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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.

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  • $\begingroup$ @AlexR 4th order Pole at $i\sqrt \alpha$, but I do not want to use residues. I want a real analytic method. Thanks $\endgroup$ – Jeff Faraci Apr 17 '14 at 20:12
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    $\begingroup$ Check this technique. $\endgroup$ – Mhenni Benghorbal Apr 17 '14 at 20:28
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    $\begingroup$ @Integrals I know this is only form of Beta function, but unfortunately I'm late Jeff. I handle another integral problem. $\endgroup$ – Tunk-Fey Apr 17 '14 at 20:28
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(#1\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{x^{4} \over \pars{\alpha + x^{2}}^4}\,\dd x ={\pi \over 32\alpha^{3/2}}\,,\quad \Re\pars{\root{\alpha}} > 0:\ {\large ?}}$.

\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}

$$ \color{#00f}{\large\int_{0}^{\infty}{x^{4} \over \pars{\alpha + x^{2}}^4}\,\dd x ={\pi \over 32\alpha^{3/2}}} $$

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  • $\begingroup$ Thanks a lot. I think you just forgot the factor of $\pi$ in the final result. +1 ! $\endgroup$ – Jeff Faraci Apr 27 '14 at 19:16
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    $\begingroup$ @Integrals Fine. I just fixed it. Thanks. $\endgroup$ – Felix Marin Apr 27 '14 at 21:46
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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, $$\int_0^\infty \frac{x^4}{(\alpha+x^2)^4}dx=\frac{\pi}{32\alpha^{3/2}}$$

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  • $\begingroup$ This is what I was looking for also. What a nice real method. Thanks a lot $\endgroup$ – Jeff Faraci May 7 '14 at 17:32
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    $\begingroup$ @Integrals Jeff, let me introduce you to Pranav. He is a friend of mine since I were an avid user on Brilliant.org. He is really good on calculus and physics problems. You can check his profile there. :) $\endgroup$ – Tunk-Fey May 9 '14 at 16:50
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    $\begingroup$ I am not even remotely close to you guys in Calculus, I am still learning. :3 $\endgroup$ – Pranav Arora May 9 '14 at 17:08
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    $\begingroup$ We are all still learning Pranav. Anyway, I also vote up almost all of your answers recently. Really nice answers. :) $$\\$$ I also always follow your posts Jeff. :) @Integrals $\endgroup$ – Tunk-Fey May 11 '14 at 16:26
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    $\begingroup$ @Tunk-Fey English is not my native tongue either. :) Thank you for making this point clear in the discussion about writing proper English and LaTex. $\endgroup$ – Jeff Faraci May 12 '14 at 7:44
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Small Hint) substituting $x=\alpha^{1/2}u$, you can pull the dependency on the parameter $\alpha$ outside the integral:

$$\int_0^\infty \frac{x^4}{(\alpha+x^2)^4}dx=\int_0^\infty \frac{\alpha^2u^4}{\alpha^4(1+u^2)^4}\alpha^{1/2}du=\frac{1}{a^{3/2}}\int_0^\infty \frac{u^4}{(1+u^2)^4}du.$$

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  • $\begingroup$ That is not really a hint at all. This is pretty standard, how would you handle the integral you have in terms of u? $\endgroup$ – Jeff Faraci Apr 17 '14 at 20:11
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    $\begingroup$ @Integrals Substitute $t=u^2$, and use $$B(x,y)=\int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}dx,$$ where $B(x,y)= \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$ is the beta function. $\endgroup$ – Meow Apr 17 '14 at 20:14
  • $\begingroup$ @Integrals Partial fractions would be my next step. I'd be curious to see if expanding in partial fractions before or after Alyosha's substitution is easier. $\endgroup$ – David H Apr 17 '14 at 20:18
  • $\begingroup$ @DavidH Thanks! $\endgroup$ – Jeff Faraci Apr 17 '14 at 20:22
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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.

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  • $\begingroup$ Before reading the name, I knew who did this post because of what you wrote:)! I guess this will suffice, but really I was looking for some clever substitutions not involving special functions. But I do not mind special functions, Thanks +1 $\endgroup$ – Jeff Faraci Apr 17 '14 at 20:42
  • $\begingroup$ See Felix solution above for what I was looking for. $\endgroup$ – Jeff Faraci Apr 28 '14 at 18:56
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    $\begingroup$ @Integrals: See Mick Jagger's reply to mathematical expectations. :-) $\endgroup$ – Lucian Apr 28 '14 at 19:31
  • $\begingroup$ If I am not mistaken, Keith Richards & Mick both wrote that song together. I do not think it was just Jagger who wrote that. :) $\endgroup$ – Jeff Faraci Apr 28 '14 at 22:11

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