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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?

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  • $\begingroup$ Partial Fraction is the algorithm that one may use to tackle this integral. Off the top of my head, I don't think there would be easier methods for this case. :) $\endgroup$ – H. R. Jun 12 '16 at 11:51
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    $\begingroup$ Perhaps an application of the Parseval theorem, but I don't know if that could be considered to be easier or quicker. $\endgroup$ – mickep Jun 12 '16 at 11:55
  • $\begingroup$ Personally, this is the fastest way I would solve this integral: By using partial fraction decomposition and then using the $\frac{1}{a^2+x^2}$ integral identity. It's a bit tedious, but integrals with quadratics in the denominator are always tedious in my opinion. $\endgroup$ – Noble Mushtak Jun 12 '16 at 12:02
  • $\begingroup$ A small note: In doing partial fraction decomposition, you don't need to have $A$ and $C$ in your ansatz, since your original function is even. Once that is noticed, the partial fraction decomposition can be done in your head. Now, I see no good reason (except, perhaps, learning new techniques) to use some other method. $\endgroup$ – mickep Jun 12 '16 at 18:26
  • $\begingroup$ You could simplify the path as mickep mentioned and also note that $$\phi^2(1+\phi^{-4}x^2)-(1+\phi^{-2}x^2)=\phi^2-1$$ $\endgroup$ – Math-fun Jun 13 '16 at 11:48
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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 $k\in\mathbb{R}\setminus\{0\}$.

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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.

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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

$$1+\tan^2 u =\sec^2 u$$

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  • $\begingroup$ After $u$-substitution, I get this integral, which gives us the correct answer if we take the integral from $u=0$ to $u=\frac \pi 2$, but I don't see how this integral is easier unless you know the identity of $\frac{a}{a+\tan^2(u)}$. $\endgroup$ – Noble Mushtak Jun 12 '16 at 12:37
  • $\begingroup$ A sorry you were asking for a more easir way. I read an alternative way. My bad :) $\endgroup$ – Tolaso Jun 12 '16 at 12:41
  • $\begingroup$ No problem. I am not the original questioner, I was just commenting here to check if I did what you were hinting at. $\endgroup$ – Noble Mushtak Jun 12 '16 at 12:50

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