Improper integral convergence $I = \int_{0^+}^{1^-}\frac{\log(x)}{1-x}dx$

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.

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 $$\zeta(2)$$.

• Any further doubt? – Robert Z Jan 10 at 16:56
• $\log(x)$ is integrable in the neighbourhood of $0$. This means that the function is finite at left end. Also, the right end limit exists and is finite. This means that the function is finite everywhere meaning the area of the function is also finite. That implies that the function converges. Right ? – Sauhard Sharma Jan 10 at 17:36
• No, the function is NOT finite at left end (it goes to $-\infty$), but its integral is finite. Hence the integral is convergent. – Robert Z Jan 10 at 17:49
• I understand now. Thanks so much !!! – Sauhard Sharma Jan 10 at 18:14