Find for which $\alpha$ the integral $\int_{0}^{1} \frac{1-x^{\alpha}}{1-x}dx$ converges 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.
 A: For any $\alpha$ , we have $\lim_{x \to 1^-} \frac{1-x^\alpha}{1-x}= \alpha$ (derivative of $x^\alpha$ type difference quotient). Hence we always have boundedness near $1$ for any $\alpha$. It is then $x=0$ which is likely to create a problem, if any.

We will now observe the function near $x=0$ and make conclusions. Note that if $\alpha \geq 0$ then the limit exists at $0$ anyway by the quotient rule.
In fact, we have for $\alpha < 0$ :
$$
\lim_{x \to 0^+} \frac{x^{-\alpha}(1-x^{\alpha})}{(1-x)} = \lim_{x \to 0^+} \frac{x^{-\alpha}-1}{1-x} = 1 \tag{*}
$$
So near $0$, $\frac{1-x^{\alpha}}{1-x}$ looks like $ x^{\alpha}$, and we know that if $\alpha \ge - 1$ that integral is not converging, while for $\alpha < -1$ it is.

Now the answer. Get rid of $\alpha \ge 0$, for which continuity holds on the interval, hence boundedness and integral convergence.
For $\alpha < 0$, we first do:
$$
\int_{0}^1 \frac{1-x^{\alpha}}{1-x} dx = \int_{0}^\delta \frac{1-x^{\alpha}}{1-x} dx + \int_{\delta} ^1 \frac{1-x^{\alpha}}{1-x} dx
$$
Provided LHS and RHS exist (for any $0<\delta < 1$). It is clear that the second integral on the RHS exists  (for all $\alpha$)due to the continuity of the integrand on the interval. It follows that the existence of the LHS integral is down to the existence of $\int_{0}^{\delta} \frac{1-x^{\alpha}}{1-x}dx$.
Now, take say $\epsilon = 0.1$. Then, there exists a $\delta > 0$ such that $0.9 < \frac{x^{-\alpha}(1-x^{\alpha})}{1-x}<1.1 $, in other words,
$$
0.9x^{\alpha} < \frac{1-x^{-\alpha}}{1-x} < 1.1x^{\alpha}
$$
for $0<x<\delta$.
If $\alpha <-1$ then the desired integral is bounded between $0.9\int_{0}^\delta x^{\alpha}dx$ and $1.1\int_{0}^\delta x^{\alpha}dx$. Otherwise the integral is unbounded, as it dominates an unbounded integral.
Hence we complete the question.

The technique to keep in mind for this problem, is that of finding asymptotically, the rate of decay of the function near the points of explosion. If the decay is not fast enough, then the integral will not converge,and vice versa, so we can use results about the decay rate to conclude.
A: 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 $1$.
A: $$\int_0^1\dfrac{1-x^{\alpha}}{1-x}dx=\int_0^1(1-x^{\alpha})\sum_{n=0}^{\infty}x^{n}dx\\=\int_0^1\ \sum_{n=0}^{+\infty}(x^n-x^{n+\alpha})dx\\=\sum_{n=0}^{+\infty} \int_0^1(x^n-x^{n+\alpha})dx\\=\sum_{n=0}^{+\infty}\left( \dfrac{1}{n+1} -  \dfrac{1}{n+\alpha+1} \right)\\=\sum_{n=0}^{+\infty}\dfrac{\alpha}{(n+1)(n+\alpha+1)} $$
N.B:  $$\int_0^1\dfrac{1-x^{\alpha}}{1-x}dx=\psi(\alpha+1)+\gamma,$$ with $\psi$ is the digamma function and $\gamma$ is the Euler constant.
For more information you can see here https://www.youtube.com/watch?v=SNUbR8lXD4M
