# Residue Theorem Question

I am having trouble with this integral. Any help would be greatly appreciated!

$$\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5}dx$$ using the keyhole'' contour.

I know that we want to obtain our domain $D$ by deleting the nonnegative reals.

• What do you mean by $2\infty$ ? – Belgi Mar 20 '13 at 7:08

The trick with problems like this is getting a consistent branch for the multivalued function - here, $\sqrt{z}$.

Consider

$$\oint_C dz \frac{\sqrt{z}}{z^2+2 z+5}$$

where $C$ is your keyhole contour about the positive real axis. Note that the integrals about the large circular arc and the small circular arc about the origin vanish in their respective limits (the large one, pretty slowly, but it does in the end). That leaves the integrals up and back along the real axis. Going back, you pick up a phase of $e^{i 2 \pi}$ which gets translated into a factor of $e^{i \pi}$, so this contour integral is then

$$\oint_C dz \frac{\sqrt{z}}{z^2+2 z+5} = 2 \int_0^{\infty} dx \frac{\sqrt{x}}{x^2+2x+5}$$

This is equal to $i 2 \pi$ times the sum of the residues within $C$. Let's evaluate those residues; first, we must identify the poles. Clearly, the poles are at $z_{\pm}=-1\pm i 2$. These are in the second and third quadrants, so it is wise to express them in the correct polar form.

This is where people make mistakes. You have to remember that the argument of these roots must lie between $0$ and $2 \pi$. This is crucial because we already made that assumption above in assigning a phase of $2 \pi$ along the other segment of the contour $C$ along the real axis. The roots are then

$$z_{\pm} = \sqrt{5} e^{i (\pi \mp \arctan{2})}$$

Note that $z_+$ is in the 2nd quadrant, while $z_-$ is in the 3rd quadrant, and the argument of $z_{\pm}$ is between $0$ and $2 \pi$.

Now we can compute the sum of the residues at these poles, which is

$$\frac{5^{1/4} e^{i(\pi + \arctan{2})/2}}{2 (-2 i)} + \frac{5^{1/4} e^{i(\pi - \arctan{2})/2}}{2 (2 i)} = -i \frac{5^{1/4}}{2} \sin{\left(\frac{1}{2} \arctan{2}\right)}$$

The contour integral about $C$ is $i 2 \pi$ times this sum, so that the integral we want is actually $i \pi$ times the sum (remember the factor of 2 above). We can also simplify the sine using a half-angle formula. The result is, after a little algebra which I leave to the reader,

$$\int_0^{\infty} dx \frac{\sqrt{x}}{x^2+2x+5} = \frac{\pi}{2} \sqrt{\frac{\sqrt{5}-1}{2}} = \frac{\pi}{2} \sqrt{\phi}$$

NB

Now of course, you could end all of this grief by substituting $x=u^2$ in the original integral. The result would be

$$\int_{-\infty}^{\infty} du \frac{u^2}{u^4+2 u^2+5}$$

To use the residue theorem, you use a semicircular contour in the upper-half plane. Of course, there are now $4$ poles, and you have to choose the correct ones that lie within that contour for the residue calculation. But this is fine for this particular integral. For others with more general powers of $x$, however, such substitutions will not work, and you will have to use the method illustrated here.