Elegant proof of $\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$?

Let $a, b > 0$ satisfy $a^2-4b^2 \geq 0$. Then: $$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$$ One way to calculate this is by computing the residues at the poles in the upper half-plane and integrating around the standard semicircle. However, the sum of the two residues becomes a complicated expression involving nested square roots, which magically simplifies to the concise expression above.

Sometimes such 'magical' cancellations indicate that there is a faster, more elegant method to reach the same result.

Is there a faster or more insightful way to compute the above integral?

Ok, I finally found a nice method. We have $$ \begin{align} \int_0^{\infty} \frac{dx}{x^4+ax^2+b^2} &= \int_0^{\infty} \frac{dx}{x^2}\frac{1}{(x-b/x)^2+2b+a} \\&= \frac{1}{b}\int_0^{\infty} \frac{dx}{(x-b/x)^2+2b+a} \\&= \frac{1}{b}\int_0^{\infty} \frac{dx}{x^2+2b+a} \\&= \frac{ \pi}{2b\sqrt{2b+a}} \end{align}$$ and the desired integral follows by symmetry.

Here the nontrivial step made use of the Cauchy-Schlömilch transformation (see e.g. here): if the integrals exist and $b > 0$, then $$\int_0^{\infty} f\left((x-b/x)^2\right)\, dx = \int_0^{\infty} f(x^2) \, dx$$ It is quite interesting that the above proof doesn't make use of the assumption that $a^2-4b^2 \geq 0$.

It's time to return the favor from here Ruben. (>‿◠)✌

First, we will prove $$\int_{-\infty}^\infty \frac{dx}{\beta^4x^4+2\beta^2\cosh(2\alpha)\,x^2+1}=\frac{\pi}{2\beta\cosh\alpha}$$ Note that $$\frac{1}{\beta^4x^4+2\beta^2\cosh(2\alpha)\,x^2+1}=\frac{1}{e^{2\alpha}-e^{-2\alpha}}\left[\frac{1}{\beta^2 x^2+e^{-2\alpha}}-\frac{1}{\beta^2 x^2+e^{2\alpha}}\right]$$ Hence \begin{align} \int_{-\infty}^\infty \frac{dx}{\beta^4x^4+2\beta^2\cosh(2\alpha)\,x^2+1}&=\frac{1}{e^{2\alpha}-e^{-2\alpha}}\left[\int_0^\infty\frac{dx}{\beta^2 x^2+e^{-2\alpha}}-\int_0^\infty\frac{dx}{\beta^2 x^2+e^{2\alpha}}\right]\\ &=\frac{1}{e^{2\alpha}-e^{-2\alpha}}\left[\frac{e^{\alpha}}{\beta}\arctan\left(e^{\alpha}x\right)-\frac{e^{-\alpha}}{\beta}\arctan\left(e^{-\alpha}x\right)\right]_{x=-\infty}^\infty\\ &=\frac{\pi}{\beta}\left[\frac{e^{\alpha}-e^{-\alpha}}{\left(e^{\alpha}-e^{-\alpha}\right)\left(e^{\alpha}+e^{-\alpha}\right)}\right]\\ &=\frac{\pi}{2\beta\cosh\alpha}\qquad\qquad\square \end{align} Now $$\int_{-\infty}^\infty \frac{dx}{x^4+\frac{2\cosh(2\alpha)}{\beta^2}\,x^2+\frac{1}{\beta^4}}=\frac{\beta^3\pi}{2\cosh\alpha}$$ Setting $a=\frac{2\cosh(2\alpha)}{\beta^2}$ and $b^2=\frac{1}{\beta^4}$, then using $\cosh\alpha=\sqrt{\frac{\cosh(2\alpha)+1}{2}}$ will give $$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}\qquad\qquad\square$$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{-\infty}^{\infty}{\dd x \over x^{4} + ax^{2} + b^{2}} ={\pi \over b\root{2b + a}}:\ {\large ?}.\qquad a, b\ >\ 0\,,\quad a^{2} - 4b^{2}\ \geq\ 0}$.

Indeed, this is essentially the @user111187 proof with the addition of some omitted details:

\begin{align}{\cal I}&\equiv\color{#66f}{\large \int_{-\infty}^{\infty}{\dd x \over x^{4} + ax^{2} + b^{2}}} =2\int_{0}^{\infty}{1 \over x^{2}}\,{\dd x \over x^{2} + a + b^{2}/x^{2}} \\[5mm]&\imp\quad \half\,{\cal I}=\int_{0}^{\infty}{1 \over x^{2}}\, {\dd x \over \pars{x - b/x}^{2} + 2b + a}\tag{1} \end{align}

With $\ds{{b \over x}\equiv t\ \imp\ x = {b \over t}}$ we'll have:

\begin{align}\half\,{\cal I}& =\int_{\infty}^{0}{t^{2} \over b^{2}}\, {-b\,\dd t/t^{2} \over \pars{b/t - t}^{2} + 2b + a} ={1 \over b}\int_{0}^{\infty}{\dd t \over \pars{b/t - t}^{2} + 2b + a}\tag{2} \end{align}

With $\pars{1}$ and $\pars{2}$:

\begin{align} \half\,b{\cal I}&=\int_{0}^{\infty}{b \over x^{2}}\, {\dd x \over \pars{x - b/x}^{2} + 2b + a} \\[5mm]\half\,b{\cal I}& =\int_{0}^{\infty}{\dd x \over \pars{x - b/x}^{2} + 2b + a} \\[5mm]\mbox{and}\ b{\cal I}&=\half\,b{\cal I} + \half\,b{\cal I} =\int_{0}^{\infty}{\pars{b/x^{2} + 1}\,\dd x \over \pars{x - b/x}^{2} + 2b + a} \end{align}

With $\ds{u \equiv x - {b \over x}\ \imp\ \dd u = \pars{1 + {b \over x^{2}}} \,\dd x}$ we'll get

\begin{align} b{\cal I}& =\int_{-\infty}^{\infty}{\dd u \over u^{2} + 2b + a} ={2 \over \root{2b + a}}\ \overbrace{\int_{0}^{\infty}{\dd u \over u^{2} + 1}} ^{\ds{\color{#c00000}{\pi \over 2}}}\ =\ {\pi \over \root{2b + a}} \\[5mm]\imp{\cal I}&\equiv\color{#66f}{\large \int_{-\infty}^{\infty}{\dd x \over x^{4} + ax^{2} + b^{2}} ={\pi \over b\root{2b + a}}} \end{align}