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I wish to evaluate

$$I=\int_{-\infty}^{\infty}\frac{dx}{(x^4+a^4)^2}$$

for a positive real number $a$. I believe the answer is $\frac{3\sqrt{2}\pi}{8a^7}$.

But I am not sure how to arrive at this. I know that $I=2\pi i\sum_{Im(s)>0}\text{res}_f(s)$ where $f=\frac{1}{(x^4+a^4)^2}$, but I am having a hard time directly computing these residues. I know that I only need to consider the residues at the points $ai(1+i)/\sqrt{2}$ and $a(1+i)/\sqrt{2}$ since these are the two poles where $Im(s)>0$, but I cannot easily compute the residues at these two points. I know that these are poles of order $2$, so I can take the holomorphic continuation $g$ of $(z-s)^2f(z)$ and compute $\text{res}_f(s)=g'(a)$, but this becomes so computationally intensive that I am certain there must a better method.

Ideally, I would like to see a residue-theoretic approach to the solution.

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I don't think residues is all that bad here. We have $x^4=-a^4=a^4e^{(2n+1)\pi i}$ so $$x_n=ae^{(2n+1)\pi i/4}=a\omega_n$$ Then to find the residue at $x_n$ we need as you said, $$\begin{align}R_n&=\lim_{x\rightarrow x_n}\frac d{dx}\left(\frac{x-x_n}{x^4+a^4}\right)^2=\lim_{x\rightarrow x_n}2\left(\frac{x-x_n}{x^4+a^4}\right)\frac{(x^4+a^4)-(x-x_n)(4x^3)}{(x^4+a^4)^2}\\ &=\lim_{x\rightarrow x_n}2\left(\frac{x-x_n}{x^4+a^4}\right)\frac{4x^3-4x^3-12x^2(x-x_n)}{2(x^4+a^4)(4x^3)}\\ &=\lim_{x\rightarrow x_n}-\frac3x\left(\frac{x-x_n}{x^4+a^4}\right)^2=\lim_{x\rightarrow x_n}-\frac3x\left(\frac1{4x^3}\right)^2=-\frac3{16a^7\omega_n^7}=-\frac{3\omega_n}{16a^7}\end{align}$$ Where we have used L'Hopital's rule on the second and third lines. So $$\begin{align}\int_{-\infty}^{\infty}\frac{dx}{(x^4+a^4)^2}&=2\pi i\left(-\frac3{16a^7}\right)\left(e^{\pi i/4}+e^{3\pi i/4}\right)\\ &=-\frac{3\pi i}{8a^7}\left(2i\sin\left(\frac{\pi}4\right)\right)=\frac{3\pi\sqrt2}{8a^7}\end{align}$$ I like @Quanto's solution up to where he got $$\begin{align}\int_{-\infty}^{\infty}\frac{dt}{t^4+1}&=2\int_0^{\infty}\frac{dt}{t^4+1}=\frac12\int_0^{\infty}\frac{u^{-3/4}du}{u+1}=\frac12\int_1^{\infty}\frac{(v-1)^{-3/4}dv}{v}\\ &=\frac12\int_0^1(1-w)^{-3/4}w^{-1/4}dw=\frac12\operatorname{B}\left(\frac14,\frac34\right)=\frac12\frac{\operatorname{\Gamma}\left(\frac14\right)\operatorname{\Gamma}\left(\frac34\right)}{\operatorname{\Gamma}(1)}\\ &=\frac12\frac{\pi}{\sin\left(\frac{\pi}4\right)}=\frac{\pi}{\sqrt2}\end{align}$$ And of course the $t$-integral is easier to do by residues than the original $x$-integral, and the $u$-integral is rather famously done by residues although it can be done without residues or infinite series.

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Use $\left(\frac{x}{x^4+a^4}\right)' = \frac{4a^4}{(x^4+a^4)^2} -\frac{3}{x^4+a^4}$ to write the integral as,

$$I=\frac1{4a^4} \left(\frac{x}{x^4+a^4}\bigg|_{-\infty}^{\infty}+ \int_{-\infty}^{\infty}\frac{3dx}{x^4+a^4} \right) =\frac3{8a^7}\int_{-\infty}^{\infty}\frac{2}{t^4+1} dt \tag 1$$

where the substitution $t= \frac xa$ is made and,

$$\int_{-\infty}^{\infty}\frac{2}{t^4+1} =\int_{-\infty}^{\infty}\frac{1+t^2}{t^4+1}dt +\int_{-\infty}^{\infty}\frac{1-t^2}{t^4+1}dt $$ $$=2\int_{0}^{\infty}\frac{d(t-\frac1t)}{(t-\frac1t)^2+2} -2\int_{0}^{\infty}\frac{d(t+\frac1t)}{(t+\frac1t)^2-2} $$ $$=\sqrt2 \tan^{-1} \frac{t-\frac1t}{\sqrt2} \bigg|_{0}^{\infty} +\sqrt2 \coth^{-1} \frac{t+\frac1t}{\sqrt2} \bigg|_{0}^{\infty} = \sqrt2 \pi + 0$$

Plug into (1) to obtain

$$I =\frac{3\sqrt2\pi}{8a^7}$$

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  • $\begingroup$ Definitely looks sound, and I upvoted as appropriate. Unfortunately, I am looking for a residue theoretic approach, so I will hold off on accepting unless there are no better answers. $\endgroup$ – user656966 Mar 5 at 3:47
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$\int\frac{1}{(x^4 + a^4)^2}dx \\= \frac{1}{32a^7} (-3 \sqrt2 log(a^2 - \sqrt2 a x + x^2) + 3 \sqrt2 log(a^2 + \sqrt2 a x + x^2) + \frac{8 a^3 x}{a^4 + x^4} - 6 \sqrt2 tan^{-1}(1 -\frac{\sqrt2 x}{a}) + 6 \sqrt2 tan^{-1}(\frac{\sqrt2 x}{a} + 1)) + constant$

Let $I=\int_{-\infty}^{\infty}\frac{dx}{(x^4+a^4)^2}$

$I= \frac{3a^{-7}\pi}{4\sqrt2} $ for $Re(a^4)\geq0 ∨ a^4$$R$

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