- Use residues to calculate the integral
\[ \int _ { 0 } ^ { \infty } { \frac { d x } { x ^ { 4 } + 4 } } . \]
Solution
The value is \(\boxed{\pi/8}\).
1. The two upper-half-plane residues have sum \(-i/8\).
The four roots of \(z^4+4\) are \(1+i\), \(-1+i\), \(-1-i\) and \(1-i\). Each is simple since \(4z^3\ne0\) there. For \(F(z)=1/(z^4+4)\), the upper-half-plane residues are \[ \operatorname{Res}_{1+i}F=\frac1{4(1+i)^3} =\frac{-1-i}{16},\qquad \operatorname{Res}_{-1+i}F=\frac1{4(-1+i)^3} =\frac{1-i}{16}. \] Their sum is \(-i/8\).
2. The upper semicircle evaluates the integral.
For \(R>2\), integrate along \([-R,R]\) and the counterclockwise upper semicircle \(\Gamma_R\). The residue theorem gives the contour integral \(2\pi i(-i/8)=\pi/4\) [@SS03]. On \(\Gamma_R\), \(|z^4+4|\geq R^4-4\), hence \[ \left|\int_{\Gamma_R}F(z)\,dz\right| \leq\frac{\pi R}{R^4-4}\longrightarrow0. \] The real integrand is bounded near zero and bounded by \(|x|^{-4}\) for \(|x|\geq1\), so the full-line integral converges absolutely. Passing to the limit gives \[ \int_{-\infty}^{\infty}\frac{dx}{x^4+4}=\frac\pi4. \] Evenness of the integrand makes the required half-line integral equal to half this value, namely \(\pi/8\).