MHB Integration using residue theorem (part 2)

[aruwin]

Hello.
I need some explanation here. I got the solution but I don't understand something.

Question:
Find the integral using Residue Theorem.

$$\int_{-\infty}^{\infty}\frac{dx}{(x^2+4)^2}$$

Here is the first part of the solution that I don't understand:

To evaluate $\int_{-\infty}^{\infty}\frac{dx}{(x^2+4)^2}$, consider $\oint_c\frac{dz}{(z^2 + 4)^2}$,
where C consists of the real axis [-R, R] with R > 2, and the upper half of Γ: |z| = R (all with counterclockwise orientation).

My question: Why is R>2?

 

[Euge]

It's an assumption, not a result, that $R > 2$. This assumption was made so that the singularity of $f$ at $z = 2i$ lies inside C.

 

[aruwin]

It's an assumption, not a result, that $R > 2$. This assumption was made so that the singularity of $f$ at $z = 2i$ lies inside C.

Ok, but why must the singularity lies inside C?

 

[Euge]

Ok, but why must the singularity lies inside C?

At some point in the calculation, you're going to let $R \to \infty$ to find the value of $\int_{-\infty}^\infty \frac{dx}{(x^2 + 4)^2}$. As $R$ increases without bound, $C$ will contain the singularity of $f$ at $z = 2i$. Knowing that, you choose $R$ large enough in the setup of the contour. Also, keep the following in mind. If $0 < R < 2$, the contour $C$ will contain no singularity of $f$, but you cannot take the limit as $R \to \infty$ as $R < 2$. Then integrating over $C$ will serve no purpose.

 

[aruwin]

At some point in the calculation, you're going to let $R \to \infty$ to find the value of $\int_{-\infty}^\infty \frac{dx}{(x^2 + 4)^2}$. As $R$ increases without bound, $C$ will contain the singularity of $f$ at $z = 2i$. Knowing that, you choose $R$ large enough in the setup of the contour. Also, keep the following in mind. If $0 < R < 2$, the contour $C$ will contain no singularity of $f$, but you cannot take the limit as $R \to \infty$ as $R < 2$. Then integrating over $C$ will serve no purpose.

Thank you for the explanation. I got it now :)