Problem 8

Proof. Consider the zeroes of the denominator lying in the top-half plane:

$\displaystyle \theta_1 = \frac{i+1}{\sqrt{2}} \quad \theta_2 = \frac{i-1}{\sqrt{2}}$

(7.42)

Enclose them in a semicircular path of radius $R$

and apply the residue theorem

$\displaystyle \int_{-R}^R f(z) + \int_\gamma f(z) = 2\pi i \operatorname{Res}(z=\theta_1,\theta_2)$

(7.43)

Setting $\gamma(t)=Re^{it}$ for $t\in[0,\pi]$ , we can see the arc integral vanishes as $R\to\infty$ :

$\displaystyle \left\vert\int_0^\pi f(\gamma(t))\gamma'(t)dt\right\vert$	$\displaystyle = \left\vert\int_0^\pi \frac{1+(Re^{it})^2}{1+(Re^{it})^4}Rie^{it} dt\right\vert$	(7.44)
	$\displaystyle \leq \int_0^\pi \frac{1+R^2}{R^4-1}Rdt$	(7.45)
	$\displaystyle = \pi \frac{1+R^2}{R^4-1}R\to 0$	(7.46)

Let us compute the residues now

$\displaystyle \operatorname{Res}(f,\theta_1)$	$\displaystyle = \lim_{z\to \theta_1} \frac{(z-\theta_1)(1+z^2)}{1+z^4} = \lim \frac{1+z^2}{4z^3}$	(7.47)
	$\displaystyle = \frac{1+i}{4i\frac{1+i}{\sqrt{2}}} = \frac{\sqrt{2}}{4i}$	(7.48)
$\displaystyle \operatorname{Res}(f,\theta_2)$	$\displaystyle = \frac{1-i}{4i\frac{1-i}{\sqrt{2}}} = \frac{\sqrt{2}}{4i}$	(7.49)

By taking limits we can see

$\displaystyle \int_{-\infty}^\infty f(z) = 2\pi i \left[\frac{\sqrt{2}}{2i}\right] =\pi\sqrt{2}$

(7.50)

The integral in question is half this because the integrand is even:

$\displaystyle \int_0^\infty \frac{1+x^2}{1+x^4} = \frac{\pi\sqrt{2}}{2}$

(7.51)

$\qedsymbol$