Evaluate \(\textstyle \int _ { - \infty } ^ { \infty } { \frac { 1 + x ^ { 2 } } { 1 + x ^ { 4 } } } d x\)
Solution
Set \[ F(z)=\frac{1+z^2}{1+z^4}. \] For \(R>1\), integrate \(F\) over the positively oriented upper semicircle with diameter \([-R,R]\).
1. The poles of \(F\) in the upper half-plane are\[ \zeta_1=e^{i\pi/4} \qquad\text{and}\qquad \zeta_2=e^{3i\pi/4}. \]
The poles are the roots of \[ z^4=-1=e^{i(\pi+2\pi k)}. \] Thus \[ z=e^{i(\pi/4+k\pi/2)}, \qquad k=0,1,2,3. \] Exactly the roots with \(k=0,1\) lie in the upper half-plane.
2. The residues at the two upper-half-plane poles are\[ \Res(F;\zeta_1) = -\frac{i}{2\sqrt2}, \qquad \Res(F;\zeta_2) = -\frac{i}{2\sqrt2}. \]
Each pole is simple, so for a root \(\zeta\) of \(1+z^4\), \[ \Res(F;\zeta) = \frac{1+\zeta^2}{4\zeta^3}. \]
For \(\zeta_1=e^{i\pi/4}\), \[ \zeta_1^2=i, \qquad \zeta_1^3=\zeta_2, \] and \(\abs{\zeta_2}=1\), so \[ \begin{aligned} \Res(F;\zeta_1) &= \frac{1+i}{4\zeta_2}\\ &= \frac{(1+i)(-1-i)}{4\sqrt2}\\ &= -\frac{i}{2\sqrt2}. \end{aligned} \]
For \(\zeta_2=e^{3i\pi/4}\), \[ \zeta_2^2=-i, \qquad \zeta_2^3=\zeta_1, \] so \[ \begin{aligned} \Res(F;\zeta_2) &= \frac{1-i}{4\zeta_1}\\ &= \frac{(1-i)^2}{4\sqrt2}\\ &= -\frac{i}{2\sqrt2}. \end{aligned} \]
3. The integral over the upper semicircular arc tends to zero as\(R\to\infty\).
On \(\abs{z}=R\), \[ \abs{1+z^2}\leq1+R^2 \] and \[ \abs{1+z^4}\geq R^4-1. \] Hence \[ \abs{F(z)} \leq \frac{1+R^2}{R^4-1}. \] The arc has length \(\pi R\), so \[ \abs{ \int_{\text{arc}}F(z)\,dz } \leq \frac{\pi R(1+R^2)}{R^4-1} \longrightarrow0. \]
4. The requested integral is\[ \boxed{ \int_{-\infty}^{\infty} \frac{1+x^2}{1+x^4}\,dx = \sqrt2\,\pi. } \]
By steps 1 and 2, the sum of the enclosed residues is \[ -\frac{i}{\sqrt2}. \] The residue theorem gives \[ \int_{C_R}F(z)\,dz = 2\pi i \left(-\frac{i}{\sqrt2}\right) = \sqrt2\,\pi. \] Split the contour into the real segment and the upper arc. Let \(R\to\infty\) and use step 3 to obtain the displayed improper integral.
5. Q.E.D.
Step 4 is the requested evaluation.