Definite Integral to Infinity of Reciprocal of x Squared plus a Squared/Proof 3

Proof
Let $C_R$ be the semicircular contour of radius $R$ situated on the upper half plane, centred at the origin, traversed anti-clockwise.

Let $\Gamma_R = C_R \cup \closedint {-R} R$.

Then, by Contour Integral of Concatenation of Contours:


 * $\ds \oint_{\Gamma_R} \frac {\d z} {z^2 + a^2} = \int_{C_R} \frac {\d z} {z^2 + a^2} + \int_{-R}^R \frac {\d x} {x^2 + a^2}$

we have:

from which:


 * $\ds \lim_{R \mathop \to \infty} \oint_{\Gamma_R} \frac {\d z} {z^2 + a^2} = \lim_{R \mathop \to \infty} \int_{-R}^R \frac {\d x} {x^2 + a^2} = \int_{-\infty}^\infty \frac {\d x} {x^2 + a^2}$

The integrand is meromorphic.

So by the Residue Theorem:


 * $\ds \lim_{R \mathop \to \infty} \oint_{\Gamma_R} \frac {\d z} {z^2 + a^2} = 2 \pi i \sum \Res {\frac 1 {z^2 + a^2} } {z_0}$

where the sum runs over the poles of $\dfrac 1 {z^2 + a^2}$.

The integrand can be written as:
 * $\dfrac 1 {\paren {z - a i} \paren {z + a i} }$

From this it can be observed that the integrand has only two simple poles: $z_0 = a i$ and $z_0 = -a i$.

Only the former of these lies in the contour, so:

So:


 * $\ds \int_{-\infty}^\infty \frac {\d x} {x^2 + a^2} = \frac \pi a$

From Definite Integral of Even Function:


 * $\ds \int_{-\infty}^\infty \frac {\d x} {x^2 + a^2} = 2 \int_0^\infty \frac {\d x} {x^2 + a^2}$

Hence:


 * $\ds \int_0^\infty \frac {\d x} {x^2 + a^2} = \frac \pi {2 a}$