<s>[INST] Integration of $\frac{1}{u^4 + (4\zeta^2-2)u^2 + 1}$ I am trying to compute
$$I(\zeta) = \int_{-\infty}^{\infty} \frac{1}{u^{4} +  \left(4 \zeta^{2} - 2\right)u^{2} + 1}\, du$$
for positive real $\zeta$. Can anyone help?
I'm way out of practice for integrals except for simple stuff like $\int 1/(1+u^2)\, du = \tan^{-1} u + C$.
Sympy fails on the definite integral and gives me this weird RootSum expression for the indefinite integral:
$$\operatorname{RootSum} {\left(t^{4} \left(4096 \zeta^{8} - 8192 \zeta^{6} + 4096 \zeta^{4}\right) + t^{2} \left(256 \zeta^{6} - 384 \zeta^{4} + 128 \zeta^{2}\right) + 1, \left( t \mapsto t \log{\left (- 512 t^{3} \zeta^{6} + 768 t^{3} \zeta^{4} - 256 t^{3} \zeta^{2} - 32 t \zeta^{4} + 32 t \zeta^{2} - 4 t + u \right )} \right)\right)}$$
Wolfram Alpha gives me the following for the indefinite integral :
$$\begin{align}
& \frac{\frac{1}{a_1}\tan^{-1} \frac{u}{a_1} - \frac{1}{a_2}\tan^{-1} \frac{u}{a_2}}{4\zeta\sqrt{\zeta^2-1}} + C \\
\\
a_1 &= \sqrt{2\zeta^2-2\zeta\sqrt{\zeta^2-1}-1} = \sqrt{b-c}\\
a_2 &= \sqrt{2\zeta^2+2\zeta\sqrt{\zeta^2-1}-1} = \sqrt{b+c}\\
\end{align}$$
(with $b=2\zeta^2-1$ and $c=2\zeta\sqrt{\zeta^2-1}$) but I'm a bit lost how it got there, and then I'm not exactly sure what to do if $\zeta \le 1$ (is the formula still valid?!)
edit: OK, partial fraction expansion is sloooowwwwly coming back to me. It looks like $a_1a_2 = 1$ and $a_1{}^2 + a_2{}^2 = 4\zeta^2-2$, so I guess they used the expansion
$$
\frac{1}{u^{4} +  \left(4 \zeta^{2} - 2\right)u^{2} + 1} = \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left(\frac{1}{u^2+a_1{}^2} - \frac{1}{u^2+a_2{}^2}\right)
$$
 [/INST] Note
\begin{align}
& \int_{-\infty}^{\infty}\frac{du}{u^4 + (4\zeta^2-2)u^2 + 1}\\
= & \int_{0}^{\infty}\left( \frac{1+\frac1{u^2}}{u^2+\frac1{u^2} + 4\zeta^2-2} -\frac{1-\frac1{u^2}}{u^2+\frac1{u^2} + 4\zeta^2-2}\right)du\\
 = & \int_{0}^{\infty}\left( \frac{d(u-\frac1{u})}{(u-\frac1{u} )^2+ 4\zeta^2} -\frac{d(u+\frac1{u})}{(u+\frac1{u} )^2+ 4\zeta^2-4}\right)\\
= & \int_{-\infty}^{\infty} \frac{dt}{t^2+ 4\zeta^2}- \int_{\infty}^{\infty} \frac{dt}{t^2+ 4\zeta^2-4}\\
=&\frac\pi{2\zeta}-0 =\frac\pi{2\zeta}
\end{align}
</s>