Friday, 15 January 2016

calculus - Prove limatoinftyleft(ln|a|sumngeq1fracann(n!)right) equals the Euler-Mascheroni constant



I want to find a series representation for li(t). Through a substitution and the comment of @user1952009 and @soke I'm managed correct my answer to get :



li(t)=limϵ00+(ϵ0exxdx+lntϵ0exxdx)



about integral logarithm




Even with the correction, everything reduces nicely with the Taylor series except that I need to find:




lima(ln|a|+n1ann(n!))



I know this is equivalently:



=lima(lna+n1an(1)nn(n!))



=lima(ln(1a)+n1an(1)n+1n(n!))




But how can I reduce this to:



lima(halna)=γ




I can confirm that it does indeed converge to γ as wolfram alpha gives me:




ln(1a)+n1an(1)n+1n(n!)=Γ(0,a)+γ





But how could I manipulate my expression to get to there?



(Question 2) is solved @user1952009



I also have another question. I found:



li(t)=γ+lnt+n=1tnn(n!)




Is this somehow resembles what is found in Wikipedia https://en.m.wikipedia.org/wiki/Logarithmic_integral_function (scroll to series representation). Yet is wrong.



I think there is some confusion in my limits of integration:



Start with



li(ex)=limϵ0+1ϵ0dtlnt+ex1+ϵdtlnt



Now the substitution t=ex so dt=exdx:




li(t)=limϵ00+(ϵ0exxdx+tϵ0exxdx)



What am I doing wrong?


Answer



Consider for t>0 : F(t)=texxdx=Ei(t)=Li(et)



for 0<a<b< : F(a)F(b)=baexxdx=ban=0(1)nxn1n!dx=n=0baxn1n!dx

=lnblna+n=1(1)nbnannn!



then use γ=0(11ex1x)exdx=lima0+ln(1ea)F(a)




so that γ=lima0+ln(1ea)+lnalnaF(a)=lima0+ln(1eaa)F(a)lna

=lima0+lnaF(a)



hence
γF(b)=lima0+F(a)+lnaF(b)=lima0+lnbn=1(1)nbnannn!=lnbn=1(1)nbnnn!


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