\[ \int_\RR {1 \over x^4 + 1} = {\pi \sqrt{2} \over 2} = {\pi \over \sqrt 2} .\]
Solution
The integrand \(f\in \bigo\qty{1\over z^4} \subseteq \bigo\qty{1\over z^{1+\eps}}\), so a semicircular contour works. Factoring the denominator: find a principal root \[ \omega^4 = -1 = e^{i\pi} \implies \omega = e^{i\pi\over 4} ,\] so \[ z^4 + 1 = (z-\omega)(z-\omega\zeta_4)(z-\omega\zeta_4^2)(z-\omega\zeta_4^3) \leadsto {\pi \over 4}, {3\pi \over 4}, {5\pi \over 4}, {7\pi \over 4} .\] So there are two simple poles in \(\HH\):
Write them as \(z_1 = e^{i\pi \over 4}\) and \(z_2 = e^{3 i \pi \over 4}\). Computing the residues: \[ \Res_{z=z_k} f(z) &= \lim_{z\to z_k} {z-z_k \over z^4 + 1} \\ &\eqLH \lim_{z\to z_k} {1\over 4z^3} \\ &= {z^{-3}\over 4 }\evalfrom_{z=z_k} ,\] so \[ 2\pi i \sum \Res_{z=z_k} &= {2\pi i \over 4}\qty{z_1^{-3} + z_2^{-3}} \\ &= {\pi i \over 2}\qty{e^{-3 i \pi \over 4} + e^{-9 i \pi \over 4}} \\ &= {\pi i \over 2}\qty{e^{-3 i \pi \over 4} + e^{-i \pi \over 4}} \\ &= {\pi i \over 2}\cdot e^{-2i\pi \over 4}\qty{e^{- i \pi \over 4} + e^{i \pi \over 4}} \\ &= \pi i\, {1\over i} \cos\qty{\pi \over 4} \\ &= {\pi \sqrt 2 \over 2} .\]