TITLE: Definite integral, $\frac 1{\ln(x)}$
QUESTION [7 upvotes]: What is $$\int_0^{\pi^{2}}\frac 1{\ln(x)}dx$$
I tried using complex residues and some identities, but no luck. Any suggestions?

REPLY [9 votes]: Divergent Integral
Using the substitution $x=e^{-t}$, we get
$$
\int_0^{\pi^2}\frac1{\log(x)}\,\mathrm{d}x
=-\int_{-2\log(\pi)}^\infty\frac{e^{-t}}t\,\mathrm{d}t\tag{1}
$$
This is not convergent near $t=0$, that is, near $x=1$.
However, we can compute the Cauchy Principal Value.

Principal Value
First, we look at an interval symmetric about $t=0$ and use the odd part of $e^{-t}$.
$$
\begin{align}
\mathrm{PV}\int_{-2\log(\pi)}^{2\log(\pi)}\frac{e^{-t}}t\,\mathrm{d}t
&=-\int_{-2\log(\pi)}^{2\log(\pi)}\frac{\sinh(t)}t\,\mathrm{d}t\\[6pt]
&=-\int_{-2\log(\pi)}^{2\log(\pi)}\sum_{k=0}^\infty\frac{t^{2k}}{(2k+1)!}\\[6pt]
&=-2\sum_{k=0}^\infty\frac{(2\log(\pi))^{2k+1}}{(2k+1)(2k+1)!}\tag{2}
\end{align}
$$

Power Series
Next, for any $x$, we can compute
$$
\begin{align}
\int_{2\log(\pi)}^x\frac{e^{-t}}{t}\,\mathrm{d}t
&=\int_{2\log(\pi)}^x\sum_{k=0}^\infty\frac{(-1)^kt^{k-1}}{k!}\,\mathrm{d}t\\[6pt]
&=\log\left(\frac{x}{2\log(\pi)}\right)+\sum_{k=1}^\infty\frac{(-x)^k}{k\,k!}-\sum_{k=1}^\infty\frac{(-2\log(\pi))^k}{k\,k!}\tag{3}
\end{align}
$$
We handle the rest with an asymptotic expansion. 

Asymptotic Expansion
Let
$$
f(x)=xe^x\int_x^\infty\frac{e^{-t}}{t}\,\mathrm{d}t\tag{4}
$$
then
$$
f'(x)=\left(1+\frac1x\right)f(x)-1\tag{5}
$$
Using $(5)$ we can compute the asymptotic expansion
$$
f(x)=1-\frac{1!}x+\frac{2!}{x^2}-\frac{3!}{x^3}+\frac{4!}{x^4}-\frac{5!}{x^5}+\dots\tag{6}
$$
Therefore, asymptotically,
$$
\int_x^\infty\frac{e^{-t}}{t}\,\mathrm{d}t\sim\frac{e^{-x}}{x}\left(1-\frac{1!}x+\frac{2!}{x^2}-\frac{3!}{x^3}+\frac{4!}{x^4}-\frac{5!}{x^5}+\dots\right)\tag{7}
$$

Synthesis
We will compute the integral to $20$ places using $(2)$ with $16$ terms, $(3)$ with $x=50$ and $180$ terms, and $(7)$ says the rest is insignificant.
$$
\begin{align}
\mathrm{PV}\int_{-2\log(\pi)}^\infty\frac{e^{-t}}t\,\mathrm{d}t
&=-6.14177296941047095756+0.03296521755189987579\\
&=-6.10880775185857108177
\end{align}
$$
Therefore
$$
\mathrm{PV}\int_0^{\pi^2}\frac1{\log(t)}\,\mathrm{d}t=6.10880775185857108177
$$