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

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

That is, we can write by expanding: