Is this integral evaluated in any book?, how is it calculated?

Question

In a book I am reading it follows from a statement that

If $\nu \in (-1/2,1/2)$

$$\int\limits_0^\infty\left((1+x)^{\nu}-x^\nu\right)^2dx=\frac{\Gamma(\nu+1)^2}{\Gamma(2\nu+2)\sin(\pi(\nu+1/2))}-\frac{1}{2\nu+1}.$$

But there is no proof of this? Is the proof easy? Is it done in any books, do you have any links or hints?

The exact statement of the book is that for $H \in (0,1)$ we have

$$\left[\int\limits_0^\infty\left((1+x)^{H-\frac{1}{2}}-x^{H-\frac{1}{2}}\right)^2dx+\frac{1}{2H}\right]^{\frac{1}{2}}=\frac{\Gamma(H+1/2)}{(\Gamma(2H+1)\sin(\pi H))^{1/2}}.$$

Answer 1

By integration by parts, $$ \int_{0}^{+\infty}\left((1+x)^\nu-x^\nu\right)^2\,dx = \int_{0}^{+\infty}2\nu x\left((1+x)^\nu-x^\nu\right)\left(x^{\nu-1}-(1+x)^{\nu-1}\right)\,dx$$ and $$\int_{0}^{+\infty}2\nu x\left((1+x)^\nu-x^\nu\right)x^{\nu-1}\,dx = -\frac{2\, \Gamma(-1-2 \nu)\,\Gamma(2+\nu)}{\Gamma(-\nu)}$$ $$\int_{0}^{+\infty}2\nu x\left((1+x)^\nu-x^\nu\right)(1+x)^{\nu-1}\,dx = \frac{1}{1+2 \nu}-\frac{2^{-1-2 \nu} \nu \Gamma\left(-\frac{1}{2}-\nu\right) \Gamma(1+\nu)}{\sqrt{\pi }}$$ follow from Euler's Beta function.