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I'm having a hard time working on this practice problem. It says:

Compute the integral: $$\int\limits_{-\infty}^{\infty}\dfrac{1}{y^4+1}\,\mathrm{d}y$$

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does the integral even include $dx$? –  Phani Raj Feb 25 '13 at 16:04
    
@PhaniRaj No, it's just as it's written. –  Tom Feb 25 '13 at 16:31
    
@Tom: So it has $x$ as a parameter? –  m0nhawk Feb 25 '13 at 16:46
    
@Tom which would imply you can take x^2-3 out? -> $$(x^2-3) \int_{-\infty}^{\infty} \frac{1}{y^4+4}dy$$ –  Bob Feb 25 '13 at 16:47
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To find the residues, you need not do Laurent expansions. But you do need to factor the denominator. –  GEdgar Feb 25 '13 at 22:27

2 Answers 2

up vote 1 down vote accepted

Don't be shed away by $y$ being the integration variable; you may call it $x$ as well. So we are told to compute $$\lim_{a\to\infty}\int_{-a}^a{dx\over x^4+1}\ .$$ Embed the $x$-axis into the complex plane and consider there the region $$\Omega:=\{z=x+iy\ |\ x^2+y^2< a^2,\ y>0\}\ .$$ Its boundary $\partial\Omega$ consists of the segment $[-a,a]$ on the real axis and a half circle of radius $a$. Apply to this situation the residue theorem $$\int_{\partial\Omega}{dz\over z^4+1}=2\pi i\sum_{\zeta\in\Omega} {\rm Res}\biggl(z\mapsto {1\over z^4+1}\biggm| \zeta\biggr)$$ and finally let $a\to\infty$.

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$$\int_{-\infty}^{\infty} \frac{1}{y^4+1}dy=\int_{-\infty}^{\infty} \frac{1}{y^4+ 2y^2 +1 - 2y^2}dy=$$

$$\int_{-\infty}^{\infty} \frac{1}{(y^4+ 2y^2 +1) - 2y^2}dy =\int_{-\infty}^{\infty} \frac{1}{(y^2 +1)^2 - 2y^2} dy$$

Can you proceed?

You should find $\frac{\pi}{\sqrt2}$

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