Check the convergence (& absolutely) of parametric integral

Question

$$\int\limits_{-1}^{1} \left(\frac{1+x}{1-x}\right)^{\alpha} \ln(2+x)dx$$

Don't know where to start..

Answer 1

Set $$ f(x):=\left(\frac{1+x}{1-x}\right)^{\alpha} \ln(2+x), \quad x \in (-1,1).$$ This function is continuous on $(-1,1)$, let's see what happens near $-1$ and near $1$.

• Near $x=-1$.

$$ f(x) \sim_{-1} \frac{1}{2^{\alpha}}(1+x)^{\alpha+1} $$ where we have used $\ln (1+u) \sim_0 u$, then the integral of $\displaystyle (1+x)^{\alpha+1} $ is convergent near $x=-1$ if and only if $\alpha >-2. $

• Near $x=1$.

$$ f(x) \sim_{1} \frac{2^{\alpha}\ln 3}{(1-x)^{\alpha}} $$ and the integral of $\displaystyle \frac{1}{(1-x)^{\alpha}} $ is convergent near $x=1$ if and only if $\alpha <1. $

Consequently, your integral is convergent for all real numbers such that $ -2<\alpha <1.$