Without using the residue formula, compute \[ \int_\RR f(x) \dx && f(x) \da {1\over x^4 + 16} .\]
Solution
Use a semicircular contour, noting the poles are at \(\pm \sqrt 2 \pm i\sqrt 2\). Write
- \(f_1(z) \da (\sqrt 2 + i\sqrt 2)f(z)\)
- \(f_2(z) \da (-\sqrt 2 + i\sqrt 2) f(z)\).
Break the curve up into two integrals \(I_1, I_2\) enclosing the poles, by Cauchy one gets
- For the loop around the right pole: \(I_1 = 2\pi i f_1(\sqrt 2 + i\sqrt 2) = {\pi \sqrt{2}(1-i) \over 32}\)
- For the loop around the left pole: \(I_2 = 2\pi i f_2(\sqrt 2 - i\sqrt 2) = {\pi \sqrt 2(1+i) \over 32}\).
Now show that \(\int_{C_R}\) vanishes: parameterize as \(\gamma(t) = Re^{it}\) and use the reverse triangle inequality: \[ \abs{ \int_{C_R} f} \leq \int_0^\pi {1\over R^4 - 16} = {\pi R \over R^4-16}\to 0 .\]