Let \(a \in \mathbb{R}\) with \(0<a<3\). Evaluate \[ \int_{0}^{\infty} \frac{x^{a-1}}{1+x^{3}} d x .\]
Solution
Write \(I\) for the integral, \(\zeta_3\da e^{2\pi i\over 3}, \omega_3 \da e^{i\pi\over 3}\). Take a indented semicircular wedge \(\Gamma\) at an angle of \(2\pi/3\), noting the pole at \(\omega_3 \da e^{i \pi \over 3}\):
Choosing a branch cut of \(\log\) along \(\theta = -\pi/2\), so \(\arg(z) \in (-\pi/2, 3\pi/2)\), this makes \(f(z) \da z^{\alpha-1}/(1+z^3)\) meromorphic on \(\Gamma\).
By the ML estimate, the integrals along \(C_\eps, C_R\) will vanish in the limit.
The contribution from the contour: parameterize \(\gamma_2\) as \(\ts{\zeta_3 t\st t\in [\eps, R]}\), then \[ \int_{\gamma_2}f(z) \dz &= \int_R^\eps {(\zeta_3 t)^{\alpha - 1} \over 1 + (\zeta_3 t)^3} \zeta_3\dt \\ &= \zeta_3 \zeta_3^{\alpha-1}\int_R^\eps {t^{\alpha-1} \over 1 + t^3}\dt \\ &\to -\zeta_3^\alpha I ,\] so the total contribution is \[ \oint_\Gamma f \to \qty{\int_{\gamma_1} + \int_{\gamma_2}}f = (1-\zeta_3^{\alpha}) I .\]
Computing the contributions from residues: \[ \Res_{z=\omega_3} f(z) &= \lim_{z\to \omega_3} {(z-\omega_3) z^{\alpha-1} \over 1+z^3} \\ &\eqLH \lim_{z\to\omega_3} {z^{\alpha-1} \over 3z^2} \\ &= {1\over 3} \omega_3^{\alpha-3} \\ &= -{1\over 3}\omega_3^\alpha .\]
Combining it all using the residue theorem: \[ 2\pi i \Res_{z=\omega_3} f(z) &= \oint_\gamma f \\ \implies I &= {-2\pi i\over 3} {\omega_3^\alpha \over 1-\zeta_3^\alpha } \\ &= {-2\pi i\over 3} {1 \over \omega_3^{-\alpha} - \omega_3^\alpha } \\ &= {-2\pi i\over 3} {1 \over -2i\sin\qty{\pi\alpha\over 3} } \\ &= {\pi\over 3}\csc\qty{\alpha\pi\over 3} .\]