# Thread: Residue Theorem evaluation of a Real Definite Integral

1. ## Residue Theorem evaluation of a Real Definite Integral

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

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.

2. I haven't checked your residue calculations, I'll assume those are correct, but you aren't done yet. The contour integral over $\displaystyle \Gamma$ isn't supposed to be equal to the real integral you're trying to evaluate... you haven't dealt with the branch cut yet. You have two pieces of the contour that lie along the real axis, the one just above and the one just below; the integral over both of them should be equal to $\displaystyle 2\pi$, but what you're trying to evaluate is just the integral along $\displaystyle \gamma_1$ as R goes to infinity. You should be able to show that the integral on $\displaystyle \gamma_1$ is some constant factor times the integral on $\displaystyle \gamma_2$, so you can solve for the integral on $\displaystyle \gamma_1$ using the fact that their sum is $\displaystyle 2\pi$.

3. That's a multi-residue calculation because the integrand is multivalued. We could write the residue calculations as:

$\displaystyle \mathop\text{mRes}\limits_{z=-1+2i}\left\{\frac{\sqrt{z}}{z^2+2z+5}\right\}=\pm1/4i\sqrt{-1+2i}$

$\displaystyle \mathop\text{mRes}\limits_{z=-1-2i}\left\{\frac{\sqrt{z}}{z^2+2z+5}\right\}=\pm1/4i\sqrt{-1-2i}$

That makes sense right? Since the square root has two sheets then the residue is going to depend on which sheet you use to integrate around the pole. So if we start the calculation on the principal sheet right above (actually right on but that's a little confusing for beginners) so right above the positive real axis (your $\displaystyle \gamma_1$), and then go around in the positive sense, then it encloses the first pole at $\displaystyle z_0=-1+2i$, $\displaystyle \sqrt{z_0}=r^{1/2}e^{i/2 \text{Arg}\,z_0}$ where Arg is is principal argument. However, as we cross the negative real axis, we leave the principal branch square root by the construction of your branch cut on the positive real axis and therefore, the residue calculation at $\displaystyle z_1=-1-2i$ then must use $\displaystyle \sqrt{z_1}=r^{1/2}e^{i/2(\text{Arg}\, z_1+2\pi)}$.

This problem should be approached from the perspective of the underlying Riemann surfaces (multisheet): draw the multisheet, identify it's principal component, the contour of integration over it, the poles, the branch-cut, and why the residue calculation requires two different residue components.