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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

$$ \frac{1}{u^{4} + \left(4 \zeta^{2} - 2\right)u^{2} + 1} = \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left(\frac{1}{u^2+a_1{}^2} - \frac{1}{u^2+a_2{}^2}\right) $$

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  • $\begingroup$ The denominator is quadratic in $u^2$. Factor the denominator, then use a partial fraction decomposition and all you have is two arctan integrals like the one you mention (the only difference is that there will be ugly constant everyone). $\endgroup$
    – User8128
    Jun 17, 2020 at 22:25
  • $\begingroup$ yeah, i just remembered PFE and came to the same conclusion. $\endgroup$
    – Jason S
    Jun 17, 2020 at 22:32
  • $\begingroup$ what is this RootSum junk? $\endgroup$
    – Jason S
    Jun 17, 2020 at 22:34
  • $\begingroup$ The definite integral will be somehow expressible as a function of the roots of that polynomial in the denominator and I think that's what that RootSum is getting at; so it is probably correct, albeit completely unhelpful since it looks very complicated. An alternate way to evaluate the integral is using the residue theorem from Complex analysis which roughly tells you in this case that the integral is $$2\pi i \times \sum \text{Res}\left(\frac{1}{u^4 + (4\zeta^2-2)u^2+1}; \text{upper half-plane}\right),$$ and the residues depend immediately on the roots of that polynomial. $\endgroup$
    – User8128
    Jun 17, 2020 at 22:39
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    $\begingroup$ @JasonS it's a sum over the roots of the polynomial. For example, RootSum[#^2 + # &, Exp[#] &] means, "add up $e^$ the roots of $x^2+x=0$", giving $1+e^{-1}$. $\endgroup$
    – Integrand
    Jun 17, 2020 at 22:39

3 Answers 3

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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}

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    $\begingroup$ how did you get to the $\frac{\pi}{2\zeta} - 0$ step? I followed you right up until there. -- OH WAIT the integral limits are the same, $[\infty,\infty]$ $\endgroup$
    – Jason S
    Jun 18, 2020 at 1:54
  • $\begingroup$ ...and there's some sneaky subtleties with the integral limits, not sure i follow how $u=[0,\infty]$ turns into $t=[-\infty,\infty]$ $\endgroup$
    – Jason S
    Jun 18, 2020 at 1:57
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    $\begingroup$ hmm i guess the open interval works: $u = (0, \infty) \mapsto t = (-\infty, \infty)$ $\endgroup$
    – Jason S
    Jun 18, 2020 at 2:00
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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*}

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  • $\begingroup$ how does that last step work to get $\pi/2|\zeta|$ ? $\endgroup$
    – Jason S
    Jun 18, 2020 at 1:50
  • $\begingroup$ oh, you have a typo, the last integral should be $dt$, not $du$ $\endgroup$
    – Jason S
    Jun 18, 2020 at 2:02
  • $\begingroup$ @JasonS, Thank you, I corrected the typo. Also, as for your first question, I thought you already know how to compute it. (Your own answer contains an integration formula for this.) $\endgroup$ Jun 18, 2020 at 12:18
  • $\begingroup$ The first question arose from my confusion in the typo; I wasn't sure how to integrate $du/(t^2+4\zeta^2)$ $\endgroup$
    – Jason S
    Jun 18, 2020 at 14:55
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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 $I(\zeta) = \frac{\pi}{2\zeta}.$

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    $\begingroup$ Note that $$ 2\zeta^2 - 1 \pm 2\zeta\sqrt{\zeta^2-1} = \left( \zeta \pm \sqrt{\zeta^2-1} \right)^2. $$ $\endgroup$ Jun 17, 2020 at 23:34
  • $\begingroup$ yeah, i got there eventually, see math.stackexchange.com/a/3724215/120 $\endgroup$
    – Jason S
    Jun 17, 2020 at 23:39

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