# What is the integral of 1/ln(lnx)?

Jun 29, 2015

This is an impossible integral to complete without resorting to calculators and evaluating at explicit bounds.

http://www.wolframalpha.com/input/?i=integral+of+1%2F%28ln%28lnx%29%29

Let's see how far we can go, though...

$\int {\left(\ln \left(\ln x\right)\right)}^{- 1} \mathrm{dx}$

Let:
$u = \ln x$
$\mathrm{du} = \frac{1}{x} \mathrm{dx}$
$x \mathrm{du} = \mathrm{dx} \to {e}^{u} \mathrm{du} = \mathrm{dx}$

$= \int {\left(\ln u\right)}^{- 1} {e}^{u} \mathrm{du}$

$= \int {e}^{u} / \left(\ln u\right) \mathrm{du}$

$s t - \int t \mathrm{ds}$

Let:
$s = {e}^{u}$
$\mathrm{ds} = {e}^{u} \mathrm{du}$
$\mathrm{dt} = \frac{1}{\ln} u \mathrm{du}$
t = ?

Detour. We need to know the integral for $\frac{1}{\ln} x$. Let:
$q = \frac{1}{\ln} u$
$\mathrm{dq} = - \ln \frac{u}{u} \mathrm{du}$
$\mathrm{dr} = \mathrm{du}$
$r = u$

$q r - \int r \mathrm{dq}$

$= \frac{u}{\ln} u + \int \ln u \mathrm{du}$

$= \frac{u}{\ln} u + u \ln | u | - u = t$

Back to $s$ and $t$:

$= {e}^{u} \left(\frac{u}{\ln} u + u \ln | u | - u\right) - \int {e}^{u} \left(\frac{u}{\ln} u + u \ln u - u\right) \mathrm{du}$

$= \frac{u {e}^{u}}{\ln} u + u {e}^{u} \ln | u | - u {e}^{u} - \int \frac{u {e}^{u}}{\ln} u \mathrm{du} - \int u {e}^{u} \ln u \mathrm{du} + \int u {e}^{u} \mathrm{du}$

The only integral we can do with real functions or standard functions is $\int u {e}^{u} \mathrm{du}$. I'm running out of variables.

$o p - \int p \mathrm{do}$

Let:
$o = u$
$\mathrm{do} = \mathrm{du}$
$\mathrm{dp} = {e}^{u} \mathrm{du}$
$p = {e}^{u}$

$u {e}^{u} - \int {e}^{u} \mathrm{du}$

$= u {e}^{u} - {e}^{u}$

So now we get:

$= \frac{u {e}^{u}}{\ln} u + u {e}^{u} \ln | u | \cancel{- u {e}^{u} + u {e}^{u}} - {e}^{u} - \int \frac{u {e}^{u}}{\ln} u \mathrm{du} - \int u {e}^{u} \ln u \mathrm{du}$

$= \frac{u {e}^{u}}{\ln} u + u {e}^{u} \ln | u | - {e}^{u} - \int \frac{u {e}^{u}}{\ln} u \mathrm{du} - \int u {e}^{u} \ln u \mathrm{du}$

$= {e}^{u} \left[\frac{u}{\ln} u + u \ln | u | - 1\right] - \int \frac{u {e}^{u}}{\ln} u \mathrm{du} - \int u {e}^{u} \ln u \mathrm{du}$

$= {e}^{\ln x} \left[\frac{\ln x}{\ln} \left(\ln x\right) + \left(\ln x\right) \ln | \ln x | - 1\right] - \int \frac{\left(\ln x\right) {e}^{\ln x}}{x \ln \left(\ln x\right)} \mathrm{dx} - \int \frac{1}{x} \left(\ln x\right) {e}^{\ln x} \ln \left(\ln x\right) \mathrm{dx}$

$= \textcolor{b l u e}{x \left[\frac{\ln x}{\ln} \left(\ln x\right) + \left(\ln x\right) \ln | \ln x | - 1\right] - \int \frac{\ln x}{\ln \left(\ln x\right)} \mathrm{dx} - \int \left(\ln x\right) \ln \left(\ln x\right) \mathrm{dx}}$