Definite Integral to Infinity of Reciprocal of x Squared plus a Squared/Proof 1
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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
| \(\ds \int_0^\infty \dfrac {\d x} {x^2 + a^2}\) | \(=\) | \(\ds \int_0^{\mathop \to +\infty} \dfrac {\d x} {x^2 + a^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{\gamma \mathop \to +\infty} \int_0^\gamma \dfrac {\d x} {x^2 + a^2}\) | 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.
$\blacksquare$