I would to ask for a logarithm integral, used for Gauss. I read that he uses it to calculate the number of primes, less than a given natural number. It is like:
$Li= \int_{0}^x(dt/lnt)$
I read that he uses this to calculate these quantities, but I don't know
which kind of expression results from this integral. Some aproximation?
thanks a lot!
Yes we can approximate it:
$$\text{li} (e^x)=\text{Pv} \int_{0}^{e^x} \frac{1}{\ln t} dt$$
$$\text{li} (e^x)=\lim_{\epsilon \to 0^+} \left(\int_{0}^{1-\epsilon} \frac{dt}{\ln t} +\int_{1+\epsilon}^{e^x} \frac{dt}{\ln t}\right)$$
Now the substitution $t=e^x$ so $dt=e^xdx$. Note I did a mistake before with the bounds as it is a bit hard to see but let's say $e^x=1=t$ then $x=\ln 1$:
$$\text{li} (t)=\lim_{\epsilon_0 \to 0^+} \left(\int_{-\infty}^{-\epsilon_0} \frac{e^x}{x}dx +\int_{\epsilon_0}^{\ln t} \frac{e^x}{x} dx\right)$$
And Taylor series yields:
$$\text{li}(t)=\gamma+\ln |\ln t|+\sum_{n=1}^{\infty} \frac{(\ln t)^n}{n(n!)}$$
Which holds for $t >0$ note $t=e^x>0$.