Evaluate $\int_{-1/2}^{1/2}\frac{\sin^4(n\pi f)}{\lvert\sin(\pi f\lvert^{2d}[\sin(\pi f)]^{2}}df$ for any $d\in(-1/2,1]$ Evaluate $$\int_{-1/2}^{1/2}\frac{\sin^4(n \,\pi f)}{\lvert\sin(\pi f)\lvert^{2d}\sin(\pi f)^{2}}df$$ for any $d\in(-1/2,1]$
This was helpful, but what if the denominator is raised to a fractional power?
 A: 
Evaluate $$\int_{-1/2}^{1/2}\frac{\sin^4(n \,\pi f)}{\lvert\sin(\pi f)\lvert^{2d}\sin(\pi f)^{2}}df$$ for any $d\in(-1/2,1]$

Let's do $x \gets f$ and write $\frac{e^{i \theta} - e^{-i \theta}}{2 i} = sin(\theta) $
Then I have
$$\int_{-1/2}^{1/2}\frac{ \left(\frac{e^{i n \pi x} -e^{-i n \pi x} }{2 i} \right)^4}{\lvert\sin(\pi x)\lvert^{2d} \left(\frac{e^{i \pi x} -e^{-i \pi x} }{2 i} \right)^2}dx = \frac{-1}{4} \int_{-1/2}^{1/2}\frac{ \left(e^{i n \pi x} -e^{-i n \pi x} \right)^4}{\lvert\sin(\pi x)\lvert^{2d} \left( e^{i \pi x} -e^{-i \pi x} \right)^2}dx$$
Where I changed most of the sine functions except the one under the absolute value. Now we can do that one too by noting the behavior of the sine in these bounds.
$$\frac{-1}{4} \int_{-1/2}^{1/2}\frac{ \left(e^{i n \pi x} -e^{-i n \pi x} \right)^4}{\lvert\sin(\pi x)\lvert^{2d} \left( e^{i \pi x} -e^{-i \pi x} \right)^2}dx $$
$$= \frac{-1}{4} \int_{0}^{1/2}\frac{ \left(e^{i n \pi x} -e^{-i n \pi x} \right)^4}{(\sin(\pi x))^{2d} \left( e^{i \pi x} -e^{-i \pi x} \right)^2}dx + \frac{-1}{4} \int_{-1/2}^{0}\frac{ \left(e^{i n \pi x} -e^{-i n \pi x} \right)^4}{(-\sin(\pi x))^{2d} \left( e^{i \pi x} -e^{-i \pi x} \right)^2}dx $$
$$= \frac{-1}{4} \int_{0}^{1/2}\frac{ \left(e^{i n \pi x} -e^{-i n \pi x} \right)^4}{(\frac{e^{i \pi x} - e^{-i \pi x}}{2 i})^{2d} \left( e^{i \pi x} -e^{-i \pi x} \right)^2}dx + \frac{-1}{4} \int_{-1/2}^{0}\frac{ \left(e^{i n \pi x} -e^{-i n \pi x} \right)^4}{(-(\frac{e^{i \pi x} - e^{-i \pi x}}{2 i}))^{2d} \left( e^{i \pi x} -e^{-i \pi x} \right)^2}$$
Now $(\pm 2i)^{2d} = 2^{2d} e^{\pm i d \pi/2}$
So now we have:
$$ = -2^d e^{i \frac{d \pi}{2}}\int_{0}^{1/2} \left(\frac{e^{i n \pi x} -e^{-i n \pi x}}{e^{i \pi x} - e^{-i \pi x}}\right)^4 (e^{i \pi x} - e^{-i \pi x})^{-d} dx$$ $$- 2^d e^{-i \frac{d \pi}{2}} \int_{-1/2}^{0}\left(\frac{e^{i n \pi x} -e^{-i n \pi x}}{e^{i \pi x} - e^{-i \pi x}}\right)^4 (e^{i \pi x} - e^{-i \pi x})^{-d} dx$$
The difference in phase goes away when we switch $ x\leftrightarrow -x$ in the second integral to combine the two into one expression:
$$-2 \times 2^d e^{i \frac{d \pi}{2}}\int_{0}^{1/2} \left(\frac{e^{i n \pi x} -e^{-i n \pi x}}{e^{i \pi x} - e^{-i \pi x}}\right)^4 (e^{i \pi x} - e^{-i \pi x})^{-d} dx$$
After working the problem to this point it is not clear to me that a closed form exists. In fact plugging values into Wolfram suggests it does not. There is no integration by parts trick to be employed and no clever change of variables. You could go on and write out explicitly the norm of the second term to get it's $d$ power written out correctly but I'm unclear that it is of much use.
I hope this helped you begin in any case.
