# 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?

• For what it's worth: using partial fractions also leads to a sum of two arctan integrals, where one has to simplify the result of the integration "by hand" afterwards to get the nice final form. – Hans Lundmark May 1 '14 at 11:02

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 $a^2-4b^2 \geq 0$.

• Very nice. I like this trick a lot. – Cameron Williams May 7 '14 at 17:11
• Indeed, you omitted several steps as compared to the cited link. – Felix Marin Dec 1 '14 at 19:41

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 $$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}\qquad\qquad\square$$

• +1 Very nice! I hope you have these results compiled somewhere. – nbubis Nov 24 '14 at 14:15


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}

Let $b=\frac{a}{2}\sin2\theta$ ($0<\theta<\frac{\pi}{2}$), then \begin{eqnarray} \int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + \frac14a^2\sin^22\theta}&=&\int_{-\infty}^{\infty}\frac{dx}{(x^2+a\sin^2\theta)(x^2+a\cos^2\theta)}\\ &=&\frac{1}{a(\cos^2\theta-\sin^2\theta)}\int_{-\infty}^{\infty}\left(\frac{1}{x^2+a\sin^2\theta}-\frac{1}{x^2+a\cos^2\theta}\right)dx\\ &=&\frac{1}{a(\cos^2\theta-\sin^2\theta)}\left(\frac{1}{\sqrt{a}\sin\theta}\arctan\frac{1}{\sqrt{a}\sin\theta}x\right.\\ &&-\left.\frac{1}{\sqrt{a}\cos\theta}\arctan\frac{1}{\sqrt{a}\cos\theta}x\right)\bigg|_{-\infty}^\infty\\ &=&\frac{1}{a(\cos^2\theta-\sin^2\theta)}\left(\frac{1}{\sqrt{a}\sin\theta}-\frac{1}{\sqrt{a}\cos\theta}\right)\pi\\ &=&\frac{\pi}{a\sqrt{a}(\cos\theta+\sin\theta)\sin\theta\cos\theta}\\ &=&\frac{\pi}{b\sqrt{2b+a}}. \end{eqnarray}