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)=limϵ0→0+(∫−ϵ0−∞exxdx+∫lntϵ0exxdx)
Even with the correction, everything reduces nicely with the Taylor series except that I need to find:
−lima→−∞(ln|a|+∑n≥1ann(n!))
I know this is equivalently:
=−lima→∞(lna+∑n≥1an(−1)nn(n!))
=lima→∞(ln(1a)+∑n≥1an(−1)n+1n(n!))
But how can I reduce this to:
lima→∞(ha−lna)=γ
I can confirm that it does indeed converge to γ as wolfram alpha gives me:
ln(1a)+∑n≥1an(−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)=limϵ→0+∫1−ϵ0dtlnt+∫ex1+ϵdtlnt
Now the substitution t=ex so dt=exdx:
li(t)=limϵ0→0+(∫−ϵ0−∞exxdx+∫tϵ0exxdx)
What am I doing wrong?
Answer
Consider for t>0 : F(t)=∫∞te−xxdx=−Ei(−t)=−Li(e−t)
for 0<a<b<∞ : F(a)−F(b)=∫bae−xxdx=∫ba∞∑n=0(−1)nxn−1n!dx=∞∑n=0∫baxn−1n!dx
then use γ=∫∞0(11−e−x−1x)e−xdx=lima→0+−ln(1−e−a)−F(a)
so that γ=lima→0+−ln(1−e−a)+lna−lna−F(a)=lima→0+−ln(1−e−aa)−F(a)−lna
hence
−γ−F(b)=lima→0+F(a)+lna−F(b)=lima→0+lnb−∞∑n=1(−1)nbn−annn!=lnb−∞∑n=1(−1)nbnnn!
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