# Show that NSK PSK\int_0^{\infty} \frac{\sin(t)}{t}dt=\frac{\pi}{2}ZSK NSK by using Laplace Transform method. I know that NSK PSKL\{\sin(t)\}=\int_0^{\infty}

Show that
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(t\right)}{t}dt=\frac{\pi }{2}$
by using Laplace Transform method. I know that
$L\left\{\mathrm{sin}\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\mathrm{sin}\left(t\right)dt=\frac{1}{{s}^{2}+1}$
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psor32

Hint:
$L\left(\frac{\mathrm{sin}t}{t}\right)={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}t}{t}{e}^{-st}dt=\mathrm{arctan}\frac{1}{s}$
EDIT: To arrive at this result, note that
$I={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}t}{t}{e}^{-st}dt$
$-\frac{\partial I}{\partial s}={\int }_{0}^{\mathrm{\infty }}\mathrm{sin}t{e}^{-st}dt$
$=\frac{1}{1+{s}^{2}}$

###### Not exactly what you’re looking for?
peterpan7117i

${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}t}{t}dt$
$={\int }_{0}^{\mathrm{\infty }}{e}^{-0\cdot t}\frac{\mathrm{sin}t}{t}dt$
$=\underset{s⇒0}{lim}{\int }_{s}^{\mathrm{\infty }}L\left[\mathrm{sin}t\right]ds$
$=\underset{s⇒0}{lim}{\int }_{s}^{\mathrm{\infty }}\frac{1}{{s}^{2}+1}ds$
$=\underset{s⇒0}{lim}{\mathrm{tan}}^{-1}\left(s\right){\mid }_{s}^{\mathrm{\infty }}$
$=\underset{s⇒0}{lim}\frac{\pi }{2}-{\mathrm{tan}}^{-1}\left(s\right)$
$=\frac{\pi }{2}$