sine integral at infinity

 $\mathcal{L}\{\int_{0}^{\infty}\frac{1}{x}\sin{tx}\,dx\}=\int_{0}^{\infty}\frac% {1}{x}\!\cdot\!\frac{x}{s^{2}+x^{2}}\,dx=\int_{0}^{\infty}\!\frac{dx}{s^{2}+x^% {2}}=\frac{1}{s}\!\operatornamewithlimits{\Big{/}}_{\!\!\!x=0}^{\,\quad\infty}% \!\arctan{\frac{x}{s}}=\frac{\pi}{2}\!\cdot\!\frac{1}{s}$

The obtained transform $\frac{\pi}{2}\!\cdot\!\frac{1}{s}$ corresponds (see the inverse Laplace transformation) to the function  $t\mapsto\frac{\pi}{2}$  because  $\mathcal{L}\{1\}=\frac{1}{s}$.  Thus we have the result

 $\displaystyle\int_{0}^{\infty}\frac{\sin{x}}{x}\,dx\;=\;\frac{\pi}{2}.$ (1)

Note 1.  Since  $x\mapsto\frac{\sin{x}}{x}$  or  $x\mapsto\operatorname{sinc}{x}$  is an even function, the result (1) may be written also

 $\int_{-\infty}^{\infty}\operatorname{sinc}{x}\,dx=\pi;$

see the $\operatorname{sinc}$-function (http://planetmath.org/SincFunction).

Note 2.  The result (1) may be easily generalised to

 $\displaystyle\int_{0}^{\infty}\frac{\sin{ax}}{x}\,dx\;=\;\frac{\pi}{2}\qquad(a% >0)$ (2)

and to

 $\displaystyle\int_{0}^{\infty}\frac{\sin{ax}}{x}\,dx\;=\;(\mbox{sgn}\,a)\frac{% \pi}{2}\qquad(a\in\mathbb{R}).$ (3)
 Title sine integral at infinity Canonical name SineIntegralAtInfinity Date of creation 2013-03-22 15:17:22 Last modified on 2013-03-22 15:17:22 Owner pahio (2872) Last modified by pahio (2872) Numerical id 18 Author pahio (2872) Entry type Derivation Classification msc 44A10 Classification msc 30A99 Synonym limit of sine integral Related topic SineIntegral Related topic SincFunction Related topic SubstitutionNotation Related topic IncompleteGammaFunction Related topic ExampleOfSummationByParts Related topic SignumFunction