The residue theorem states that, if C is a closed path and a function f(*) has a finite number n of

*simple* poles inside C, is...

$\displaystyle \int_{C} f(z) = 2\pi i \sum_{i=1}^{n} r_{i}$ (1)

... where...

$\displaystyle r_{i} = \lim_{z \rightarrow z_{i}} (z-z_{i}) f(z) $ (2)

In our example is...

$\displaystyle f(z)= \frac{z^{2}}{(z^{2}+1)(z^{2}+4)}$ (3)

... C is the gray path in figure...

... for the Jordan's lermma is...

$\displaystyle \lim_{R \rightarrow \infty} \int_{ABC} f(z)\cdot dz= 0$ (4)

... so that, combining (1),(2),(3) and (4), is...

$\displaystyle \int_{-\infty}^{\infty}\frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)}= 2\pi i\cdot \sum_{i=1}^{n} r_{i}$ (5)

In our case f(*) has only two poles inside C : $\displaystyle z_{1}=i$, $\displaystyle z_{2}=2i$, so that is...

$\displaystyle r_{1}= \lim_{z \rightarrow i} (z-i)\cdot f(z)= \frac{i}{6}$

$\displaystyle r_{2}= \lim_{z \rightarrow 2i} (z-2i)\cdot f(z)= -\frac{i}{3}$ (6)

By (1) and (5) we have finally...

$\displaystyle \int_{-\infty}^{\infty} \frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)} = \frac{\pi}{3}$ (7)

... so that...

$\displaystyle \int_{0}^{\infty} \frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)} = \frac{\pi}{6}$ (8)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$