Paper 2, Section II, AComplex Analysis or Complex MethodsPart IB, 2012By a suitable choice of contour show that, for −1<α<1-1<\alpha<1−1<α<1,∫0∞xα1+x2 dx=π2cos(απ/2)\int_{0}^{\infty} \frac{x^{\alpha}}{1+x^{2}} \mathrm{~d} x=\frac{\pi}{2 \cos (\alpha \pi / 2)}∫0∞1+x2xα dx=2cos(απ/2)π