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