Logarithmic Integral as Non-Convergent Series

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

That is:

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


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:

\(\displaystyle \operatorname {li} \left({z}\right)\) \(=\) \(\displaystyle \left[{\frac t {\ln t} }\right]_0^z - \int_0^t t \, \mathrm d (\ln^{-1} t)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac z {\ln z} + \int_0^z \frac {\mathrm d t} {\ln^2 t}\) $\quad$ $\dfrac 0 {\ln 0} = 0$ and Derivative of Function to Power of Function $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac z {\ln z} + \frac z {\ln^2 z} + \int_0^z \frac {2 \, \mathrm d t} {\ln^3 x}\) $\quad$ Integration by Parts, $\dfrac 0 {\ln^2 0} = 0$, $t \, \mathrm d \left({ln^{-2} t}\right) = -2 \,\ln^{-3} t$ $\quad$

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:

\(\displaystyle \operatorname {li} \left({z}\right)\) \(=\) \(\displaystyle u_n + \left[{t \, v_n}\right]_0^z - \int_0^z t \, \mathrm d \left({v_n}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle u_n + \left[{t \, v_n}\right]_0^z + \int_0^z -t \, \mathrm d \left({v_n}\right)\) $\quad$ $\quad$

which gives us the recurrence relations:

\((1):\quad\) \(\displaystyle u_{n + 1}\) \(=\) \(\displaystyle u_n + \left[{t \, v_n}\right]_0^z\) $\quad$ $\quad$
\((2):\quad\) \(\displaystyle v_{n + 1}\) \(=\) \(\displaystyle -t \cdot \frac {\mathrm d} {\mathrm d t} \left({v_n}\right)\) $\quad$ $\quad$
  • By recurrence on $n$, with the following recurrence hypothesis:
\((\text{R.H.}):\quad\) \(\displaystyle v_n\) \(=\) \(\displaystyle \frac {n!} {\ln^{n + 1} t}\) $\quad$ $\quad$

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 $n+1$:

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

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 $n=0$):

\((3):\quad\) \(\displaystyle v_n\) \(=\) \(\displaystyle \frac {n!} {\ln^{n+1} t}\) $\quad$ $\quad$

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

\(\displaystyle u_{n + 1}\) \(=\) \(\displaystyle u_n + \left[{t \cdot \frac {n!} {\ln^{n + 1} t} }\right]_0^z\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle u_n + \frac {z \, n!} {\ln^{n + 1} z} - \frac {0 \cdot n!} {\ln^{n + 1} 0}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle u_n + \frac {z \, n!} {\ln^{n + 1} z}\) $\quad$ $\quad$

That is, we can write by expanding:

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