I am stuck in the following problem: I have the following Fourier series: $$\cos (\alpha x)= \frac{1}{2}a_0+ \sum_{k=1}^{\infty}a_k\cos(kx)$$ $-\pi\leq x \leq \pi$, and $\alpha>0$ and not an integer. The question is to find explicit expressions for the Fourier coefficients $a_k$. I started with the standard series: $$a_k=\frac{1}{\pi}\int_{-\pi}^{\pi}\cos(\alpha x) \cdot \cos (kx) dx$$ Now I know that $\cos a \cos b= \frac{1}{2}\big[\cos(a-b)+\cos(a+b)\big]$ Plugging in this values to obtain: $$a_k=\frac{1}{2\pi}\int_{-\pi}^{\pi}\cos(\alpha -k)x +\cos(\alpha +k)x dx$$ Which can be changed to because I prefer to put in zero's: $$a_k=\frac{1}{\pi}\int_{o}^{\pi}\cos(\alpha -k)x +\cos(\alpha +k)x dx$$ Now I'll do the integration to obtain: $$a_k=\frac{1}{\pi}\bigg[\frac{1}{\alpha -k}\sin(\alpha-k)x+\frac{1}{a+k}\sin(\alpha +k)x\bigg]_{0}^{\pi}$$ Putting in the values for ${\pi}$ and $0$ giving: $$a_k=\frac{1}{\pi}\bigg[\frac{1}{a-k}\sin(a-k)\pi+\frac{1}{a+k}\sin(a+k)\pi\bigg]$$ Again using a trick:$sin(a+/-b)=\sin a\cos b +/-\cos a \sin b$ which result in: $$a_k=\frac{1}{\pi}\bigg[\frac{1}{\alpha-k}\sin(\alpha \pi)\cos k\pi -\cos \alpha \pi \sin k\pi+\frac{1}{\alpha +k}\sin(\alpha \pi)\cos k\pi+ \cos \alpha \pi \sin k \pi)\bigg]$$ Knowing that: $\sin k\pi = 0$ and $\cos k \pi= (-1)^k$, to obtain: $$a_k=\frac{1}{\pi}\bigg[\frac{1}{\alpha-k}\sin(\alpha \pi)(-1)^k +\frac{1}{\alpha +k}\sin(\alpha \pi)(-1)^k \bigg]$$ Which can be expressed: $$a_k=\frac{1}{\pi} (-1)^k \sin \alpha \pi \bigg[\frac{1}{\alpha-k} +\frac{1}{\alpha +k} \bigg]$$ This can be rearranged to:$$a_k=\frac{1}{\pi} (-1)^{k-1} \sin \alpha \pi \bigg[\frac{2\alpha}{k^2 - \alpha^2} \bigg]$$
But this is totally not the answer of the book which says:
$$\cos \alpha x= \frac {\sin \alpha \pi}{\alpha \pi}+ \frac{2 \alpha}{\pi}\sin \alpha \pi \sum_{k=1}^{\infty}(-1)^{k-1}\frac{\cos kx}{k^2-\alpha^2}$$
Did I start wrong?
So by correcting the mistake: $$a_k=\frac{2\alpha}{\pi} (-1)^{k-1} \sin \alpha \pi \bigg[\frac{1}{k^2 - \alpha^2} \bigg]$$ so:
$$\cos (\alpha x)=\frac{1}{2}a_0+\frac{2\alpha}{\pi}\sin\alpha \pi\sum_{k=1}^{\infty}(-1)^{k-1}\frac{\cos (kx)}{k^2-\alpha^2}$$
for $k=0$ we obtain:$\frac{\ sin\alpha \pi}{\alpha \pi}$
Hence the required relation:
$$\cos \alpha x= \frac {\sin \alpha \pi}{\alpha \pi}+ \frac{2 \alpha}{\pi}\sin \alpha \pi \sum_{k=1}^{\infty}(-1)^{k-1}\frac{\cos kx}{k^2-\alpha^2}$$