Prove that some function is the solution of some equation Show that $$x(t)=\sum_{n=0}^{\infty}\frac{(-1)^n(t/2)^{2n}}{(n!)^2}$$ is the solution of $$x*x=\int_{0}^t x(u)x(t-u)du=\sin t$$
My approach:
I suppose that I have to use the Laplace transform. I tried to calculate
$$\mathcal{L}(\sin t)=\int_{0}^{\infty}e^{-\lambda t}\sin tdt=\frac{1}{\lambda^2+1}$$
But how do I continue?
 A: By the uniqueness of the Laplace transform it suffices to show that:
$$\mathcal{L}\left(\int_{0}^t x(u)x(t-u)du\right)=\frac{1}{\lambda^2+1} \tag{1}$$
To do so, insert the sum definition:
$$x \left( t \right) =\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{
n} \left(\dfrac{t}{2} \right) ^{2\,n}}{ \left( n! \right) ^{2}}} \tag{2}$$
into the integral and note that term by term integration is justified as the sum is absolutely convergent. Each term can be integrated by comparison with the Beta function:
$$\begin{aligned}
\int _{0}^{t}\! \left( \dfrac{u}{2} \right) ^{2\,n} \left( \dfrac{t}{2}-\dfrac{u}{2}
 \right) ^{2\,m}{du}&=2\, \left( \dfrac{t}{2} \right) ^{2\,m+1+2\,n}\int _{0}
^{1}\!{y}^{2\,n} \left( 1-y \right) ^{2\,m}{dy}\quad:y=ut\\
&=2\,{\frac { \left(\dfrac{t}{2}\right) ^{2\,m+1+2\,n}\Gamma  \left( 2\,m+1
 \right) \Gamma  \left( 2\,n+1 \right) }{\Gamma  \left( 2\,m+2+2\,n
 \right) }}
\end{aligned} \tag{3}
$$
and the Laplace transform can be applied to each term by comparison with the integral definition of the Gamma function:
$$ \int _{0}^{\infty }\!2\, \left( \dfrac{t}{2}\right) ^{2\,m+1+2\,n}{{\rm e}^
{-\lambda\,t}}{dt}={\frac { \left(\dfrac{1}{2}\right) ^{2\,n+2\,m}\Gamma 
 \left( 2\,m+2+2\,n \right) }{{\lambda}^{2\,m+2+2\,n}}} \tag{4}$$
and therefore:
$$ 
\begin{aligned}
\mathcal{L}\left(\int_{0}^t x(u)x(t-u)du\right)&=\sum _{m=0}^{\infty } \sum _{n=0}^{\infty }4\,{\frac { \left( -
1 \right) ^{m+n}\,\Gamma 
 \left( 2\,n+1 \right) \Gamma  \left( 2\,m+1 \right) }{ \left( \Gamma 
 \left( n+1 \right)  \right) ^{2} \left( \Gamma  \left( 1+m \right) 
 \right) ^{2}}} \left( \dfrac{\lambda}{2} \right) ^{-2\,m-2\,n-2} \\
&= \left( \sum _{m=0}^{\infty }{\frac { \left( -1 \right) ^{m}{2}^{-2\,m
}{\lambda}^{-2\,m-1}\Gamma  \left( 2\,m+1 \right) }{ \left( \Gamma 
 \left( 1+m \right)  \right) ^{2}}} \right) ^{2}\\
&=\left(\dfrac{1}{\lambda} \sum _{m=0}^{\infty }{-1/2\choose m}{\left(\lambda^2\right)}^{-m}
 \right) ^{2}\\
&=\left(\dfrac{1}{\lambda}\dfrac{1}{\sqrt{1+\dfrac{1}{\lambda^2}}}\right)^2\\
&=\frac{1}{\lambda^2+1} 
\end{aligned}
$$
A: $\mathcal{L}_{t\to\lambda}\left\{\int_0^tx(u)x(t-u)~du\right\}=\mathcal{L}_{t\to\lambda}\{\sin t\}$
$(X(\lambda))^2=\dfrac{1}{\lambda^2+1}$
$X(\lambda)=\pm\dfrac{1}{\sqrt{\lambda^2+1}}$
$x(t)=\mathcal{L}^{-1}_{\lambda\to t}\biggl\{\pm\dfrac{1}{\sqrt{\lambda^2+1}}\biggr\}=\mathcal{L}^{-1}_{\lambda\to t}\biggl\{\pm\dfrac{1}{\lambda\sqrt{1+\dfrac{1}{\lambda^2}}}\biggr\}=\mathcal{L}^{-1}_{\lambda\to t}\biggl\{\pm\dfrac{1}{\lambda}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2\lambda^{2n}}\biggr\}=\mathcal{L}^{-1}_{\lambda\to t}\biggl\{\pm\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2\lambda^{2n+1}}\biggr\}=\pm\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n}}{4^n(n!)^2}$
$x(t)=\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n}}{4^n(n!)^2}$ is one of the solutions.
