How would I go about evaluating the following integral? Let $d\in [-1/2,1/2]$. Its related to the spectral density function of a fractionally differenced processes.
$$\int_\frac{-1}{2}^\frac{1}{2} \vert \sin(\pi f)\vert^{2d} df$$
I know that the integral is finite, since it is bounded by unity over an interval of finite length, but I'm not exactly sure how to proceed.
The integral does not converge for $d = -\frac{1}{2}$, but it does converge for all the other $d$. Let $d \in (-\frac{1}{2},\frac{1}{2}]$. Since the integrand is an even function, you can write your integral as
$$2\int_0^{1/2}|\sin(\pi f)|^{2d}\, df.$$
Via the $u$-substitution $u = \pi f$, we get
$$ 2\int_0^{\pi/2} |\sin u|^{2d} \frac{du}{\pi} = \frac{2}{\pi}\int_0^{\pi/2}\sin^{2d}(u)\, du = \frac{2}{\pi}\frac{\Gamma\left(d + \frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(d + 1)},$$
using the identity
$$\int_0^{\pi/2} \sin^{2x-1}(u)\cos^{2y-1}(u)\, du = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)},\quad x,y > 0.$$
Since $\Gamma(\tfrac{1}{2}) = \sqrt{\pi}$,
$$\frac{2}{\pi}\frac{\Gamma\left(d + \frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(d + 1)} = \frac{2}{\sqrt{\pi}}\frac{\Gamma\left(d + \frac{1}{2}\right)}{\Gamma(d + 1)}.$$
So for $d\in (-\frac{1}{2},\frac{1}{2}]$,
$$\int_{-1/2}^{1/2}|\sin(\pi f)|^{2d}\, df = \frac{2}{\sqrt{\pi}} \frac{\Gamma\left(d + \frac{1}{2}\right)}{\Gamma(d + 1)}.$$
