Prove that the integral $\int_0^1\frac{\log x}{x-1}dx$ exists How do you show that \begin{equation*}\int_{0}^{1}\frac{\log x}{x-1}\mathop{}\!\mathrm{d}x\end{equation*} exists? (without using the $\text{Li}_2$ function etc. - I just want to show existence, not calculate the value).
 A: $\frac{\log x}{x-1}$ can be extended to a continuous function in 1 as $\lim_{x \to 1} \frac{\log x}{x-1} = 1$ so no problem for convergence of integral there.
For $x \to 0^+$ it's easy to show that $\frac{\log x}{x-1} < 2 \log x$ in a right neighbourhood of $0$ and $ 2 \log x$ converges, so the given integral is convergent.
A: With $y:=1-x$ the integral becomes $-\int_0^1\frac{\ln(1-y)}{y}dy=\int_0^1\sum_{n\ge1}\frac{y^{n-1}}{n}dy=\sum_{n\ge1}\frac{1}{n^2}$, which converges by comparison with $\frac1n-\frac{1}{n+1}$. (In particular, the monotone convergence theorem justifies the sum-integral exchange.) I contend that's a proof of convergence that doesn't compute the integral. If you evaluate its value, I can't be held responsible.
A: Note
\begin{align*}
\int_{0}^{1}\frac{\log x}{x-1}dx
 &=  \int_{0}^{1}\frac{\log x} {x^2-1}dx+ \int_{0}^{1}\frac{x\log x}{x^2-1}dx\\
 &
= \frac12 \int_{0}^{1} \left( \frac{\log x}{x-1}- \frac{\log x}{x+1} \right)dx + \frac14\int_{0}^{1}\frac{\log t}{t-1}dt\\
\end{align*}
which leads to 
$$ \int_{0}^{1}\frac{\log x}{x-1}dx=-2 \int_{0}^{1}\frac{\log x}{x+1}dx< -2 \int_{0}^{1}\log x dx=2
$$
Thus, the integral is bounded, hence existing.
