Proof of the existence of the Ramanujan–Soldner constant

Question

I know that the Ramanujan–Soldner constant is the positive zero of the Logarithmic integral defined as $li\left(x\right) = \displaystyle\int_{0}^{x} \frac{dt}{\log t}$ and that it is equal to: $\mu \approx 1.4513692348\cdots$ but i can't seem to find a proof of its existence or value. Does someone know how to prove this constant exists?

Answer 1

Given that $li(x)$ is continuous on the interval $(1,\infty)$, we could test values of $x$ that are to the left and right of $\mu$. Lets have $x_0=e^{\frac{1}{4}}\approx1.284$. Using the series expansion of the logarithmic integral

$$li(x)=\gamma+\ln(\ln(x))+\sum_{n=1}^\infty \frac{(\ln(x))^n}{n!n}$$ we get $$li(e^{\frac{1}{4}})=\gamma-2\ln2+\sum_{n=1}^\infty \frac{4^{-n}}{n!n}$$

We have to show that this value is less than zero.

We have that $$\gamma-2\ln2+\sum_{n=1}^\infty \frac{4^{-n}}{n!n}<\gamma-2\ln2+e^{\frac{1}{4}}-1\approx-0.525$$ So $li(e^{\frac{1}{4}})<-0.525$. Next lets have $x_1=\sqrt{e}\approx 1.649$. $$li(\sqrt{e})=\gamma-\ln2+\sum_{n=1}^\infty \frac{2^{-n}}{n!n}>\gamma-\ln2+\sinh\left(\frac{1}{2}\right)\approx 0.405$$ Since the logarithmic integral is continuous on this interval and it goes from negative to positive, it is garunteed that there is a zero in the interval $(e^{\frac{1}{4}}, e^{\frac{1}{2}})$ and that $1.284<\mu<1.649$. By choosing $x_0$ and $x_1$ that are closer to each other in the interval and making the right comparisons, we can get better approximations of $\mu$.