calculus - Prove $lim_{a to -infty} left(-ln |a| -sum_{n geq 1} frac{a^n}{n(n!)}right)$ equals the Euler-Mascheroni constant

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

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

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