The function


Has the following poles of order 2:

$$ z(k)=a \exp\left( \frac{\left(2k+1\right)}4 i\pi \right)$$

$f$ is even, therefore: $$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx =\frac{1}{2}\int _{-\infty }^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx$$

$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=i\pi \sum _k\:Res\left(f,\:z\left(k\right)\right)$$

$$Res\left(f,\:z\left(k\right)\right)=\lim _{z\to z\left(k\right)}\left(\frac{1}{\left(2-1\right)!}\left(\frac{d}{dz}\right)^{2-1}\frac{z^6\left(z-z\left(k\right)\right)^2}{\left(z^4+a^4\right)^2}\right)$$

$$z^4+a^4=z^4-z_k^4\implies\dfrac{z^6(z-z_k)^2}{(z^4+a^4)^2}=\dfrac{z^6}{(z^3+z_k z^2+z_k^2 z+z_k^3)^2}$$

$$Res\left(f,\:z_k\right)=\lim _{z\to \:z_k}\left(\frac{d}{dz}\left(\frac{z^6}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^2}\right)\right)$$

$$Res\left(f,\:z_k\right)=\frac{2z_kz^5\left(z^2+2z_kz+3z_k^2\right)}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^3}=\frac{2z_k^6\cdot 6z_k^2}{\left(4z_k^3\right)^3}$$


$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi }{16a}\sum _{k=0}^n\:e^{-\frac{\left(2k+1\right)}{4}i\pi }$$

We consider only the residues within the upper half plane, that is to say those corresponding to $k=0$ and $k=1$.

$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(e^{-\frac{i\pi }{4}\:\:}+e^{-\frac{3i\pi \:}{4}\:\:}\right)$$

$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(\frac{\sqrt{2}}{2}\:-i\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)$$

$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3\pi \sqrt{2}\:}{16a}$$

  • $\begingroup$ In case it helps wolframalpha.com/input/… $\endgroup$ – Syrtis Major Apr 29 at 10:00
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    $\begingroup$ I'm afraid whipping out the result from an online calculator won't satisfy my maths teacher. $\endgroup$ – Velyth Apr 29 at 10:02
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    $\begingroup$ I hope and wish that this would not satisfy yourself ! Cheers $\endgroup$ – Claude Leibovici Apr 29 at 10:11
  • $\begingroup$ Are you open to solutions that don't use contour integration? $\endgroup$ – J.G. Apr 29 at 10:19
  • $\begingroup$ I swear there was this exact same question in the review queue within the past couple of days. However I cannot find it via Approach$0$. Maybe it's closed and deleted.... or how about this older post? $\endgroup$ – Lee David Chung Lin Apr 29 at 10:26

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{x^{6} \over \pars{x^{4} + a^{4}}^{2}}\,\dd x} \,\,\,\stackrel{x/\verts{a}\ \mapsto\ x}{=}\,\,\, {1 \over \verts{a}}\int_{0}^{\infty}{x^{6} \over \pars{x^{4} + 1}^{2}}\,\dd x \\[5mm] \stackrel{\mrm{I.B.P.}}{=}\,\,\,&\ -\,{1 \over 4\verts{a}}\int_{x\ =\ 0}^{x\ \to\ \infty}x^{3} \,\dd\pars{1 \over x^{4} + 1} \\[5mm] = &\ {1 \over 4\verts{a}}\int_{0}^{\infty} {1 \over x^{4} + 1}\,\pars{3x^{2}}\,\dd x \\[5mm] = &\ {3 \over 4\verts{a}}\int_{0}^{\infty} {\dd x \over x^{2} + 1/x^{2}}\,\dd x = {3 \over 4\verts{a}}\int_{0}^{\infty} {\dd x \over \pars{x - 1/x}^{2} + 2} \\[5mm] = &\ {3 \over 8\verts{a}}\bracks{\int_{0}^{\infty} {\dd x \over \pars{x - 1/x}^{2} + 2} + \int_{\infty}^{0} {-\dd x/x^{2} \over \pars{1/x - x}^{2} + 2}} \end{align}

In the last line, the last integral is equivalent to the first one: It just arises from a $\ds{x \mapsto 1/x}$ change of variable.

Then, \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{x^{6} \over \pars{x^{4} + a^{4}}^{2}}\,\dd x} = {3 \over 8\verts{a}}\int_{0}^{\infty} {1 + 1/x^{2} \over \pars{x - 1/x}^{2} + 2}\,\dd x \\[5mm] \stackrel{x - 1/x\ \mapsto\ x}{=}\,\,\, &\ {3 \over 8\verts{a}}\int_{-\infty}^{\infty} {\dd x \over x^{2} + 2} = {3 \over 8\verts{a}}\,{1 \over 2}\,\root{2} \int_{-\infty}^{\infty}{\dd x/\root{2} \over \pars{x/\root{2}}^{2} + 1} \\[5mm] \stackrel{x/\root{2}\ \mapsto\ x}{=}\,\,\, &\ {3\root{2} \over 16\verts{a}}\ \underbrace{\int_{-\infty}^{\infty}{\dd x \over x^{2} + 1}} _{\ds{=\ \pi}}\ =\ \bbx{{3\root{2}\pi \over 16}\,{1 \over \verts{a}}} \end{align}

  • $\begingroup$ nice work!!!!!!! $\endgroup$ – logo Apr 30 at 21:02

If you want an approach that doesn't require as much differentiation, first use partial fractions to write $\dfrac{z^3}{z^4+a^4}=\sum_{k=0}^3\dfrac{c_k}{z-z(k)}$ so $$\frac{z^6}{(z^4+a^4)^2}=\sum_k\frac{c_k^2}{(z-z(k))^2}+2\sum_{k<l}\frac{c_kc_l}{(z-z(k))(z-z(l))}\\=\sum_k\frac{c_k^2}{(z-z(k))^2}+2\sum_{k<l}\frac{c_kc_l}{z(k)-z(l)}\left(\frac{1}{z-z(k)}-\frac{1}{z-z(l)}\right).$$The first sum doesn't contribute to $\int_{\Bbb R}\dfrac{x^6 dx}{(x^4+a^4)^2}$, but some of the latter sum's terms do, viz. $$\oint\frac{dz}{(z-w)^{n+1}}=2\pi i\delta_{n0}$$for enclosed $w$. Hence$$\int_0^\infty\frac{x^6 dx}{(x^4+a^4)^2}=\pi i\sum_{k<l}\frac{c_kc_l}{z(k)-z(l)}\left([k\in S]-[l\in S]\right),$$where $\{z(k)|k\in S\}$ is the set of residues your contour encloses and $[]$ is the Iverson bracket, i.e. $[k\in S]$ is $1$ if $k\in S$ or $0$ otherwise. If you're experienced with Beta functions, you should try calculating the integral separately with $x=a\tan^{1/2}t$ to double-check you get the same answer twice. (I get $\frac{3\pi}{8a\sqrt{2}}$.)

  • $\begingroup$ Thanks for putting forth this approach but I'm really supposed to proceed as I already did, can you tell me if my work up there is close to the answer? $\endgroup$ – Velyth Apr 29 at 14:26
  • $\begingroup$ I haven't checked your residues yet, but the approach I describe is basically the same; the reason residues have the formula you know they do is because of the loop integral I gave. If on the other hand you're sure your teacher doesn't want such a "first principles" approach, I guess you'll have to do things the way you already do them. $\endgroup$ – J.G. Apr 29 at 15:03

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