Computing the Fourier series of $f(\theta)=\frac{\pi}{\sin\pi\alpha}e^{i(\pi-\theta)\alpha}$ for $x\in[0, 2\pi]$

Question

Just like the previous question I asked, this one has a solution on p. 36 of Stein and Shakarchi's Fourier Analysis: $\sum_{n=-\infty}^{\infty}\frac{e^{in\theta}}{n+\alpha}$.

But when I tried computing the Fourier coefficients I got $\hat{f}(n)=\frac{e^{i\pi\alpha}\sin(\pi(\pi+n))}{(\pi+n)\sin\pi\alpha}$, so the Fourier series would be $\sum_{n=-\infty}^{\infty}\hat{f}(n)=\frac{e^{i\pi\alpha}\sin(\pi(\pi+n))}{(\pi+n)\sin\pi\alpha}e^{in\theta}$ . . .

Any help would be much appreciated.

Answer 1

Recall: We have $$ \frac{e^{i\pi\alpha}-e^{-i\pi\alpha}}{2i}=\sin(\pi\alpha) $$

Now it's just a matter of computation: \begin{align} \hat{f}(n) &= \frac{1}{2\pi}\int_0^{2\pi}\frac{\pi}{\sin(\pi­\alpha)}e^{i(\pi-t)\alpha}e^{-int}\,dt \\ &=\frac{1}{\sin(\pi­\alpha)}\cdot\frac{e^{i\pi\alpha}}{2}\int_0^{2\pi}e^{-it(n+\alpha)}\,dt \\ &=\frac{1}{\sin(\pi­\alpha)}\cdot\frac{e^{i\pi\alpha}}{2}\cdot\bigg(\frac{1-e^{-2i\pi(n+\alpha)}}{i(n+\alpha)}\bigg) \\ &=\frac{1}{\sin(\pi­\alpha)}\cdot\frac{e^{i\pi\alpha}(1-e^{-2i\pi\alpha})}{2i}\cdot\frac{1}{n+\alpha} \\ &=\frac{1}{\sin(\pi­\alpha)}\cdot\frac{e^{i\pi\alpha}-e^{-i\pi\alpha}}{2i}\cdot\frac{1}{n+\alpha}\\ &=\frac{1}{\sin(\pi­\alpha)}\cdot\sin(\pi\alpha)\cdot\frac{1}{n+\alpha}\\ &=\frac{1}{n+\alpha} \end{align} By the way it is assumed that $\alpha\not\in\mathbb{Z}$ otherwise $f$ would not make sense.