I would like to evaluate the integral: $$\int_{-\infty}^{\infty}\left|\frac{1}{(a+i\omega)^{n+1}}\right|^2\,\mathrm{d}\omega,$$ where $i$ is the imaginary number and $a\in\mathbb{R}$ and $n\in\mathbb{N}$ are constants.
I have tried to perform a binomial expansion of the denominator where the even indices will represent the real part and the odd indices the imaginary part. From that I can calculate the real and imaginary parts of the fraction but then I have also to calculate the squared term which makes the integrand very messy and I cannot see anyway of evaluating the integral.
I have also tried to use partial fractions, such as, $$\left|\frac{1}{(a+i\omega)^{n+1}}\right|^2 = \frac{1}{(a+i\omega)^{n+1}(a-i\omega)^{n+1}} = \frac{A}{(a+i\omega)^{1}}+\frac{B\omega+C}{(a+i\omega)^{2}}+...$$ but again things start to get super-messy and very difficult to solve.
Question:
Is there are any trick to solve this integral? Or, should I still follow one of the two methods I have described?
The Fourier Transform of $f(t)=e^{-a\omega}u(t)$ is $F(\omega)=\dfrac{1}{a+j\omega}$. Also we know that $$FT((-jt)^nf(t))=F^{(n)}(\omega)$$from the other side$$F^{(n)}(\omega)=\dfrac{(-j)^nn!}{(a+j\omega)^{n+1}}$$therefore according to Parseval's identity we have $$\int_{\Bbb R}|f(t)|^2dt=\dfrac{1}{2\pi}\int_{\Bbb R}|F(\omega)|^2d\omega$$hence by substitution we obtain:$$\int_{0}^{\infty}t^{2n}e^{-2at}dt=\dfrac{1}{2\pi}\int_{\Bbb R}|F(\omega)|^2d\omega=\dfrac{(n!)^2}{2\pi}\int_{\Bbb R}|\dfrac{1}{(a+j\omega)^{n+1}}|^2d\omega$$which means that$$\int_{\Bbb R}|\dfrac{1}{(a+j\omega)^{n+1}}|^2d\omega=\dfrac{2\pi\Gamma(2n+1)}{(2a)^{2n+1}\Gamma^2(n+1)}$$here is a sketch of integral for $a=1$
where red square markers indicate on when $n\in\Bbb Z$