Definite Integral to Infinity of Reciprocal of x Squared plus a Squared

From ProofWiki
Jump to navigation Jump to search

Theorem

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

for $a \ne 0$.


Corollary

$\ds \int_0^\infty \dfrac {\d x} {1 + x^2} = \frac \pi 2$

for $a \ne 0$.


Proof 1

\(\ds \int_0^\infty \dfrac {\d x} {x^2 + a^2}\) \(=\) \(\ds \int_0^{\mathop \to +\infty} \dfrac {\d x} {x^2 + a^2}\)
\(\ds \) \(=\) \(\ds \lim_{\gamma \mathop \to +\infty} \int_0^\gamma \dfrac {\d x} {x^2 + a^2}\) Definition of Improper Integral on Closed Interval Unbounded Above
\(\ds \) \(=\) \(\ds \lim_{\gamma \mathop \to +\infty} \intlimits {\frac 1 a \arctan \frac x a} 0 \gamma\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \frac 1 a \lim_{\gamma \mathop \to +\infty} \paren {\arctan \frac \gamma a - \arctan 0}\)
\(\ds \) \(=\) \(\ds \frac 1 a \lim_{\gamma \mathop \to +\infty} \arctan \frac \gamma a\) Arctangent of Zero is Zero
\(\ds \) \(=\) \(\ds \frac 1 a \frac \pi 2\) Arctangent Tends to Half Pi as Argument Tends to Infinity

Hence the result.

$\blacksquare$


Proof 2

\(\ds \int_0^\infty \dfrac {\d x} {x^2 + a^2}\) \(=\) \(\ds \frac \pi { 2 a^{2 - 1} } \csc \left({\frac \pi 2}\right)\) Definite Integral to Infinity of $\dfrac 1 {1 + x^n}$: Corollary
\(\ds \) \(=\) \(\ds \frac \pi {2 a}\) Sine of Right Angle


$\blacksquare$


Proof 3

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:

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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Definite_Integral_to_Infinity_of_Reciprocal_of_x_Squared_plus_a_Squared&oldid=495408"