Evaluar

Question

La función

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

Tiene los siguientes polos de orden 2:

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

$f$ es aún, por lo tanto: $$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx =\frac{1}{2}\int _{-\infty }^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx$$

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

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

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

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

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

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

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

Sólo tenemos en cuenta los residuos dentro de la mitad superior del plano, es decir, los correspondientes a $k=0$ e $k=1$.

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

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

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

Answer 1

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

En la última línea, la última integral es equivalente a la primera: sólo surge a partir de una $\ds{x \mapsto 1/x}$ cambio de variable.

A continuación, \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}

Answer 2

Si desea un enfoque que no requiera tanta diferenciación, primero use fracciones parciales para escribir $\dfrac{z^3}{z^4+a^4}=\sum_{k=0}^3\dfrac{c_k}{z-z(k)}$ para que $$\frac{z^6}{(z^4+a^4)^2}=\sum_k\frac{c_k^2}{(z-z(k))^2}+2\sum_{k<l}\frac{c_kc_l}{(z-z(k))(z-z(l))}\\=\sum_k\frac{c_k^2}{(z-z(k))^2}+2\sum_{k<l}\frac{c_kc_l}{z(k)-z(l)}\left(\frac{1}{z-z(k)}-\frac{1}{z-z(l)}\right).$$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 \ en S \}$ is the set of residues your contour encloses and $ []$ is the Iverson bracket, i.e. $ [k \ en S]$ is $ 1$ if $ k \ en 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}} $ .)