 to - 
            pi f(1+rexp.iphi)irexp i.phi=0

            by cauchys theorem.
            r=radius of half circles at 1
            phi=argument z

            i

            Now we expand z.expa/(z-1)² in a laurent serie.

            f(z)=1/(x-1)²+a/(x-1)+g(z) (analytic) function

            integrate on the upper half circle

            we have 2/r + pi.i.a +0 (cauchy theorem-analytic function g(z)

            integrate on the lower cut

            we have exp 2.i.a.pi (-2/r+i.a.pi+o) 

            In the limit r->0 we get the standard expression (I) above

            Now the problem is the ter 2/r in (I)

            Now problem for the calculation of an integral as in my example they 
            insert

            a convergence factor:let us take without loss of generality x.expb 
            and then
            in teresult we take the lim b->0.

            for my example -1/4(xexp 2ia -2 +xexp-2ia)

            remember cotg(ix)=-i coth(x)

            cotg(-x)=-cotg(x)

            We use now expression (I) with a=2ia and a=-2ia

            then we have:

            -1/'4pi{-2ia+b)cotg(2iab)+2bcotg b-(-2ia+b)cotg(-2ia+b)}in this 
            expr. you have -2/r+4/r-2/r see formule (I) the singular terms 
            cancel in the limit

            take lim b->0

            =-1/4 pi{-4acoth(2pi.a) +2}

            =pi.a coth (2.pi.a) - 1/2 what the desired result is.

            Ik know its a bit difficult to write the mathematical expressions.

            How can I type math expressions:software or math keyboard.

            I think the method can be used for higher order poles on the real 
            axis.