Definite Integral to Infinity of Reciprocal of x Squared plus a Squared/Proof 3
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Theorem
- $\ds \int_0^\infty \dfrac {\d x} {x^2 + a^2} = \frac \pi {2 a}$
for $a \ne 0$.
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:
\(\ds \size {\int_{C_R} \frac {\d z} {z^2 + a^2} }\) | \(\le\) | \(\ds \pi R \cdot \max_{0 \mathop \le \theta \mathop \le \pi} \size {\frac 1 {R^2 e^{2 i \theta} + a^2} }\) | Estimation Lemma for Contour Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\pi R} {R^2 + a^2}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac {\pi R} {R^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi R\) | ||||||||||||
\(\ds \) | \(\to\) | \(\ds 0\) | as $R \to \infty$ |
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 Cauchy's 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 summation 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:
\(\ds \lim_{R \mathop \to \infty} \oint_{\Gamma_R} \frac {\d z} {z^2 + a^2}\) | \(=\) | \(\ds 2 \pi i \Res {\frac 1 {z^2 + a^2} } {a i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi i \cdot \frac 1 {\map {\frac \d {\d z} } {z^2 + a^2}_{z \mathop = a i} }\) | Residue at Simple Pole | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi i \cdot \frac 1 {2 a i}\) | Derivative of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi a\) |
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:
- $\ds \int_0^\infty \frac {\d x} {x^2 + a^2} = \frac \pi {2 a}$
$\blacksquare$