Evaluate $\int_0^1\frac{x^a-x^{-a}}{x-1}dx$ I have heard that:
$$\int_0^1\frac{x^a-x^{-a}}{x-1}dx=\frac1 a-\pi\cot(\pi a)$$
when $-1<a<1$. How would I prove this?
That doesn't have an elementary indefinite integral, but the definite integral is quite simple.
Someone suggested I use complex analysis to prove it, but I am relatively new to the subject. (I have gotten up to contour integrals, but I'm not sure how to use them to evaluate this particular integral.) I also tried expanding it with a series, but it didn't help.
 A: It could be a way to use the digamma function $\psi$ which verify 
$$ \psi(a+1)=-\gamma+\int_0^1 \frac{1-x^a}{1-x} \mathrm{d}x, \quad |a|<1,$$
since $$\psi(a)-\psi(1-a)=-\pi \cot (\pi a),$$
$$\psi(1+a)-\psi(a)=\frac{1}{a}.$$
But this is not elementary.
I doubt that you have to prove all from the beginning.
A: 
I also tried expanding it with a series, but it didn't help.

It should have helped. If we write
$$\frac{x^{-a}-x^a}{1-x} = (x^{-a}-x^a)\sum_{n=0}^\infty x^n,$$
by the monotone convergence theorem, we have
$$\begin{align}
\int_0^1 \frac{x^{-a}-x^a}{1-x}\,dx
&= \sum_{n=0}^\infty \int_0^1 (x^{-a}-x^a)x^n\,dx\\
&= \sum_{n=0}^\infty \left(\frac{1}{n+1-a} - \frac{1}{n+1+a}\right)\\
&= - \sum_{n=1}^\infty \left(\frac{1}{a-n}+\frac{1}{a+n}\right)\\
&= \frac{1}{a} - \pi \cot (\pi a)
\end{align}$$
by the well-known partial fraction decomposition of the cotangent.
A: HINT: Denote the integral by $ I $. Since $$ x^a-x^{-a}=e^{a\ln(x)}-e^{-a\ln(x)}=2\sinh(a\ln x) .$$ 
Substituting $\ln x= u $, we get
$$ I=\int_{-\infty}^0 \frac{ 2\sinh au}{e^u-1  }e^udu $$
To change the limits of the integral to more convenient one, again substitue $ u=-t$ we have 
\begin{align}
I&=\int_0^{\infty}\frac {2\sinh at}{e^{-t }-1} e^{-t} dt
\\
&=\int_0^{\infty}\frac{2\sinh at}{e^{t/2}(e^{-t/2}-e^{t/2})}dt
\\
&=\int_0^{ \infty }e^{-t/2}\frac {\sinh at}{\sinh (t/2)} dz
\end{align}
Now consider the function $$ f (z)=\frac{e^{(a-\frac{1}{2})z}}{\sinh (z/2)} $$ and integrate over the closed rectangle $C: \pm R, i R$. Don't forget to jump by a small contours on $0, i\pi $.  Clearly, $ f (z) $ is analytic on and inside the contour $ C $. Hence by Cauchy Goursat Thm.  $\int_{c } f (z) dz=0$. . . 
