sine integral at infinity

The value of the improper integral (one of the Dirichlet integrals)

$${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}𝑑x=\underset{x\to \mathrm{\infty }}{lim}\mathrm{Si}x,$$

where Si means the sine integral (http://planetmath.org/SineIntegral) function, is most simply determined by using Laplace transform which may be aimed to the integrand (see integration of Laplace transform with respect to parameter).  Therefore the integrand must be equipped with an additional parametre $t$:

$$ℒ\{{\int }_0^{\mathrm{\infty }}\frac{1}{x}\sin txdx\}={\int }_0^{\mathrm{\infty }}\frac{1}{x}\cdot \frac{x}{s^2+x^2}𝑑x={\int }_0^{\mathrm{\infty }}\frac{dx}{s^2+x^2}=\frac{1}{s}\underset{x=0}{\overset{\mathrm{\infty }}{/}}\arctan \frac{x}{s}=\frac{\pi }{2}\cdot \frac{1}{s}$$

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 $ℒ\{1\}=\frac{1}{s}$.  Thus we have the result

${\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{\sin x}{x}}𝑑x={\displaystyle \frac{\pi }{2}}.$(1)

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

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{sinc}xdx=\pi ;$$

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

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

${\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{\sin ax}{x}}dx={\displaystyle \frac{\pi }{2}}\mathit{\hspace{1em}\hspace{1em}}(a>0)$(2)

and to

${\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{\sin ax}{x}}dx=(\text{sgn}a){\displaystyle \frac{\pi }{2}}\mathit{\hspace{1em}\hspace{1em}}(a\in ℝ).$(3)