Evaluating Contour Integral using Residue Theorem

[haruspex]

No-one seems to like my approach, but I shall persist anyway.:shy:
$I_1 = \int_{r=0}^\infty f(r).dr$
$I_2 = \lim_{R→\infty} \int_{\theta=0}^{2\pi/3} f(R e^{i\theta}).dR e^{i\theta} = 0$
$I_3 = \int_{r=\infty}^0 f(r e^{2\pi i/3}).d(r e^{2\pi i/3}) = \int_{r=\infty}^0 \frac{(r e^{2\pi i/3})^{\frac 1 2}}{1+(r e^{2\pi i/3})^3}e^{2\pi i/3}.dr= -\int_{r=0}^\infty \frac{\sqrt r e^{\pi i/3}e^{2\pi i/3}}{1+r^3}.dr = \int_{r=0}^\infty \frac{\sqrt r }{1+r^3}.dr = I_1$
Since the residue at the only pole inside the contour is -i/3, 2 I₁ = I₁+I₂ + I₃ = 2πi * (-i/3), so I₁ = π/3.

 

[jackmell]

No-one seems to like my approach, but I shall persist anyway.:shy:
$I_1 = \int_{r=0}^\infty f(r).dr$
$I_2 = \lim_{R→\infty} \int_{\theta=0}^{2\pi/3} f(R e^{i\theta}).dR e^{i\theta} = 0$
$I_3 = \int_{r=\infty}^0 f(r e^{2\pi i/3}).d(r e^{2\pi i/3}) = \int_{r=\infty}^0 \frac{(r e^{2\pi i/3})^{\frac 1 2}}{1+(r e^{2\pi i/3})^3}e^{2\pi i/3}.dr= -\int_{r=0}^\infty \frac{\sqrt r e^{\pi i/3}e^{2\pi i/3}}{1+r^3}.dr = \int_{r=0}^\infty \frac{\sqrt r }{1+r^3}.dr = I_1$
Since the residue at the only pole inside the contour is -i/3, 2 I₁ = I₁+I₂ + I₃ = 2πi * (-i/3), so I₁ = π/3.

It's an easier, simpler approach haruspex than what I initially came up with (the keyhole contour). Good though to see it solved several ways. That comes in handy too later on with other more difficult problems. :)