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
$$
\frac{1}{u^{4} +  \left(4 \zeta^{2} - 2\right)u^{2} + 1} = \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left(\frac{1}{u^2+a_1{}^2} - \frac{1}{u^2+a_2{}^2}\right)
$$
 A: 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}
A: Let $I$ denote the integral. Then using the identity
$$ u^4 + (4\zeta^2-2)u^2 + 1 = u^2 \left( ( u - u^{-1} )^2 + 4\zeta^2 \right), $$
we may write
$$ I = 2 \int_{0}^{\infty} \frac{u^{-2}}{( u - u^{-1} )^2 + 4\zeta^2} \, \mathrm{d}u. \tag{1} $$
Now applying the substitution $u \mapsto u^{-1}$,
$$ I = 2 \int_{0}^{\infty} \frac{1}{( u - u^{-1} )^2 + 4\zeta^2} \, \mathrm{d}u. \tag{2} $$
Averaging $\text{(1)}$ and $\text{(2)}$, we get
\begin{align*}
I
&= \int_{0}^{\infty} \frac{1 + u^{-2}}{( u - u^{-1} )^2 + 4\zeta^2} \, \mathrm{d}u \\
&= \int_{-\infty}^{\infty} \frac{1}{t^2 + 4\zeta^2} \, \mathrm{d}t \tag{$t=u-u^{-1}$} \\
&= \frac{\pi}{2|\zeta|}.
\end{align*}
A: Partial fraction expansion gives
$$
\frac{1}{u^{4} +  \left(4 \zeta^{2} - 2\right)u^{2} + 1} = \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left(\frac{1}{u^2+a_1{}^2} - \frac{1}{u^2+a_2{}^2}\right)
$$
and since
$$\int \frac{du}{u^2+a^2}\, = \frac{1}{a}\int \frac{a\, du}{u^2+a^2}\,= \frac{1}{a}\int \frac{a^2\, dx}{a^2x^2+a^2}\, = \frac{1}{a}\tan^{-1}x+C = \frac{1}{a}\tan^{-1}\frac{u}{a}+C$$
(with $u = ax$) then I get
$$\begin{align}
I(\zeta) &= \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left[\frac{1}{a_1}\tan^{-1}\frac{u}{a_1} - \frac{1}{a_2}\tan^{-1}\frac{u}{a_2}\right]_{-\infty}^{\infty} \\
&= \frac{\pi}{4\zeta\sqrt{\zeta^2-1}}\left[\frac{1}{a_1} - \frac{1}{a_2}\right] \\
&= \frac{\pi(a_2 - a_1)}{4\zeta\sqrt{\zeta^2-1}} \end{align}$$
I can simplify further since $a_1= \sqrt{b-c}$ and $a_2=\sqrt{b+c}$ with  $b=2\zeta^2-1$ and $c=2\zeta\sqrt{\zeta^2-1}$:
$\sqrt{b+c}-\sqrt{b-c} = \frac{(b+c) - (b-c)}{\sqrt{b+c}+\sqrt{b-c}} = \frac{2c}{\sqrt{b+c}+\sqrt{b-c}}$
so we have
$$\begin{align}
I(\zeta) &=  \frac{\pi}{4\zeta\sqrt{\zeta^2-1}}\cdot\frac{2c}{\sqrt{b+c}+\sqrt{b-c}} \\
&= \frac{\pi}{\sqrt{b+c}+\sqrt{b-c}} \\
&= \frac{\pi}{\sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}}. \\
\end{align}$$
For positive real $\zeta$ the denominator can be simplified to $2\zeta$ (see https://math.stackexchange.com/a/3724215/120)
so we have $I(\zeta) = \frac{\pi}{2\zeta}.$
