I need to show :
$$f(\theta)=\cos(n\theta)\int_{0}^{2\pi}w(\theta')\cos(n\theta')d\theta' =\int_{0}^{2\pi}w(\theta-\theta')\cos(n\theta')d\theta'$$
The author says, Fourier expanding the integrand of the right hand side leads to the result. I also know $w$ is $2\pi$ periodic and symmetric. I have been stuck on this for a long while now. Any help would be appreciated!
Thanks in advance.
On
Performing the change of variables $\theta'\mapsto \theta-\theta'$ and using periodicity of the integrand to adjust the integration region ($w$ can be extended from a function over $[0,2\pi)$ to be a periodic function over all the real axis), we get $$ \int_0^{2\pi} w(\theta-\theta')\cos(n\theta') d\theta' =\int_{\theta-2\pi}^{\theta}w(\theta')\cos(n\theta-n\theta') d\theta' =\int_{0}^{2\pi}w(\theta')\cos(n\theta-n\theta') d\theta'. $$ Now, let us expand the last expression using the formula $\cos(n\theta-n\theta')=\cos(n\theta)\cos(n\theta')+\sin(n\theta)\sin(n\theta')$. By assumption, $w(\theta')$ is symmetric, while $\sin(n\theta)$ is antisymmetric on the interval $[0,2\pi)$, hence $$ \sin(n\theta)\int_0^{2\pi}w(\theta')\sin(n\theta')d\theta'=0. $$ Therefore $$ \int_0^{2\pi} w(\theta-\theta')\cos(n\theta') d\theta' = \cos(n\theta) \int_0^{2\pi}w(\theta')\cos(n\theta')d\theta'. $$ Alternatively we can follow the hint given in the text and expand $w$ in its Fourier series (since it is symmetric, only the cosine series is needed) $$ w(\theta-\theta')=\frac{1}{\pi}\sum_{m=0}^\infty \cos(m(\theta-\theta'))\int_0^{2\pi} w(\tau)\cos(m\tau)d\tau. $$ Substituting into $$ \int_0^{2\pi}w(\theta-\theta')\cos(n\theta')d\theta' $$ and using $$ \int_0^{2\pi} \cos(n\theta')\cos(m(\theta-\theta'))d\theta' =\cos(m\theta)\int_0^{2\pi}\cos(n\theta') \cos(m\theta')d\theta' = \pi \cos(n\theta) \delta_{nm} $$ where $\delta_{nm}$ is the Kronecker symbol, again yields $$ \int_0^{2\pi}w(\theta-\theta')\cos(n\theta')d\theta'=\cos(n\theta')\int_0^{2\pi}w(\tau)\cos(n\tau)d\tau $$
Starting with the RHS, use the substitution $x = \theta - \theta'$ to obtain $$\int_{0}^{2\pi}w(\theta - \theta')\cos(n \theta')\ d\theta' = \int_{\theta - 2\pi}^{\theta} w(x)\cos(n(x - \theta))\ dx$$ Since the integrand is $2\pi$-periodic, we can integrate over any interval of length $2\pi$ without changing the result, so the above is the same as $$\int_{-\pi}^{\pi}w(x)\cos(n(x - \theta))\ dx$$ Applying the trig identity $\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$ gives us $$\cos(n\theta)\int_{-\pi}^{\pi} w(x)\cos(nx)\ dx + \sin(n\theta)\int_{-\pi}^{\pi} w(x)\sin(nx)\ dx$$ As $w$ is even and $\sin$ is odd, the integrand in the second integral is odd, so the second integral is zero. This leaves us with $$\cos(n\theta)\int_{-\pi}^{\pi} w(x)\cos(nx)\ dx$$ which is the same as $$\cos(n\theta)\int_{0}^{2\pi} w(x)\cos(nx)\ dx$$ since as above, we can integrate over any interval of length $2\pi$.