# 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:

 $\ds \map {\operatorname {li} } z$ $=$ $\ds \intlimits {\frac t {\ln t} } 0 z - \int_0^t t \map \rd {\ln^{-1} t}$ $\ds$ $=$ $\ds \frac z {\ln z} + \int_0^z \frac {\d t} {\ln^2 t}$ $\dfrac 0 {\ln 0} = 0$ and Derivative of Function to Power of Function $\ds$ $=$ $\ds \frac z {\ln z} + \frac z {\ln^2 z} + \int_0^z \frac {2 \rd t} {\ln^3 x}$ Integration by Parts, $\dfrac 0 {\ln^2 0} = 0$, $t \map \rd {ln^{-2} t} = -2 \ln^{-3} t$

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:

 $\ds \map {\operatorname {li} } z$ $=$ $\ds u_n + \bigintlimits {t \, v_n} 0 z - \int_0^z t \map \rd {v_n}$ $\ds$ $=$ $\ds u_n + \bigintlimits {t \, v_n} 0 z + \int_0^z -t \map \rd {v_n}$

which gives us the recurrence relations:

 $\text {(1)}: \quad$ $\ds u_{n + 1}$ $=$ $\ds u_n + \bigintlimits {t \, v_n} 0 z$ $\text {(2)}: \quad$ $\ds v_{n + 1}$ $=$ $\ds -t \cdot \map {\frac \d {\d t} } {v_n}$

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 $n + 1$:

 $\ds v_{n + 1}$ $=$ $\ds -t \cdot \map {\frac \d {\d t} } {v_n}$ from $(2)$ $\ds$ $=$ $\ds -t \cdot \map {\frac \d {\d t} } {\frac {n!} {\ln^{n + 1} t} }$ from $(\text{R.H.})$ $\ds$ $=$ $\ds -t \cdot n! \cdot \paren {-\paren {n + 1} } \cdot \frac 1 t \cdot \ln^{-\paren {n + 1} - 1} t$ Derivative of Function to Power of Function, Derivative of Natural Logarithm Function $\ds$ $=$ $\ds \frac {\paren {n + 1}!} {\ln^{n + 2} t}$ Definition of Factorial

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):

 $\ds u_{n + 1}$ $=$ $\ds u_n + \intlimits {t \cdot \frac {n!} {\ln^{n + 1} t} } 0 z$ $\ds$ $=$ $\ds u_n + \frac {z \, n!} {\ln^{n + 1} z} - \frac {0 \cdot n!} {\ln^{n + 1} 0}$ $\ds$ $=$ $\ds u_n + \frac {z \, n!} {\ln^{n + 1} z}$

That is, we can write by expanding:

$(4) \quad u_{n + 1} = \ds \sum_{i \mathop = 0}^n \frac {z \, i!} {\ln^{i + 1} z}$

$\blacksquare$