Let $ \Psi(n) $ denote the Digamma function, how to prove that
$$\Psi\left(n + \dfrac{1}{2}\right) - \Psi(n + 1) = -2\int_0^1\dfrac{x^{2n}}{1 + x}\,dx $$
I have tried using the integral representations of the Digamma function and was able to simplify the LHS, however I'm not sure on how to get the integral on the RHS.
Your help would be appreciated. Thanks.
From $$\Psi(z+1)=-\gamma+\int_0^1{\frac{1-t^z}{1-t}\,dt}$$ (c.f. Wikipedia), we have \begin{align*} \Psi\left(n+\frac{1}{2}\right)-\Psi(n+1)&=\int_0^1{\frac{(1-t^{n-1/2})-(1-t^n)}{1-t}\,dt} \\ &=\int_0^1{\frac{t^n-t^{n-1/2}}{1-t}\,dt} \\ &=\int_0^1{\frac{\sqrt{t}-1}{1-t}\cdot t^{n-1/2}\,dt} \\ &=-\int_0^1{\frac{t^n}{1+\sqrt{t}}\cdot\frac{dt}{\sqrt{t}}} \\ &=-\int_0^1{\frac{x^{2n}}{1+x}\cdot2\,dx} \end{align*} where $x=\sqrt{t}$.