There are similar questions out there, but I was hoping someone could show how to would evaluate the following integral

$$\int_{-1/2}^{1/2}\bigg(\frac{\sin(n\pi f)}{\sin(\pi f)}\bigg)^4 df$$

I've seen the approach of using Euler's formula quite a bit, but I've never been able to wrap my head around it. Most of the answers that I've use that use it simply state it and skip some of the details. Any help?


Assuming $n \gt 0$. I would rewrite as

$$\frac1{2 \pi} \int_{-\pi}^{\pi} dt \frac{\sin^4{n t}}{\sin^4{t}} $$

We can express this as an integral over the unit circle in the complex plane and use the residue theorem. Thus, the integral is equal to, upon subbing $z=e^{i t}$:

$$\frac1{i 2 \pi} \oint_{|z|=1} \frac{dz}{z} \left (\frac{z^n-z^{-n}}{z-z^{-1}}\right )^4 = \frac1{i 2 \pi} \oint_{|z|=1} dz \frac1{z^{4 n-3}} \left (\frac{z^{2 n}-1}{z^2-1} \right )^4$$

The only pole to speak of is at $z=0$. The residue at this pole is the coefficient of $z^{-1}$ in the Laurent expansion of the integrand. Note that

$$\frac1{z^{4 n-3}} \left (\frac{z^{2 n}-1}{z^2-1} \right )^4 = \frac1{z^{4 n-3}} \left (1+z^2+z^4+\cdots+z^{2 n-2} \right )^4$$

So, really, the residue is simply the coefficient of $z^{4 (n-1)}$ in the expansion of the fourth power of the sum in the parentheses. This is not as hard as it looks. Note that

$$\left (1+z^2+z^4+\cdots+z^{2 n-2} \right )^2 = 1+2 z^2 + 3 z^4 + \cdots + n z^{2 (n-1)}+\cdots+3 z^{4 n-8} + 2 z^{4 n-6} + z^{4 n-4}$$

So, therefore, the coefficient of $z^{4 (n-1)}$ in the square of the above sum is

$$n^2 + 2 \sum_{k=1}^{n-1} k^2 = n^2 + \frac13 (n-1)(n)(2 n-1) = \frac13 n (2 n^2+1)$$

By the residue theorem, the integral we seek is equal to that residue. Thus, for $n \gt 0$,

$$\frac1{2 \pi} \int_{-\pi}^{\pi} dt \frac{\sin^4{n t}}{\sin^4{t}} = \frac13 n (2 n^2+1)$$

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