convergence of $\int _0^1\:\frac{\ln\left(1-x\right)}{\left(1+x\right)^2}dx$

Question

How do I prove convergence of $$\int _0^1\:\frac{\ln\left(1-x\right)}{\left(1+x\right)^2}dx$$ and if it's convergent, calculate the value of the integral?

I noticed that the values that the function under the integral takes will not be positive on the integration domain $(0,1)$, so I cannot use the convergence criterion with limits.

I tried proving absolute convergence but $\left|\ln\left(1-x\right)\right|\rightarrow \infty $ when $x\rightarrow 1$, so I can't use the comparison criterion either. I'm left with Abel's criterion but I've no idea how to use that, or if it will do me any good.

Answer 1

If you want to calculate the value of the integral (in case of convergence) anyway, you could just use the definition of the improper integral and see if the following limit exists: $$\lim_{b \to 1^-} \int_0^b \frac{\ln(1-x)}{(1+x)^2} \,\mbox{d}x$$ If this limit exists, you have not only proven convergence of the improper integral, you also have its value. For the integral, proceed with integration by parts: $$\int \frac{\ln(1-x)}{(1+x)^2} \,\mbox{d}x = -\frac{\ln(1-x)}{1+x}-\int\frac{1}{(1-x)(1+x)}\,\mbox{d}x$$ where the last integral can be done via partial fractions. Can you take it from here?

Answer 2

Notice that $ - \frac{\ln (1-x)}{ 1+x^2} \leq -\ln (1-x)$ and prove the convergence of $$\int _0^1 \ln (1-x)d x$$

Answer 3

$\frac{1}{(1+x)^2}$ is a non-negative and bounded function on the integration range, hence the given integral is convergent since $$ \int_{0}^{1}\log(1-x)\,dx = \int_{0}^{1}\log(x)\,dx = -1. \tag{1}$$ Through the substitution $x=1-e^{-t}$, the original integral equals: $$ \int_{0}^{+\infty}\frac{-te^{-t}}{(2-e^{-t})^2}\,dt=-\sum_{n\geq 0}\int_{0}^{+\infty}\frac{(n+1)t e^{-(n+1)t}}{2^{n+2}}\,dt=-\sum_{n\geq 0}\frac{1}{2^{n+2}(n+1)}\tag{2} $$ i.e. $\color{red}{\large -\frac{\log 2}{2}}$.