Logarithmic Integral as Non-Convergent Series

Theorem
The logarithmic integral can be defined in terms of a non-convergent series.

That is:
 * $\displaystyle \operatorname {li} \left({z}\right) = \sum_{i \mathop = 0}^{+\infty} \frac {i! \, z} {\ln^{i + 1} z}

= \frac z {\ln z} \left({\sum_{i \mathop = 0}^{+\infty} \frac {i!} {\ln^i z} }\right)$

Proof
From the definition of the logarithmic integral:
 * $\displaystyle \operatorname {li} \left({z}\right) = \int_0^z \frac {\mathrm d t}{\ln t}$

Using Integration by Parts:

This sequence can be continued indefinitely.

We will consider the nature of the terms outside and inside the integral, after a number $n$ of iterations of integration by parts.

Let $u_n$ be the term outside the integral.

Let $v_n$ be the term inside the integral.

After $n$ iterations of Integration by Parts as above, we have:


 * $\displaystyle \operatorname {li} \left({z}\right) = u_n + \int_0^z v_n \, \mathrm d t$


 * $\displaystyle u_0 = 0$


 * $\displaystyle v_0 = \frac 1 {\ln t}$

It follows that:

which gives us the recurrence relations:


 * By recurrence on $n$, with the following recurrence hypothesis:


 * When $n = 0$, we have:
 * $\displaystyle v_0 = \frac{1}{\ln t} = \frac{0!}{\ln^{0+1} t}$
 * Which verifies the hypothesis.
 * By supposing true at $n$, we have at $n+1$:


 * So $\left({\rm R.H.}\right)$ is verified at $n+1$ if it is verified at $n$, so it is proved for every $n\in\N$ (since it is true at $n=0$):


 * By taking $\left({\rm 1}\right)$, and inserting $\left({3}\right)$ in, a new expression for $u_{n+1}$ in function of $u_n$ (recursive expression):
 * $\displaystyle u_{n+1} = u_n + \left[{t\cdot\frac{n!}{\ln^{n+1} t}}\right]_0^z$
 * $\displaystyle u_{n+1} = u_n + \frac{z\,n!}{\ln^{n+1} z} - \frac{0\cdot n!}{\ln^{n+1} 0}

= u_n + \frac{z\,n!}{\ln^{n+1} z}$
 * That is, we can write by expanding: