Billateral laplace transform sin(t) / t u (t) Hello everyone at my course I have problem solving Laplace transform of 
$\frac{\sin(t)}{t}$  $u{(t)}$
I have no idea I tried by definiton but get integral which cant be solved I already took a look at Finding the Laplace Transform of sin(t)/t
But It doesnt help me at all becouse there is used Taylor series expansion
becouse I m still begginer is there any easier way to solve it
Thanks in advante 
 A: The hint is very useful on that linked problem! Using it, we have 
$$\displaystyle f(s) = \int_0^\infty\dfrac{\sin t}{t}e^{-st}~ds$$
hence
$$f'(s) = \int_0^\infty \sin t e^{-st}~ds = -\dfrac{d}{ds} \arctan s$$
Can you now find $F(s)$ and solve for the constant?
A: 
In THIS ANSWER, I showed that the inverse Laplace Transform of $\arctan(s)-\pi/2$ is $-\frac{\sin(t)}{t}$ by carrying out the integral 
$$\mathscr{L}^{-1}(\arctan(s)-\pi/2)=\int_{\sigma -i\infty}^{\sigma+i\infty}e^{st}(\arctan(s)-\pi/2)\,ds$$


Herein, we carry out the forward Laplace Transform of the sinc function.  Proceeding, we have
$$\begin{align}
F(s)&=\int_0^\infty e^{-st}\frac{\sin(t)}{t}\,dt
\end{align}$$
Now, differentiating we have
$$\begin{align}
F'(s)&=-\int_0^\infty e^{-st}\sin(t)\,dt\\\\
&=-\frac1{s^2+1}
\end{align}$$
whereupon integrating and using $\lim_{s\to \infty }F(s)=0$ yields

$$F(s)=\pi/2-\arctan(s)$$

A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\infty}{\sin\pars{t} \over t}\,\expo{-st}\,\dd t & =
\int_{0}^{\infty}\
\overbrace{\pars{{1 \over 2}\int_{-1}^{1}\expo{-\ic \omega t}\,\dd \omega}}^{\ds{\ds{\sin\pars{t} \over t} \atop}}\
\,\expo{-st}\,\dd t =
{1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}\expo{-\pars{s + \ic\omega}t}
\,\,\dd t\,\dd\omega
\\[5mm] & = 
{1 \over 2}\int_{-1}^{1}{\dd\omega \over s + \ic\omega} =
{1 \over 2}\int_{-1}^{1}{s \over \omega^{2} + s^{2}}\,\dd\omega =
\int_{0}^{1/s}{\dd\omega \over \omega^{2} + 1}
=
\bbx{\ds{\arctan\pars{1 \over s}}}
\end{align}
