Find for which $\alpha$ the integral $\int_{0}^{1} \frac{1-x^{\alpha}}{1-x}dx$ converges.
My Attempt:
suppose $f(x) = \frac{1-x^{\alpha}}{1-x}$.
I think that the integral converges for $\alpha > -1$.
First I've tried to use the linearity of integrals, such that:
$$\int_{0}^{1} \frac{1-x^{\alpha}}{1-x}dx = \int_{0}^{1} \frac{1}{1-x}dx – \int_{0}^{1} \frac{x^{\alpha}}{1-x}dx$$
as $$\int_{0}^{1} \frac{1}{1-x}dx = |_{0}^{1} \ln(1-x) $$
but because the first part of the integral diverges and is not dependent on $\alpha$, then it's not helpful.
the reason why I was foucsed in the point $x=1$ is because for $x=0$, I'll apply the $\int_{0}^{1} \frac{1-x^{\alpha}}{1-x}dx$ limit comparison test with the function $g(x) = \frac{1}{1-x}$, and lim$_{x\to 0} \frac{f(x)}{g(x)} = $ lim$_{x\to 0} 1-x^{\alpha} = 1$, then for every $0<t<1: \ \int_{0}^{t}f(x)$ converges.
I've checked different values of $\alpha$ and I'm wondering if the answer is connected to the fact that given $b>0$, the integral $\int_{0}^{b} \frac{1}{x^\alpha}dx$ converges if and only if $\alpha < 1$.
I suppose that the easiest way to prove for which $\alpha$ the integral converges is by using the limit comparison test for improper integrals, but I can't find a function that will prove/disprove my hypothesis.
Best Answer
Let's check separately convergence at $0$ and $1$, by splitting the integral at $c\in(0,1)$.
If $\alpha\ge0$ there is no question about convergence at $0$. If $\alpha<0$, you can consider $$ \int_0^c \frac{1-x^\alpha}{1-x}\,dx=\int_0^c \frac{1}{1-x}\,dx-\int_0^c \frac{x^\alpha}{1-x}\,dx $$ The first integral is not a problem, so we tackle the second one with the substitution $t=1/x$ to get $$ \int_{1/c}^\infty \frac{t^{\beta-1}}{t-1}\,dt $$ where $\beta=-\alpha>0$. This is asymptotic to $t^{\beta-2}$ and we have convergence if and only if $\beta-2<-1$, hence $\alpha>-1$.
Hence our integral converges at $0$ if and only if $\alpha>-1$.
Now note that $$ \lim_{x\to1}\frac{1-x^\alpha}{1-x}=\alpha $$ so there is no real problem with convergence at $1$.