Integrating a two complex functions I need to evaluate the norm $||G(s)||_2$ where $G(s) = \frac{\alpha_1 s+\alpha_o}{s^2+\beta_1s+\beta_o }$ is a transfer function.
And
$||G(s)||^2_2$ = $\int_{-\infty}^{\infty} trace {G(j\omega) G^*(j\omega)} \,d\omega$
I am able to get to this point
$||G(s)||^2_2$ = $\int_{-\infty}^{\infty}  \frac{\alpha_1^2\omega^2 + \alpha_0^2}{\omega^4+(\beta_1^2-2\beta_o)\omega^2+\beta_o^2} \,d\omega$
The above expression is analogous to $\int_{-\infty}^{\infty}  \frac{ax^2 + b}{x^4+cx^2+d^2} \,dx$
I tired splitting the functions and use square method but it doesn't seem to let me anywhere. Any suggestions in solving this integral? Thanks
 A: I am assuming that all the constants are $\gt 0$.
We start by noticing that the integrand is even, thus
$$I =2 \int\limits_0^\infty\frac{\alpha_1^2x^2+\alpha_0^2}{x^4+(\beta_1^2-2\beta_0)x^2+\beta_0^2 }\, \mathrm dx $$
Now, using change of variables $x^4=\beta_0^2 t^4$, we get
$$\begin{align} I &= \frac2{\beta_0^{3/2}}\int\limits_0^\infty \frac{\alpha_1^2\beta_0t^2+\alpha_0^2}{t^4+\frac{t^2}{\beta_0}(\beta_1^2-2\beta_0)+1}\,\mathrm dt \\ &=  \frac{2\alpha_1^2\beta_0}{\beta_0^{3/2}}\underbrace{\int\limits_0^\infty \frac{t^2}{t^4+\frac{t^2}{\beta_0}(\beta_1^2-2\beta_0)+1}\,\mathrm dt}_{t\mapsto\frac1t}+\frac{2\alpha_0^2}{\beta_0^{3/2}} \int\limits_0^\infty  \frac{\mathrm dt} {t^4+\frac{t^2}{\beta_0}(\beta_1^2-2\beta_0)+1} \\ I &= \frac{2(\alpha_1^2\beta_0+\alpha_0^2)}{\beta_0^{3/2}} \int\limits_0^\infty \frac{\mathrm dt} {t^4+\frac{t^2}{\beta_0}(\beta_1^2-2\beta_0)+1}\end{align}$$
Now, again using $t\mapsto \frac1t$ and adding the result, we get
$$\begin{align}I &= \frac{\alpha_1^2\beta_0+\alpha_0^2}{\beta_0^{3/2}} 
\int\limits_0^\infty\frac{1+\frac1{t^2}}{t^2+\frac1{t^2}+\frac{\beta_1^2-2\beta_0}{\beta_0}}\,\mathrm dt \\ &= \frac{\alpha_1^2\beta_0+\alpha_0^2}{\beta_0^{3/2}} 
\int\limits_{-\infty}^\infty\frac{\mathrm dy}{y^2+\frac{\beta_1^2}{\beta_0} }\ , t-\frac1t=y \end{align}$$
Thus, we conclude that
$$\color{red}{\boxed{\boxed{\,\int\limits_{-\infty}^\infty\frac{\alpha_1^2x^2+\alpha_0^2}{x^4+x^2(\beta_1^2-2\beta_0)+\beta_0^2}\,\mathrm dx = \frac{\pi(\alpha_1^2\beta_0+\alpha_0^2)}{\beta_0\beta_1}}}}$$
