I need to calculate the integral
$$\int_{a=-\pi}^{\pi}\frac{e^{ins}}{1- \frac{1}{2} \, e^{i(t-s)}}ds$$
which can be interperted as Fourier coefficient.
I didn't succeed to calculate this integral although I tried many attempts. Any suggestions as to how to calculate the integral?
On
$$I=\int_{[-\pi, \pi]}\frac{e^{ins}}{1-\frac{1}{2}e^{i(t-s)}}ds$$ Let $$e^{is}=z$$ Then $$I=\frac{1}{i}\oint_{|z|=1}\frac{z^{n}}{z-\frac{1}{2}e^{it}}dz$$ There is a pole at $z=\frac{1}{2}e^{it}$ inside the contour. The residue is $$Res\Big{\{}\frac{z^{n}}{z-\frac{1}{2}e^{it}}; \ z=\frac{1}{2}e^{it}\Big{\}}=\lim_{z\rightarrow\frac{1}{2}e^{it}}\Big(z-\frac{1}{2}e^{it}\Big)\frac{z^{n}}{z-\frac{1}{2}e^{it}}=\frac{e^{int}}{2^{n}}$$ Thus $$I=2\pi{i}\Big(\frac{1}{i}Res\Big{\{}\frac{z^{n}}{z-\frac{1}{2}e^{it}}; \ z=\frac{1}{2}e^{it}\Big{\}}\Big)=\frac{\pi{e^{int}}}{2^{n-1}}$$
Using the geometric progression, we can write $$\dfrac{1}{1- \frac{1}{2} \, e^{i(t-s)}} = \sum\limits_{k=0}^{\infty}{\dfrac{1}{2^k} \, e^{ik(t-s)}}.$$ This series is absolutely convergent, so we can integrate it term by term: $$\int\limits_{-\pi}^{\pi}\frac{e^{ins}}{1- \frac{1}{2} \, e^{i(t-s)}}ds = \int\limits_{-\pi}^{\pi} e^{ins} \sum\limits_{k=0}^{\infty}{{\dfrac{1}{2^k} \, e^{ik(t-s)}}} \ ds = \\ = \sum\limits_{k=0}^{\infty}{\dfrac{e^{ikt}}{2^k} \int\limits_{-\pi}^{\pi}} e^{i(n-k)s} \, ds = 2 \pi \dfrac{e^{int}}{2^n} = \dfrac{\pi e^{int}}{2^{n-1}},$$ since $$\int\limits_{-\pi}^{\pi} e^{i(n-k)s} \, ds = \begin{cases} 0, & k \ne n, \\ 2\pi, & k=n. \end{cases}$$