Analytic formula for $f(\alpha):= \int_{-\infty}^\infty\frac{|e^{is}-1|}{|s|^{2\alpha+1}}\,ds$

Question

For any $\alpha \in (0,1)$, define $\displaystyle f(\alpha):= \int_{-\infty}^\infty\frac{|e^{is}-1|}{|s|^{2\alpha+1}}\,ds$.

Question. Is there an analytic formula for $f(\alpha)$, say, in terms of special functions ?

Note. I'm also fine with good upper-bounds for $f(\alpha)$ in terms of simpler expressions in $\alpha$.

Answer 1

The integral converges only when $0 < \alpha < \frac{1}{2}$. For such $\alpha$, we have

\begin{align*} f(\alpha) &= 4\int_{0}^{\infty} \frac{\left|\sin(s/2)\right|}{s^{2\alpha+1}} \, \mathrm{d}s \\ &= \frac{4}{\Gamma(2\alpha+1)}\int_{0}^{\infty}\int_{0}^{\infty} \left|\sin(s/2)\right|t^{2\alpha}e^{-st} \, \mathrm{d}t\mathrm{d}s \\ &= \frac{4}{\Gamma(2\alpha+1)} \int_{0}^{\infty} t^{2\alpha} \Biggl( \sum_{n=0}^{\infty}\int_{0}^{2\pi} \sin(s/2) e^{-(s+2n\pi)t} \, \mathrm{d}s \Biggr) \, \mathrm{d}t \\ &= \frac{8}{\Gamma(2\alpha+1)} \int_{0}^{\infty} \frac{t^{2\alpha}}{4t^2 + 1}\biggl(\frac{1+e^{-2\pi t}}{1 - e^{-2\pi t}}\biggr) \, \mathrm{d}t \\ &= \frac{8}{\Gamma(2\alpha+1)\sin(\pi \alpha)} \sum_{n=1}^{\infty} \frac{n^{2\alpha}}{4n^2 - 1}. \end{align*}

Here, the last step follows from the residue computation. However, I am not sure if the last sum simplifies.