Solve integration using Laplace Transform Method

Question

Show that $$\int_{0}^{\infty}\frac{\sin (t)}{t}dt=\frac{\pi }{2}$$ by using Laplace Transform method. I know that $$L\left \{\sin(t) \right \} = \int_{0}^{\infty}e^{-st}\sin(t)dt=\frac{1}{s^{2}+1}$$ how to proceed further ?

Answer 1

Hint:

$$\mathcal L (\sin t/t) = \int_0^\infty \frac{\sin t}{t} e^{-st} \ dt = \arctan\frac{1}{s}.$$

EDIT: To arrive at this result, note that \begin{align*} I & = \int_0^\infty \frac{\sin t}{t} e^{-st} \ dt \\ -\frac{\partial I}{\partial s} & = \int_0^\infty \sin t e^{-st} \ dt\\ & = \frac{1}{1+s^2}. \end{align*} Can you finish from here?

Answer 2

$\int_{0}^{\infty}\frac{\sin (t)}{t}dt$

$=\int_0^{\infty}e^{-0\cdot t}\frac{\sin t}{t} dt$

$=\lim_{s\to 0}\int_s^{\infty}L[\sin t] ds$

$=\lim_{s\to 0}\int_s^{\infty}\frac{1}{s^2+1} ds$

$=\lim_{s\to 0}\left.\tan^{-1}(s)\right|_s^\infty$

$=\lim_{s\to 0}\frac{\pi}{2}-\tan^{-1}(s)$

$=\frac{\pi}{2}$

Answer 3

Using $$ \int_0^\infty f(x)g(x)dx = \int_0^\infty (Lf)(s)(L^{-1}g)(s)ds $$ one has \begin{eqnarray} \int_{0}^{\infty}\frac{\sin (t)}{t}dt&=&\int_0^\infty \sin t \cdot\frac1tdt=\int_0^\infty L\{\sin t\}L^{-1}\{\frac1t\}ds\\ &=&\int_0^\infty \frac{1}{s^2+1}\cdot 1ds=\frac{\pi }{2} \end{eqnarray}