I was solving a few problems regarding convergence and divergence when I ran into this one. I tried searching on the internet but couldn't find an exact match to the problem. The task is to determine whether the following integral diverges or converges.
$$I = \int_{0^+}^{1^-}\frac{\log(x)}{1-x}dx$$
I tried solving it using comparison test but could not find an appropriate function to test it against. No other tests would fit. I also tried integrating the function using by parts but only reached till here (not entirely sure whether it is right or wrong) ($\log(x)$ - first part and $(1-x)$ - second part) $$I=-\log(x)\cdot \log(1-x) + \int \frac{\log(1-x)}{x}dx$$ Putting $y=1-x$ in second part makes it equal to the original integral. So $$I=-\frac12\log(x)\cdot \log(1-x)$$
Here, I face problems putting in the limits.
Please help.
Hint. The function $\frac{\log(x)}{1-x}$ is continuous and negative in $(0,1)$ so we can apply the comparison test.
As $x\to 0^+$ then $$\frac{\log(x)}{1-x}\sim \log(x)$$ Is $\log(x)$ integrable in a right neighbourhood of $0$?
Moreover, as $x\to 1^-$, $$\frac{\log(x)}{1-x}=\frac{\log(1-(1-x))}{1-x}\sim \frac{-(1-x)}{1-x}=-1.$$
What may we conclude?
P.S. The value of such integral is quite famous: see integral representations of $\zeta(2)$.