How to prove that $\int_0^1 \frac{x^{1-p}(1-x)^pdx}{x^2+1}=\frac{\pi}{\sin px}(2^{\frac{p}{2}}\cos {\frac{p\pi}{4}}-1)$

Question

$$\int_0^1 \frac{x^{1-p}(1-x)^pdx}{x^2+1}=\frac{\pi}{\sin px}(2^{\frac{p}{2}}\cos {\frac{p\pi}{4}}-1),(-1<p<2)$$

I don't know how to start. It's a beautiful integral, can someone help me,and any hint or trick will be appreciated.

Answer 1

$$\int_{0}^{1}\frac{x}{x^2+1}\left(1-\frac{1}{x}\right)^p\,dx = \int_{1}^{+\infty}\frac{(x-1)^p}{x(x^2+1)}\,dx = \int_{0}^{+\infty}\frac{x^p}{(x+1)(x^2+2x+2)}\,dx$$ since $\mathcal{L}^{-1}\left(\frac{1}{(x+1)(x^2+2x+2)}\right)=e^{-s}(1-\cos s)$ and $\mathcal{L}(x^p)=\frac{\Gamma(p+1)}{s^{p+1}}$ can be turned into $$ \Gamma(p+1)\int_{0}^{+\infty}s^{-1-p}e^{-s}(1-\cos s)\,ds\qquad (\cos s=\text{Re}\,e^{is}) $$ hence it is enough to exploit the integral representation for the $\Gamma$ function together with its reflection formula.