Logarithmic Integral and Eulerian Logarithmic Integral Differ by Constant

Theorem

Let $x \in \R$ be a real number such that $x > 2$.

Let $\map \li x$ denote the logarithmic integral of $x$.

Let $\map \Li x$ denote the Eulerian logarithmic integral of $x$.

Then $\map \li x - \map \Li x$ is a constant.

Proof

\(\ds \map \li x - \map \Li x\)

\(=\)

\(\ds \PV_0^x \frac {\d t} {\ln t} - \int_2^x \frac {\d t} {\ln t}\)

Definition of Logarithmic Integral, Definition of Eulerian Logarithmic Integral

\(\ds \)

\(=\)

\(\ds \lim_{\varepsilon \mathop \to 0^+} \paren {\int_\varepsilon^{1 - \varepsilon} \frac {\rd t} {\ln t} + \int_{1 + \varepsilon}^x \frac {\rd t} {\ln t} } - \int_2^x \frac {\d t} {\ln t}\)

Definition of Cauchy Principal Value

\(\ds \)

\(=\)

\(\ds \lim_{\varepsilon \mathop \to 0^+} \paren {\int_\varepsilon^{1 - \varepsilon} \frac {\rd t} {\ln t} + \int_{1 + \varepsilon}^x \frac {\rd t} {\ln t} } - \int_2^x \frac {\d t} {\ln t}\)

\(\ds \)

\(=\)

\(\ds \lim_{\varepsilon \mathop \to 0^+} \paren {\int_\varepsilon^{1 - \varepsilon} \frac {\rd t} {\ln t} + \int_{1 + \varepsilon}^2 \frac {\rd t} {\ln t} } + \int_2^x \frac {\rd t} {\ln t} - \int_2^x \frac {\d t} {\ln t}\)

Sum of Integrals on Adjacent Intervals for Integrable Functions

\(\ds \)

\(=\)

\(\ds \lim_{\varepsilon \mathop \to 0^+} \paren {\int_\varepsilon^{1 - \varepsilon} \frac {\rd t} {\ln t} + \int_{1 + \varepsilon}^2 \frac {\rd t} {\ln t} }\)

This needs considerable tedious hard slog to complete it. In particular: Is it sufficient just to state "which is obviously constant" or would more work be needed to establish what that constant is? To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): logarithmic integral