Logarithmic Integral as Non-Convergent Series

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

That is:
 * $\ds \map {\operatorname {li} } z = \sum_{i \mathop = 0}^{+\infty} \frac {i! \, z} {\ln^{i + 1} z} = \frac z {\ln z} \paren {\sum_{i \mathop = 0}^{+\infty} \frac {i!} {\ln^i z} }$

Proof
From the definition of the logarithmic integral:
 * $\ds \map {\operatorname {li} } z = \int_0^z \frac {\d t} {\ln t}$

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:


 * $\ds \map {\operatorname {li} } z = u_n + \int_0^z v_n \rd t$


 * $u_0 = 0$


 * $v_0 = \dfrac 1 {\ln t}$

It follows that:

which gives us the recurrence relations:

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


 * $\text{R.H.}: \quad v_n = \dfrac {n!} {\ln^{n + 1} t}$

When $n = 0$, we have:


 * $v_0 = \dfrac 1 {\ln t} = \dfrac {0!} {\ln^{0 + 1} t}$

which verifies the hypothesis.

By supposing true at $n$, we have at $n + 1$:

So $(\text{R.H.})$ 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 $n=0$):


 * $(3) \quad v_n = \dfrac {n!} {\ln^{n + 1} t}$

By taking $(1)$, and substituting from $(3)$, a new expression for $u_{n + 1}$ in function of $u_n$ (recursive expression):

That is, we can write by expanding:
 * $(4) \quad u_{n + 1} = \ds \sum_{i \mathop = 0}^n \frac {z \, i!} {\ln^{i + 1} z}$