Laplace transformation of $ \frac{\sin^2{t}}{t^2} $? I tried convolution theorem also tried this process..
$$Laplace(\frac{f(t)}{t}) = \int_s^\infty F(u) du $$
So,
$$Laplace(\frac{\sin{t}}{t} * \frac{\sin{t}}{t}) = \int_s^\infty F(u) du  * \int_s^\infty F(u) du $$
where F(u) = Laplace(f(t))
I know this look's funny as there's no rule like the last equation.
Please help...
 A: \begin{align*}
\mathcal L  \left( \frac{\sin^2(x)}{x^2} \right ) &= \int_0^\infty e^{-sx}  \frac{\sin^2(x)}{x^2} dx \\
&= \left[e^{-sx} \sin^2(x) \int \frac 1{x^2}dx  \right ]_0^\infty - \int_0^\infty \left( 2 e^{-s x} \sin (x) \cos (x)-s e^{-s x} \sin ^2(x) \right ) \frac{(-1)}{x} dx  \\
&= \int_0^\infty e^{-sx} \frac{\sin(2x)}{x} dx -  s \int_0^\infty e^{-sx} \frac{\sin^2(x)}{x} dx \\ &= \arctan \left( \frac 2 s \right )- s\frac{1}{4} \left(\log \left(s^2+4\right)-2 \log (s)\right)
\end{align*}
To find $\displaystyle \int_0^\infty e^{-sx} \frac{\sin(2x)}{x} dx $
\begin{align*}
\int_0^\infty e^{-sx} \frac{\sin(2 x)}{x} dx &= 
\int_0^\infty e^{- \frac s 2 (2x)} \frac{\sin(2x)}{2x} d(2x)\\ 
 &= \int_0^\infty e^{- \frac s 2 u} \frac{\sin (u)}{u}du\\ 
 &= \int_0^\infty e^{- \frac s 2 u} \sin (u) \int_0^\infty e^{- u t} dt du \\ 
 &= \int_0^\infty \int_0^\infty e^{ - \left( \frac s 2 + t \right )u } \sin(u) du dt\\ 
 &= \int_0^\infty \frac{4}{(4 + (s + 2 t)^2)} dt \\ 
 &= \frac \pi 2 - \arctan \left( \frac s 2  \right ) \\ 
 &= \arctan \left( \frac 2 s \right )
\end{align*}
The relation $\displaystyle \arctan(x) + \arctan \left( \frac 1 x\right) = \frac \pi 2$ has been used above
To find $\displaystyle \int_0^\infty e^{-sx} \frac{\sin^2(x)}{x} dx$ we proceed in the same way.
\begin{align*}
\int_0^\infty e^{-sx} \frac{\sin^2(x)}{x}dx &= \int_0^\infty \int_0^\infty e^{-(s+t)x} \sin^2(x)dx \\ 
 &= \int_0^\infty \int_0^\infty  e^{-(s+t)x} \frac{1 - \cos(2x)}{2}dx dt \\ 
 &= \frac 1 4 \log \left( 1 + \frac 4 {s^2} \right )\\ 
\end{align*}
A: \begin{align}
\phi\left(\mu\right)
&\equiv
\int_{0}^{\infty}{\rm e}^{-st}\,{\sin^{2}\left(\mu t\right) \over t^{2}}\,{\rm d}t
\\[3mm]
\phi'\left(\mu\right)
&=
\int_{0}^{\infty}{\rm e}^{-st}\,{\sin\left(2\mu t\right) \over t}\,{\rm d}t\,,
\quad
\phi''\left(\mu\right)
=
2\int_{0}^{\infty}{\rm e}^{-st}\cos\left(2\mu t\right)\,{\rm d}t
\\[3mm]
\phi''\left(\mu\right)
&=
2\,\Re\int_{0}^{\infty}{\rm e}^{\left(-s + 2{\rm i}\mu\right)t}\,{\rm d}t
=
2\,{s \over s^{2} + 4\mu^{2}}
\\[3mm]
\phi'\left(\mu\right)\ -\ \overbrace{\phi'\left(0\right)}^{=\ 0}
&=
2\,\int_{0}^{\mu}{s\,{\rm d}\mu' \over s^{2} + 4\mu'^{2}}
=
\int_{0}^{2\mu/s}{{\rm d}\mu' \over 1 +\mu'^{2}}
=
\arctan\left(2\mu \over s\right)
\\[3mm]
\phi\left(\mu\right)\ -\ \overbrace{\phi\left(0\right)}^{=\ 0}
&=
\int_{0}^{\mu}\arctan\left(2\mu' \over s\right)\,{\rm d}\mu'
=
\mu\arctan\left(2\mu \over s\right)
-
\int_{0}^{\mu}\mu'\,{2/s \over \left(2\mu'/s\right)^{2} + 1}\,{\rm d}\mu'
\\[3mm]&=
\mu\arctan\left(2\mu \over s\right)
-
{1 \over 4}\,s\ln\left(\left[2\mu \over s\right]^{2} + 1\right)
\end{align}
$$
\phi\left(1\right)
\equiv
\color{#ff0000}{\large%
\int_{0}^{\infty}{\rm e}^{-st}\,{\sin^{2}\left(t\right) \over t^{2}}\,{\rm d}t
\color{#000000}{\ =\ }
\arctan\left(2 \over s\right)
-
{1 \over 4}\,s\ln\left({4 \over s^{2}} + 1\right)}
$$
