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
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, $a \in (-1,1)$.
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
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{#c00000}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}
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.
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 $a \in (-1,1)$
$$\large\int_0^1 \frac{x^{-a}-x^a}{1-x}\,\mathrm dx= \frac{1}{a} \, - \, \pi \cot(a\pi)$$
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 $$ H_{a}-H_{-a}=-\pi\cot(\pi a)+\frac1a. $$
