Hi I am trying to show $$ \int_0^\infty \frac{x^4}{(\alpha+x^2)^4}dx=\frac{\pi}{32\alpha^{3/2}},\quad \Re(\sqrt \alpha)> 0. $$ I am looking for a solution to this NOT using contour integration, but real analysis methods. Thanks.
For the complex case, we have poles at $\pm i\sqrt\alpha$, 4th order.
Use the substitution $x=\sqrt{\alpha}\tan\theta$, to get: $$\int_0^{\pi/2} \frac{\tan^4\theta \sec^2\theta}{\alpha^{3/2} \sec^8\theta}\,d\theta=\frac{1}{\alpha^{3/2}}\int_0^{\pi/2}\sin^4\theta \cos^2\theta\,d\theta$$
It can be easily shown that: $$\int_0^{\pi/2}\sin^4\theta \cos^2\theta\,d\theta=\frac{\pi}{32}$$ Hence, $$\int_0^\infty \frac{x^4}{(\alpha+x^2)^4}dx=\frac{\pi}{32\alpha^{3/2}}$$
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(#1\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{x^{4} \over \pars{\alpha + x^{2}}^4}\,\dd x ={\pi \over 32\alpha^{3/2}}\,,\quad \Re\pars{\root{\alpha}} > 0:\ {\large ?}}$.
\begin{align} \color{#c00000}{\int_{0}^{\infty}{x^{4} \over \pars{\alpha + x^{2}}^4}\,\dd x} &=\half\int_{0}^{\infty}{x^{3/2} \over \pars{\alpha + x}^4}\,\dd x ={1 \over 4}\int_{0}^{\infty}{x^{1/2} \over \pars{\alpha + x}^{3}}\,\dd x \\[3mm]&={1 \over 16}\int_{0}^{\infty}{x^{-1/2} \over \pars{\alpha + x}^{2}} \,\dd x \end{align}
With $\ds{t \equiv x^{1/2}\quad\imp\quad x = t^{2}\,,\quad \dd x = 2t\,\dd t}$: \begin{align} \color{#c00000}{\int_{0}^{\infty}{x^{4} \over \pars{\alpha + x^{2}}^4}\,\dd x} &={1 \over 8}\int_{0}^{\infty}{\dd t \over \pars{\alpha + t^{2}}^{2}} ={1 \over 8\alpha^{3/2}}\lim_{\mu \to \infty}\int_{0}^{\mu/\root{\alpha}} {\dd t \over \pars{1 + t^{2}}^{2}} \\[3mm]&={1 \over 8\alpha^{3/2}}\lim_{\mu \to \infty} \int_{0}^{\mu\pars{\root{\alpha}}^{*}} {\dd t \over \pars{1 + t^{2}}^{2}} ={1 \over 8\alpha^{3/2}}\int_{0}^{\infty}{\dd t \over \pars{1 + t^{2}}^{2}} \end{align}
We'll replace $\ds{t = \root{\alpha}\tan\pars{\theta}}$: \begin{align} \color{#c00000}{\int_{0}^{\infty}{x^{4} \over \pars{\alpha + x^{2}}^4}\,\dd x} &={1 \over 8\alpha^{3/2}}\int_{0}^{\pi/2}\cos^{2}\pars{\theta}\,\dd\theta ={1 \over 8\alpha^{3/2}}\int_{0}^{\pi/2}{1 + \cos\pars{2\theta} \over 2}\,\dd\theta \\[3mm]&={1 \over 8\alpha^{3/2}}\,\pars{\half\,{\pi \over 2}} \end{align}
$$ \color{#00f}{\large\int_{0}^{\infty}{x^{4} \over \pars{\alpha + x^{2}}^4}\,\dd x ={\pi \over 32\alpha^{3/2}}} $$
Small Hint) substituting $x=\alpha^{1/2}u$, you can pull the dependency on the parameter $\alpha$ outside the integral:
$$\int_0^\infty \frac{x^4}{(\alpha+x^2)^4}dx=\int_0^\infty \frac{\alpha^2u^4}{\alpha^4(1+u^2)^4}\alpha^{1/2}du=\frac{1}{a^{3/2}}\int_0^\infty \frac{u^4}{(1+u^2)^4}du.$$
where $B(x,y)= \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$ is the beta function.
– Meow Apr 17 '14 at 20:14I am looking for a solution to this NOT using contour integration, but real analysis methods.
All integrals of the form $\displaystyle\int_0^\infty\frac{x^n}{(1+x^m)^p}$ are solved by letting $t=\dfrac1{(1+x^m)^p}$ , then recognizing the expression of the beta function in the new integral, and using Euler's reflection formula for the $\Gamma$ function. In this case, we first have to factor $\alpha$ outside of the parenthesis, and let $y^m=\dfrac{x^m}\alpha$. But you already knew this, since you said that you are familiar with many of my other answers on the same topic.
