Theorem: Logarithmic Integral as Non-Convergent Series

The logarithmic integral can be defined in terms of a non-convergent series.
That is:
:$\ds \map \li z = \sum_{i \mathop = 0}^{+\infty} \frac {i! \, z} {\ln^{i + 1} z} = \frac z {\ln z} \paren {\sum_{i \mathop = 0}^{+\infty} \frac {i!} {\ln^i z} }$

Proof:
From the definition of the logarithmic integral:
:$\ds \map \li z = \int_0^z \frac {\d t} {\ln t}$
Using Integration by Parts:
{{begin-eqn}}
{{eqn | l = \map \li z
      | r = \intlimits {\frac t {\ln t} } 0 z - \int_0^t t \map \rd {\ln^{-1} t}
}}
{{eqn | r = \frac z {\ln z} + \int_0^z \frac {\d t} {\ln^2 t}
      | c = $\dfrac 0 {\ln 0} = 0$ and Derivative of Function to Power of Function
}}
{{eqn | r = \frac z {\ln z} + \frac z {\ln^2 z} + \int_0^z \frac {2 \rd t} {\ln^3 x}
      | c = Integration by Parts, $\dfrac 0 {\ln^2 0} = 0$, $t \map \rd {ln^{-2} t} = -2 \ln^{-3} t$
}}
{{end-eqn}}
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:
:$\ds \map \li z = u_n + \int_0^z v_n \rd t$
:$u_0 = 0$
:$v_0 = \dfrac 1 {\ln t}$
It follows that:
{{begin-eqn}}
{{eqn | l = \map \li z
      | r = u_n + \bigintlimits {t \, v_n} 0 z - \int_0^z t \map \rd {v_n}
      | c = 
}}
{{eqn | r = u_n + \bigintlimits {t \, v_n} 0 z + \int_0^z -t \map \rd {v_n}
      | c = 
}}
{{end-eqn}}
{{improve|This is actually a proof by induction using different language.}}
which gives us the recurrence relations:
{{begin-eqn}}
{{eqn | n = 1
      | l = u_{n + 1}
      | r = u_n + \bigintlimits {t \, v_n} 0 z
}}
{{eqn | n = 2
      | l = v_{n + 1}
      | r = -t \cdot \map {\frac \d {\d t} } {v_n}
}}
{{end-eqn}}
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 $n + 1$:
{{begin-eqn}}
{{eqn | l = v_{n + 1}
      | r = -t \cdot \map {\frac \d {\d t} } {v_n}
      | c = from $(2)$
}}
{{eqn | r = -t \cdot \map {\frac \d {\d t} } {\frac {n!} {\ln^{n + 1} t} }
      | c = from $(\text{R.H.})$
}}
{{eqn | r = -t \cdot n! \cdot \paren {-\paren {n + 1} } \cdot \frac 1 t \cdot \ln^{-\paren {n + 1} - 1} t
      | c = Derivative of Function to Power of Function, Derivative of Natural Logarithm Function
}}
{{eqn | r = \frac {\paren {n + 1}!} {\ln^{n + 2} t}
      | c = {{Defof|Factorial}}
}}
{{end-eqn}}
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):
{{begin-eqn}}
{{eqn | l = u_{n + 1}
      | r = u_n + \intlimits {t \cdot \frac {n!} {\ln^{n + 1} t} } 0 z
      | c = 
}}
{{eqn | r = u_n + \frac {z \, n!} {\ln^{n + 1} z} - \frac {0 \cdot n!} {\ln^{n + 1} 0}
      | c = 
}}
{{eqn | r = u_n + \frac {z \, n!} {\ln^{n + 1} z}
      | c = 
}}
{{end-eqn}}
That is, we can write by expanding:
:$(4): \quad u_{n + 1} = \ds \sum_{i \mathop = 0}^n \frac {z \, i!} {\ln^{i + 1} z}$
{{qed}}
Category:Logarithmic Integral
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