4
$\begingroup$

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 $k=1$.

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

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

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

$\endgroup$
  • $\begingroup$ In case it helps wolframalpha.com/input/… $\endgroup$ – Syrtis Major Apr 29 at 10:00
  • 2
    $\begingroup$ I'm afraid whipping out the result from an online calculator won't satisfy my maths teacher. $\endgroup$ – Velyth Apr 29 at 10:02
  • 3
    $\begingroup$ 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
2
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

$\endgroup$
  • $\begingroup$ nice work!!!!!!! $\endgroup$ – logo Apr 30 at 21:02
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.