Definite integral, $\frac 1{\ln(x)}$ What is $$\int_0^{\pi^{2}}\frac 1{\ln(x)}dx$$
I tried using complex residues and some identities, but no luck. Any suggestions?
 A: 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
$$
\mathrm{PV}\int_0^{\pi^2}\frac1{\log(t)}\,\mathrm{d}t=6.10880775185857108177
$$
A: This type of integral is called a logarithmic integral function $\mathrm{li}(x)$. A logarithmic integral function is used in physics and number theory. 
In the case of having $0$ and $\pi^2$ as your bounds, the function would be $\mathrm{li}(\pi^2) - \mathrm{li}(0)$ which simplifies to $\mathrm{li}(\pi^2)$. Calculating it would be tedious and you would definitely not need to learn the long equation from the integral table, so you may just refer to it as $\mathrm{li}(\pi^2)$ instead of figuring out the actual answer. It's acceptable in mathematics.
EDIT: using a logarithmic integral calculator evaluated at an approximation of $\pi^2$, the solution to this integral is approximately $6.019$
EDIT 2: incase you wanted to see the equation, because it is equal to the exponential integral $\mathrm{Ei}(\ln x)$, your integral would be
\begin{equation}
\int_{-\ln \pi^2}^{\infty}\frac{e^{-t}}{t}\mathrm{dt}
\end{equation}
However, because when $t$ is $0$, the $\mathrm{Ei}$ function is undefined at $0$, you must make it the sum of two integrals approaching 0.
\begin{equation}
\underset{h \to 0}{\lim} \int_{-\ln \pi^2}^{-h}\frac{e^{-t}}{t}\mathrm{dt} + \int_{h}^{\infty}\frac{e^{-t}}{t}\mathrm{dt}
\end{equation}
Which in the end equals approximately $6.019$
