Definite Integral to Infinity of Sine x over Root x
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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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.53$