Integral $I=\int_0^\infty \frac{x^4}{(\alpha+x^2)^4}dx$ 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.
 A: 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,
$$\int_0^\infty \frac{x^4}{(\alpha+x^2)^4}dx=\frac{\pi}{32\alpha^{3/2}}$$
A: $\newcommand{\+}{^{\dagger}}
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$\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}

$$
\color{#00f}{\large\int_{0}^{\infty}{x^{4} \over \pars{\alpha + x^{2}}^4}\,\dd x
={\pi \over 32\alpha^{3/2}}}
$$
A: Small Hint) substituting $x=\alpha^{1/2}u$, you can pull the dependency on the parameter $\alpha$ outside the integral:
$$\int_0^\infty \frac{x^4}{(\alpha+x^2)^4}dx=\int_0^\infty \frac{\alpha^2u^4}{\alpha^4(1+u^2)^4}\alpha^{1/2}du=\frac{1}{a^{3/2}}\int_0^\infty \frac{u^4}{(1+u^2)^4}du.$$
A: 
I 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.
