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)=lim



about integral logarithm




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




-\lim_{a \to -\infty} \left(\ln |a| +\sum_{n \geq 1} \frac{a^n}{n(n!)}\right)



I know this is equivalently:



=-\lim_{a \to \infty} \left(\ln a +\sum_{n \geq 1} \frac{a^n(-1)^n}{n(n!)}\right)



=\lim_{a \to \infty} \left(\ln (\frac{1}{a}) +\sum_{n \geq 1} \frac{a^n(-1)^{n+1}}{n(n!)}\right)




But how can I reduce this to:



\lim_{a \to \infty} (h_a-\ln a)=\gamma




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




\ln (\frac{1}{a}) +\sum_{n \geq 1} \frac{a^n(-1)^{n+1}}{n(n!)}=\Gamma (0,a)+\gamma





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



(Question 2) is solved @user1952009



I also have another question. I found:



\text{li}(t)=\gamma+\ln t+\sum_{n=1}^{\infty} \frac{t^n}{n(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



\text{li} (e^x)=\lim_{\epsilon \to 0^+} \int_{0}^{1-\epsilon} \frac{dt}{\ln t} +\int_{1+\epsilon}^{e^x} \frac{dt}{\ln t}



Now the substitution t=e^x so dt=e^xdx:




\text{li} (t)=\lim_{\epsilon_0 \to 0^+} (\int_{-\infty}^{-\epsilon_0} \frac{e^x}{x}dx +\int_{\epsilon_0}^{t} \frac{e^x}{x} dx)



What am I doing wrong?


Answer



Consider for t > 0 : F(t) = \int_{t}^\infty \frac{e^{-x}}{x} dx = -Ei(-t)= -Li(e^{-t})



for 0 < a < b < \infty : F(a)-F(b) = \int_a^b \frac{e^{-x}}{x} dx = \int_a^b \sum_{n=0}^\infty (-1)^n\frac{x^{n-1}}{n!} dx = \sum_{n=0}^\infty \int_a^b \frac{x^{n-1}}{n!} dx = \ln b - \ln a + \sum_{n=1}^\infty (-1)^n\frac{b^n - a^n }{n \, n!}



then use \gamma = \int_0^\infty (\frac{1}{1-e^{-x}}- \frac{1}{x}) e^{-x} dx = \lim_{a \to 0^+} -\ln(1-e^{-a}) - F(a)




so that \gamma = \lim_{a \to 0^+} - \ln(1-e^{-a})+\ln a - \ln a- F(a) = \lim_{a \to 0^+} -\ln(\frac{1-e^{-a}}{a})-F(a) - \ln a = \lim_{a \to 0^+} -\ln a-F(a)



hence
-\gamma- F(b) = \lim_{a \to 0^+} F(a)+ \ln a - F(b) =\lim_{a \to 0^+} \ln b - \sum_{n=1}^\infty (-1)^n\frac{b^n - a^n }{n \, n!} = \ln b - \sum_{n=1}^\infty (-1)^n \frac{b^n }{n \, n!}


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