Question
For $a > 0$ and any $n\in\mathbb{Z}$, what is the value of $$ c_n := \int_{-\pi}^{\pi} \frac{a^2}{(a^2 + x^2)^2} \cos(nx) dx? $$
My Attempt
CORRECTION Letting $f_a(x) := \frac{a^2}{(a^2 + x^2)^2}$ for any $a>0$, we can compute that the Fourier transform of $f_a$ is given by $$ \hat{f}_a(\xi) = \int_{-\infty}^{\infty} f_a(x) e^{i \xi x} dx =\frac{\pi}{2a}(1+a|\xi|)e^{-|\xi|a} \qquad \forall \xi \in \mathbb{R}. $$
Well, if we let $\chi_{b}(x):=\cases{1 & $x\in[-b,b]$\\ 0 & otherwise}$ for any $b > 0$, we also know that $$ \begin{align} \hat{\chi}_{b}(\xi) &= \int_{-\infty}^{\infty} \chi_{b}(x) e^{i\xi x} dx\\ &= \int_{-b}^{b} e^{i\xi x} dx\\ &= \frac{1}{i\xi}\left(e^{ib\xi} - e^{-ib\xi}\right)\\ &= \frac{2\sin(b\xi)}{\xi} = 2b\, \text{sinc}(b\xi). \end{align} $$ Furthermore, letting $g_b(\xi) := 2b\, \text{sinc}(bx)$, we can see that for $x\in\mathbb{R}$, by the Fourier inversion theorem, the inverse Fourier transform of $g_b$ is $$ \begin{align} \check{g}_b(x) &:= \frac{1}{2\pi}\int_{-\infty}^{\infty}g_b(\xi) e^{-i\xi x} d\xi = \chi_{b}(x). \end{align} $$
Returning back to our original question, we see that for $n \in \mathbb{Z}$ $$ \begin{align} c_n &= \text{Re}\left[\int_{-\infty}^{\infty}f_a(x) \chi_{\pi}(x)e^{inx} dx\right] \\ &= \text{Re}\left[\widehat{f_a \chi_{\pi}}(n)\right]\\ &= \frac{1}{2\pi}\text{Re}\left[(\hat{f}_a \ast \hat{\xi}_{\pi})(n)\right] \end{align} $$ where $\ast$ denotes the convolution between two functions
i.e. (brief description) for two functions $f,g$ $$ (f\ast g)(x) = \int_{-\infty}^{\infty} f(y) g(x-y) dy = \int_{-\infty}^{\infty} f(x-y) g(y) dy $$
NOTE: There was an error in an initial calculation which rendered the other work incorrect. See history of post to see other work.
