Definite Integral to Infinity of Cosine of a x^2
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Theorem
- $\ds \int_0^\infty \map \cos {a x^2} \rd x = \frac 1 2 \sqrt {\frac \pi {2 a} }$
where $a$ is a positive real number.
Proof
We have, by Euler's Formula: Corollary:
- $\map \exp {-i a x^2} = -i \map \sin {a x^2} + \map \cos {a x^2}$
As $\map \sin {a x^2}$ and $\map \cos {a x^2}$ are both real for real $a, x$, we therefore have:
\(\ds \int_0^\infty \map \cos {a x^2} \rd x\) | \(=\) | \(\ds \int_0^\infty \map \Re {\map \exp {-i a x^2} \rd x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \Re {\int_0^\infty \map \exp {-i a x^2} \rd x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt a} \map \Re {\int_0^\infty \map \exp {-i t^2} \rd t}\) | substituting $\sqrt a x = t$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt a} \map \Re {\frac 1 2 \sqrt {\frac \pi 2} \paren {1 - i} }\) | Definite Integral to Infinity of $\map \exp {-i x^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \sqrt {\frac \pi {2 a} }\) |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.50$