Is there an alternative proof for the following identity? $\forall \text{ }0 < \alpha < 1; a,b$ $$\int_{0}^{\infty} x^{\alpha - 1} \Big[\psi(b+x) - \psi(a+x)\Big] \mathrm dx = \frac{\pi}{\sin(\alpha \pi)}\Big[\zeta(1-\alpha, a) - \zeta(1-\alpha, b))\Big]$$
I am interested if anyone has any other techniques/methods to solve this integral without the series expansion of the digamma function (the method I used). The result of this integral is included in this post of mine requesting for more polygamma integrals (I don't want the original post to be an essay, so I am deriving this result here).
I considered the series formula of the digamma function, giving the following
\begin{align*} \psi (b+x) - \psi(a+x)&=-\gamma +\sum _{n=0}^{\infty }\left({\frac {1}{n+1}}-{\frac {1}{n+b+x}}\right) - \Big[-\gamma +\sum _{n=0}^{\infty }\left({\frac {1}{n+1}}-{\frac {1}{n+a+x}}\right)\Big]\\\\ &= \sum _{n=0}^{\infty }\left({\frac {1}{n+1}}-{\frac {1}{n+b+x}}\right) - \sum _{n=0}^{\infty }\left({\frac {1}{n+1}}-{\frac {1}{n+a+x}}\right)\\\\ &= \sum _{n=0}^{\infty }\left({\frac {1}{n+a+x}}-{\frac {1}{n+b+x}}\right) \end{align*}
Substituting this into the original integral gives
\begin{align*} \int_{0}^{\infty} x^{\alpha - 1} \Big[\psi(b+x) - \psi(a+x)\Big] \mathrm dx &= \int_{0}^{\infty} x^{\alpha - 1} \Big[\sum _{n=0}^{\infty }\left({\frac {1}{n+a+x}}-{\frac {1}{n+b+x}}\right)\Big] \mathrm dx\\\\ &= \sum _{n=0}^{\infty}\int_{0}^{\infty} x^{\alpha - 1} \Big[\left({\frac {1}{n+a+x}}-{\frac {1}{n+b+x}}\right)\Big] \mathrm dx \\\\ &= \frac{\pi}{\sin(\alpha \pi)} \sum_{n=0}^{\infty} \Big[(n+a)^{\alpha - 1} - (n+b)^{\alpha - 1}\Big]\\\\ &= \frac{\pi}{\sin(\alpha \pi)} \Big[\zeta(1-\alpha, a) - \zeta(1-\alpha, b)\Big] \end{align*}
I used the series formula obtained at the beginning, then swapped the sum and integral signs, integrated the two terms ($\int_{0}^{\infty} x^{\alpha - 1} \frac{1}{n+q+x} \mathrm dx = (n+q)^{\alpha - 1} \frac{\pi}{\sin(\alpha \pi)}$ I used the result (obtained through the substitution $u = x/(n+q)$ and using this result), and used the definition of the Hurwitz zeta function to obtain the final form.
Feel free to share any techniques you may have. Thanks :)
