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I want to show that: $$\int_{0}^{\infty} \frac{1}{(x^2+1)^2(x^2+4)}=\frac{\pi}{18}$$ so considering: $$\int_{\gamma} \frac{1}{(z^2+1)^2(z^2+4)}$$ where gamma is the curve going from $0$ to $-R$ along the real axis, from $-R$ to R via a semi-circle in the upper plane and then from $R$ to 0 along the real axis.

Using the residue theorem we have that: $$\int_{\gamma} \frac{1}{(z^2+1)^2(z^2+4)}=2\pi i \sum Res$$ so re-writing the integrand as $\displaystyle\frac{1}{(z-2i)(z+2i)(z+i)^2(z-i)^2}$

we can see that there is two simple poles at $2i$,$-2i$ and two poles of order 2 at $i$,$-i$. Calculating the residues: $$Res_{z=2i}=\lim_{z\rightarrow 2i} \displaystyle\frac{1}{(z+2i)(z+i)^2(z-i)^2}=\frac{1}{36i}$$

$$Res_{z=-2i}=\lim_{z\rightarrow 2i} \displaystyle\frac{1}{(z-2i)(z+i)^2(z-i)^2}=\frac{-1}{36i}$$

$$Res_{z=i}\lim_{z\rightarrow i} \frac{d}{dz} \frac{1}{(z-2i)(z+2i)(z+i)^2}=\frac{2i}{36}+\frac{2}{24i}$$

$$Res_{z=-i}\lim_{z\rightarrow -i} \frac{d}{dz} \frac{1}{(z-2i)(z+2i)(z-i)^2}=\frac{-2i}{36}+\frac{-2}{24i}$$

But now the sum of the residues is 0 and so when I integrate over my curve letting R go to $\infty$ (and the integral over top semi-circle goes to 0) I will just get 0?

Not sure what I've done wrong? Thanks very much for any help

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Note that only $2i$ and $i$ lie in the region bounded by $\gamma$. Therefore, the sum of the residue on the right hand side is the sum of the residue at $2i$ and $i$ only. – Paul Apr 26 '12 at 9:00
Of course, thanks very much! I was staring at my calculations of those residues for ages! That will give me $\frac{-i}{18}\times 2\pi i$ then the 2's cancel when I change the direction of one of the integrals and I get my result. Thanks very much again! – hmmmm Apr 26 '12 at 9:09
@Paul Maybe you can expand your comment as an answer. – Davide Giraudo Apr 26 '12 at 11:48
Since your coutour confusingly has winding number -1 around the two residues, you have to put a minus sign in the residue theorem. – GEdgar Jun 8 '12 at 19:11
Also note that you don't need contour integration to solve this integral. Partial fractions-like techniques would work. – bartgol Jun 19 '12 at 22:09
up vote 6 down vote accepted

Consider the contour $C$ that spans along $-R$ to $R$ and around the arc $Re^{i\theta}$ for $0\le\theta\le \pi$.



and we see the poles are located at $\pm i$ and $\pm 2i$. Letting $R \to \infty$, it is very clear that the denominator explodes, causing the integral around the arc to disappear. Then

$$\oint_C f(z)\, dz = 2\pi i(\operatorname*{Res}_{z = i}f(z) + \operatorname*{Res}_{z = 2i}f(z))$$

because $2i$ and $i$ are the only poles in $C$.
The pole of $i$ is of order 2:

$$ \operatorname*{Res}_{z = i}f(z) = \lim_{z \to i} \frac{1}{1!}\frac{d}{dz} (z-i)^2 f(z)= \lim_{z \to i} \frac{d}{dz}\frac{1}{(z+i)^2(z^2+4)}= \lim_{z \to i} \frac{2(2z^2 +iz+4)}{(i+z)^3(4+z^2)^2}=-\frac{i}{36} $$

The pole of $2i$ is simple:

$$ \operatorname*{Res}_{z = 2i}f(z) = \lim_{z \to 2i} (z-2i)f(z) = \frac{1}{(-4+1)^2(2i+2i)}=-\frac{i}{36} $$

So finally

$$ \int_0^\infty f(x)\, dx = \frac{1}{2}\int_{-\infty}^\infty f(x)\, dx = \pi i\left(-\frac{i}{36}-\frac{i}{36}\right) = \frac{\pi}{18} $$

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When using the residue theorem, you only consider residues enclosed by the path you are integrating over. In your case, you only consider residues in the upper half plane, as any point in the upper half plane will be enclosed by the semicircle as $R$ goes to infinity. Only $2i$ and $i$ lie in the upper half plane, out of the four poles of the function, so you only consider residues at those points. Your error comes from summing all of the residues, even ones that don't lie in the region bounded by the contour of integration.

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