Evaluate $\int _0^{\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx$

Question

The function

$$f\left(z\right)=\frac{z^6}{\left(z^4+a^4\right)^2}$$

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

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

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

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

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

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

$$Res\left(f,\:z_k\right)=\frac{12z_k^8}{64z_k^9}=\frac{3}{16z_k}$$

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

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

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

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

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

Answer 1

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 $\frac{3\pi}{8a\sqrt{2}}$.)

Answer 2

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

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{6}\over \pars{x^{4} + a^{4}}^{2}}\,\dd x} \\[5mm] \stackrel{x\ =\ \verts{a}t^{1/4}}{=}\,\,\,&\ {1 \over 4\verts{a}}\int_{0}^{\infty}{t^{\color{red}{7/4} - 1} \over \pars{1 + t}^{2}}\,\dd t \end{align} Note that \begin{align} {1 \over \pars{1 + t}^{2}} & = \sum_{k = 0}^{\infty}{-2 \choose k}t^{k} = \sum_{k = 0}^{\infty}{k + 1 \choose k} \pars{-1}^{k}\,t^{k} \\[5mm] & = \sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{2 + k}} {\pars{-t}^{k} \over k!} \end{align} With the Ramanujan's Master Theorem: \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{6}\over \pars{x^{4} + a^{4}}^{2}}\,\dd x} = {1 \over 4\verts{a}}\Gamma\pars{7 \over 4} \Gamma\pars{2 - {7 \over 4}} \\[5mm] = &\ {1 \over 4\verts{a}}{3 \over 4}\Gamma\pars{3 \over 4} \Gamma\pars{1 \over 4} = {3 \over 16\verts{a}}\,{\pi \over \sin\pars{\pi/4}} \\[5mm] = &\ \bbx{{3\root{2}\pi \over 16}\,{1 \over \verts{a}}} \\ & \end{align}

Answer 4

Consider $x=a \tan^{1/2}u$ to obtain$$\frac{1}{2|a|}\int_0^{\pi/2}\cos^{-1/2}u\sin^{5/2}udu.$$ Now recall the Beta fucntion.