$\displaystyle \int_{0}^{\infty}\frac{\sqrt{x}dx}{x^2+2x+5}

$

Step 1: Write in terms of z, and define the closed curve to integrate over.

$\displaystyle f(z) = \frac{\sqrt{z}dz}{z^2+2z+5}

$

Let D be the Complex Plane with the line $\displaystyle x \ge 0$ deleted.

$\displaystyle \gamma_1 = te^{i\delta}$ with $\displaystyle \varepsilon \le t \le R$

$\displaystyle \gamma_2 = Re^{i\theta}$ with $\displaystyle \delta \le \theta \le (2\pi - \delta )$

$\displaystyle \gamma_3 = te^{i(2\pi -\delta )}$ with $\displaystyle -R \le \theta \le -\varepsilon$

$\displaystyle \gamma_4 = Re^{i\theta}$ with $\displaystyle (2\pi - \delta ) \le \theta \le \delta$

Step 2: Find the poles of f(z) and calculate their residues.

$\displaystyle z_0 = -1+2i$

$\displaystyle Res(f;z_0) = \frac{\sqrt{(-1+2i)}}{4i} = \frac{\sqrt{5}e^{i\theta_0}}{4i}$

where $\displaystyle \theta_0 = Arg(-1 + 2i) = \pi - tan^{-1}(2)$

$\displaystyle z_1 = -1-2i$

$\displaystyle Res(f;z_0) = \frac{\sqrt{(-1+2i)}}{-4i} = \frac{\sqrt{5}e^{i\theta_1}}{-4i}$

where $\displaystyle \theta_1 = Arg(-1 - 2i) = tan^{-1}(2) - \pi$

Step 3: Apply the Reside Theorem

Okay... so let $\displaystyle \Gamma = \gamma_1 + \gamma_2 + \gamma_3 + \gamma_4$, then

$\displaystyle \int_{\Gamma}f(z)dz = 2i\pi \sum_{i=0}^{1}Res(f;z_i) = \frac{\sqrt{5}\pi}{2}(e^{i\theta_0}-e^{i\theta_1})$

and since $\displaystyle \theta_0 = -\theta_1$

$\displaystyle (e^{i\theta_0}-e^{i\theta_1}) = (cos\theta_0 + i sin\theta 0) - (cos\theta_0 - i sin\theta_0)$

$\displaystyle cos(\pi -tan^{-1}(2)) + i sin(\pi -tan^{-1}(2)) - cos(\pi -tan^{-1}(2)) + i sin(\pi -tan^{-1}(2)) =$

$\displaystyle 2i sin(\pi -tan^{-1}(2)) = -sin(-tan^{-1}(2)) = sin(tan^{-1}(2)) = \frac{4}{\sqrt{5}}$

Therefore,

$\displaystyle \int_{\Gamma}f(z)dz = 2\pi$

But this answer is very wrong.