Definite Integral to Infinity of Sine x over Root x

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int_0^\infty \frac {\sin x} {\sqrt x} \rd x = \sqrt {\frac \pi 2}$


Proof

\(\ds \int_0^\infty \frac {\sin x} {\sqrt x} \rd x\) \(=\) \(\ds \int_0^\infty 2 u \frac {\sin u^2} {\sqrt {u^2} } \rd u\) substituting $x = u^2$
\(\ds \) \(=\) \(\ds 2 \int_0^\infty \sin u^2 \rd u\)
\(\ds \) \(=\) \(\ds \sqrt {\frac \pi 2}\) Definite Integral to Infinity of $\sin a x^2$

$\blacksquare$


Sources