Prove $\int_{0}^{\infty}{1\over (1+{\phi^{-2}}x^2)(1+{\phi^{-4}}x^2)}dx={\pi\over 2}$ Prove
$\phi$; golden ratio

$$I=\int_{0}^{\infty}{1\over (1+{\phi^{-2}}x^2)(1+{\phi^{-4}}x^2)}dx={\pi\over 2}\tag1$$

Let $q=\phi^{-2}$ and $p=\phi^{-4}$
$${1\over (1+qx^2)(1+px^2)}={Ax+B\over 1+qx^2}+{Cx+D\over 1+px^2}\tag2$$
Result
$A=0$, $B=\phi$, $C=0$ and $D=-{1\over  \phi}$
Sub the result into $(2)$ and we got
$${1\over (1+\phi^{-2}x^2)(1+\phi^{-4}x^2)}={\phi\over 1+\phi^{-2}x^2}-{\phi^{-1}\over 1+\phi^{-4}x^2}$$
Hence
$$I=\int_{0}^{\infty}\left({\phi\over 1+\phi^{-2}x^2}-{\phi^{-1}\over 1+\phi^{-4}x^2}\right)dx\tag3$$
Simplified $(3)\rightarrow (4)$
$$I=\phi^3\int_{0}^{\infty}{1\over \phi^2+x^2}dx-\phi^3\int_{0}^{\infty}{1\over \phi^4+x^2}dx\tag4$$
Recall
$$\int_{0}^{\infty}{1\over a^2+x^2}dx={\pi \over 2a}\tag5$$
Utilise $(5)$ to brings $(4)$ to $(6)$
$$I=\phi^3\cdot{\pi\over 2\phi}-\phi^3{\pi\over 2\phi^2}\tag6$$
$$I={\pi\over 2}(\phi^2-\phi)={\pi\over 2}\tag7$$
This method it is a bit tedious, can someone tackle integral (1) with a quicker method?
 A: Through the residue theorem, by setting $f(x)=\frac{1}{\left(1+\frac{x^2}{\varphi^2}\right)\left(1+\frac{x^2}{\varphi^4}\right)}$ we have:

$$\begin{eqnarray*} I = \pi i \cdot\!\!\!\!\!\! \sum_{z\in\left\{i\varphi,i\varphi^2\right\}}\!\!\!\text{Res}(f(x),x=z)&=&\pi i\left(\frac{i\varphi^5}{2(\varphi^2-\varphi^4)}+\frac{i\varphi^4}{2(\varphi^4-\varphi^2)}\right)\\&=&\frac{\pi}{2}\cdot(\varphi^2-\varphi)\cdot\frac{\varphi^3}{\varphi^4-\varphi^2}\\&=&\frac{\pi}{2}\cdot 1\cdot 1 = \color{red}{\frac{\pi}{2}}.\end{eqnarray*} $$

An alternative is to prove a more general statement:
$$ \forall A,B>0,\qquad I(A,B)=\int_{0}^{+\infty}\frac{dx}{\left(1+\frac{x^2}{A^2}\right)\left(1+\frac{x^2}{B^2}\right)}=\frac{\pi}{2}\cdot\frac{AB}{A+B}$$
through Lagrange's identity and Glasser's master theorem, or the fact that the Fourier transform of a Cauchy distribution is a Laplace distribution. For instance, Lagrange and Glasser give:
$$\begin{eqnarray*}I(A,B)=\int_{0}^{+\infty}\frac{dx}{\left(\frac{A+B}{AB}x\right)^2+\left(1-\frac{x^2}{AB}\right)^2}&=&\frac{AB}{A+B}\int_{0}^{+\infty}\frac{dx}{x^2+\left(1-\frac{AB x^2}{(A+B)^2}\right)^2}\end{eqnarray*}$$
where $\int_{0}^{+\infty}\frac{dx}{x^2+(1-k^2 x^2)^2}$ constantly equals $\frac{\pi}{2}$ for any $k\in\mathbb{R}\setminus\{0\}$.
A: I think a quite simple approach is with the residue theorem. We have $$\begin{align}
\int_{0}^{\infty}\frac{1}{\left(1+\left(\frac{x}{\phi}\right)^{2}\right)\left(1+\left(\frac{x}{\phi^{2}}\right)^{2}\right)}dx= & \phi^{2}\int_{0}^{\infty}\frac{1}{\left(1+\phi^{2}x^{2}\right)\left(1+x^{2}\right)}dx \\
 = & \phi^{2}\int_{0}^{\infty}f\left(x\right)dx
\end{align}
 $$ so if we take as contour the semicircle with $\textrm{Im}\left(x\right)>0
 $ and with the diameter on the real axis from $-R
 $ to $R
 $. It is not difficult to see that the integral on the semicircle go to zero if $R\rightarrow\infty
 $. So $$\begin{align}
\int_{-\infty}^{\infty}\frac{1}{\left(1+\phi^{2}x^{2}\right)\left(1+x^{2}\right)}dx= &2\pi i\left(\textrm{Res}_{x=i}f\left(x\right)+\textrm{Res}_{x=\frac{i}{\phi}}f\left(x\right)\right) \\
 = & \pi\left(\frac{1}{1-\phi^{2}}+\frac{\phi}{\phi^{2}-1}\right)
\end{align}
 $$ then, by symmetry, $$\phi^{2}\int_{0}^{\infty}\frac{1}{\left(1+\phi^{2}x^{2}\right)\left(1+x^{2}\right)}dx=\frac{\pi}{2}\left(\frac{\phi^{2}}{1-\phi^{2}}+\frac{\phi^{3}}{\phi^{2}-1}\right)=\frac{\pi}{2}.$$
Addendum: Note that the above argument can be easily generalized.
A: Here is a hint. You may begin by substituing $x=\varphi \tan u$. Then play algebraically at the denominator (by factoring). Don't forget that
$$1+\tan^2 u =\sec^2 u$$
