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$$\int_{-\infty}^{\infty} \frac{x^2}{x^6+9}dx$$ I'm a bit puzzled as how to go about solving this integral. I can see that it isn't undefined on -infinity to infinity. But I just need maybe a hint on how to go about solving the problem.

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In general, I'm opposed to editing people's questions to correct them on issues like adding a $dx$; that's something that can be pointed out (e.g. in the comments) so they're aware of the change. Not a big problem though. –  Zev Chonoles Mar 2 '13 at 22:38
    
@ZevChonoles If you are talking to, ok, I will say next time that I add $dx$ (for example) in OP question/title...Maybe you say something different, sorry, my English is poor –  Cortizol Mar 2 '13 at 22:45

3 Answers 3

Hint: put $\;u = x^3$, so $\,du \;=\; 3x^2\, dx \;\implies\; x^2\, dx \;= \;\frac 13\, du$

This gives you $$\frac 13 \int_{-\infty}^\infty \frac {du}{u^2 + (3)^2} $$

Look familiar?:

Using one more substitution, let $\quad u =3\tan(\theta),\quad du =3\sec^2(\theta)\,d\theta, \quad \theta=\arctan\left(\frac{u}{3}\right)$

And determine the corresponding bounds for integrating.

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if u=x^3 dont you mean du=3x^2dx? –  EhBabay Mar 2 '13 at 22:53
    
Thanks, @JohnCarpenter: typo fixed!! –  amWhy Mar 2 '13 at 23:02
    
Does this make sense? –  amWhy Mar 2 '13 at 23:30
    
It does but I'm hazy on how I need to fix my new upper and lower limits. –  EhBabay Mar 4 '13 at 14:09

It is natural to let $u=x^3$. But note that the substitution $x^3=3u$ is more efficient.

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With complex analysis: define the complex function

$$f(z)=\frac{z^2}{z^6+9}\;\;,\;\;\text{with simple poles at}\,\,\,z_k:=\sqrt[6] 9\,\,e^{\frac{\pi i}{6}(1+2k)}\;\;,\;k=0,1,2,3,4,5$$

Note that the only poles on the upper half plane are the first three ones $\,z_0,\,z_1,\,z_2\,$ , with residues

$$Res_{z=z_0}(f)=Res_{z=z_2}(f)=-\frac{i}{18}\;\;,\;\;\;Res_{z=z_1}(f)=\frac{i}{18}$$

Choosing the contour

$$C_R:=[-R,R]\cup\gamma_R:=\{z\in\Bbb C\;;\;|z|=R\,,\,\Im (z)\ge 0\}$$

we get, by Cauchy's Integral Theorem, that

$$(**)\;\;\;\;\;\;\;2\pi i\left(\sum_{k=0}^2 Res_{z=z_k}(f)\right)=\frac{\pi}{9}=\oint_{C_R}f(z)\,dz=\int\limits_{-R}^Rf(x)\,dx+\int\limits_{\gamma_R} f(z)\,dz$$

But

$$\left|\int\limits_{\gamma_R} f(z)\,dz\right|\le\max_{z\in \gamma_R}\frac{|z|^2}{|z^6+9|}\pi R\le\frac{\pi R^3}{R^6-9}\xrightarrow[R\to\infty]{}0$$

Thus passing to the limit in (**) above we get

$$\int\limits_{-\infty}^\infty\frac{x^2}{x^6+9}dx=\frac{\pi}{9}$$

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Nice job, +1. You miss a $\mbox{Im}\;z\geq 0$ in the definition of your $\gamma_R$. I think. –  1015 Mar 3 '13 at 3:52
    
Indeed so, @julien . Thanks. –  DonAntonio Mar 3 '13 at 3:53

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