Evaluating $\int_0^1\frac{x^{-a}-x^{a}}{1-x}\,\mathrm dx$ How to evaluate following integral

$$\int_0^1\frac{x^{-a}-x^{a}}{1-x}\,\mathrm dx$$

I tried the feynman way 
$$\begin{align} 
I'(a)&=\int_0^1\frac{-x^{-a}\ln x-x^{a}\ln x}{1-x}\,\mathrm dx\\
&= \int_0^1\ln x\left(\frac{-x^{-a}-x^{a}}{1-x}\right)\,\mathrm dx\\
\end{align}$$
I don't have any idea about how to proceed!
Some help is appreciated
 A: Write
$$x^a-x^{-a} = \log{x} \int_{-a}^a dy \, x^y $$
Reverse the order of integration to get
$$-\int_{-a}^a dy \, \int_0^1 dx \frac{x^y \log{x}}{1-x} $$
Now,
$$\int_0^1 dx \frac{x^y \log{x}}{1-x} = \sum_{k=0}^{\infty} \int_0^1 dx \, x^{y+k} \log{x} = -\sum_{k=0}^{\infty} \frac1{(y+k+1)^2} $$
So we now have, as the integral,
$$\sum_{k=1}^{\infty} \int_{-a}^a \frac{dy}{(y+k)^2} = 2 a \sum_{k=1}^{\infty} \frac1{k^2-a^2} = \frac1{a} \left (1-\pi a \cot{\pi a}  \right ) $$
This last sum is a well-known result which may be proven in any number of ways, for example by the residue theorem.
Note that, for convergence, $a \in (-1,1)$.
A: If integral be on the interval $[0,\infty)$ we have formula 
$$\int_0^\infty\frac{x^{a-1}-x^{b-1}}{1-x}dx=\pi (cot(\pi a)-cot(\pi b))$$ for $0<a,b<1$ . because
$$\int_0^\infty\frac{x^{a-1}-x^{b-1}}{1-x}dx=\int_0^1\frac{x^{a-1}-x^{b-1}}{1-x}dx+\int_1^\infty\frac{x^{a-1}-x^{b-1}}{1-x}dx$$ and $$\int_1^\infty\frac{x^{a-1}-x^{b-1}}{1-x}dx=\int_0^1\frac{t^{-b}-t^{-a}}{1-t}dt$$ by substituation $x=\frac{1}{t}$
now since $$\sum_{n=0,1,2,..} (t^{n+a-1}-t^{n-a})=\frac{t^{a-1}-t^{-a}}{1-t}$$
$$\sum_{n=0,1,2,..} (t^{n+b-1}-t^{n-b})=\frac{t^{b-1}-t^{-b}}{1-t}$$
reorder integral and infinite sum and use the 
$$\pi cot(\pi a)=\frac{1}{a}+\sum_{n\in \mathbb{N}}(\frac{1}{n+a}-\frac{1}{n-a})$$
but for your question similarly we have $$\int_0^1\frac{x^{-a}-x^{a}}{1-x}dx=-(\frac{1}{-a}+\frac{1}{a+1} )-\sum_{n\in \mathbb{N}}(\frac{1}{n+(a+1)}-\frac{1}{n-(a-1)})$$ now use $$\pi cot(\pi a)=\frac{1}{a}+\sum_{n\in \mathbb{N}}(\frac{1}{n+a}-\frac{1}{n-a})$$ which can be proved by fourier series. for example use fourier serie of $f(x)=cos(mx)$ in interval $[0,\pi]$
also you could use $$\pi cot(\pi x)= \sum_{n\in\mathbb{Z}}\frac{1}{n+x}$$
which can be proved by fourier series or Residue theory in complex function 
A: Notice that $x^b-x^a=\Big(1-x^a\Big)-\Big(1-x^b\Big)$. Thus, our integral becomes $H_a-H_b$, where $H_n$ 
is the $n^{th}$ generalized harmonic number. At the same time, it can be shown, using telescoping  series, that $H_n=\displaystyle\sum_{k=1}^\infty\frac n{k~\big(k+n\big)}.~$ Putting the two together, and taking into consideration the 
fact that $b=-a$, we have $I=2a~\displaystyle\sum_{k=1}^\infty\frac1{k^2-a^2}$, which can ultimately be proven to equal $\dfrac1a-$ 
$-~\pi\cot\big(a\pi\big)$ by differentiating the natural logarithm of Euler's infinite product expression for  the sine function.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large\int_{0}^{1}{x^{-a} - x^{a} \over 1 - x}\,\dd x}
=\int_{0}^{1}{1 - x^{a} \over 1 - x}\,\dd x
-\int_{0}^{1}{1 - x^{-a} \over 1 - x}\,\dd x
=\Psi\pars{a + 1} - \Psi\pars{-a + 1}
\\[5mm]&=\bracks{\Psi\pars{a} + {1 \over a}} - \Psi\pars{1 - a}
={1 \over a} - \bracks{\Psi\pars{1 - a} - \Psi\pars{a}}
\\[5mm]&=\color{#66f}{\large{1 \over a} - \pi\cot\pars{\pi a}}\,,\qquad
\verts{\Re\pars{a}}\ <\ 1
\end{align}
A: We can do it without using The Feynman Way
$$\begin{align}
\int_0^1 \frac{x^{-a}-x^a}{1-x}\,\mathrm dx
&=\int_0^1 \left[\frac{x^{-a}}{1-x}-\frac{-x^a}{1-x}\right]\,\mathrm dx\tag{1}\\
&=\int_0^1\sum_{k=1}^\infty\left(x^{k-a-1}-x^{k+a-1}\right)\,\mathrm dx\tag{2}\\
&=\sum_{k=1}^\infty\int_0^1\left(x^{k-a-1}-x^{k+a-1}\right)\,\mathrm dx\tag{3}\\
&= \sum_{n=1}^\infty \left(\frac{1}{k-a} - \frac{1}{k+a}\right)\tag{4}\\
&= \sum_{n=1}^\infty \left(\frac{k+a-k+a}{(k-a)(k+a)}\right)\tag{5}\\
&= \sum_{n=1}^\infty \left(\frac{2a}{k^2-a^2}\right)\tag{6}\\
&= 2a\sum_{n=1}^\infty\frac{1}{k^2-a^2}\tag{7}\\
&= 2a\left[\frac{1}{2a} \left( \frac{1}{a} \, - \, \pi \cot(a\pi) \right)\right]\tag{8}\\
&= \frac{1}{a} \, - \, \pi \cot(a\pi)\tag{9}\\
\end{align}$$
Note that for convergence $a \in (-1,1)$

$$\large\int_0^1 \frac{x^{-a}-x^a}{1-x}\,\mathrm dx= \frac{1}{a} \, - \, \pi \cot(a\pi)$$

A: Here is another solution by use the Harmonic Number. Note
$$ H_a=\int_0^1\frac{1-x^a}{1-x}dx $$
and hence
$$ \int_0^1\frac{x^{-a}-x^a}{1-x}dx=H_a-H_{-a}. $$
From this, we have
$$ H_{1-a}-H_a=\pi\cot(\pi a)-\frac1a+\frac{1}{1-a}, H_a=H_{a-1}+\frac{1}{a}. $$
Replacing $-a$ by $a$ in the first identity gives
$$ H_{1+a}-H_{-a}=-\pi\cot(\pi a)+\frac1a+\frac{1}{1+a}. $$
Replacing $a$ by $1+a$ in the second identity gives
$$H_{1+a}=H_{a}+\frac{1}{1+a}. $$
Thus we obtain
$$ H_{a}-H_{-a}=-\pi\cot(\pi a)+\frac1a. $$
