Integral recurrence relation Let $f(t)$ be a continuous function on $\mathbb{R}$,and function sequence $\{u_{n}(t)\}$$(n=1,2,\cdots )$
defined by $u_{1}(t) = f(t),u_{n+1}(t)=\int_{0}^{t}\sin(t-s)u_{n}(s)ds \ (t\in \mathbb{R})$
(1)$u(t)=\sum_{n=1}^{\infty}u_{n}(t)$ is uniformly convergent on $[-T,T]$ $(\forall T>0)$.
(2)Express $u(t)$ by using  $f(t)$.
(3)if $u(t)=1$ $\forall t\in \mathbb{R}$ , find $f(t)$.
(1) is easy(Weierstrass M-test),but I have no idea about (2)(3).
How to solve (2)(3)?
Thanks!
 A: Given (1), let $g_n(x)=\sin(t-x)u_n(x)$, for every $x,t\in\mathbb{R}$. We have that:
$$g_n(x)\leq u_n(x)$$
for every $x\in\mathbb{R}$. Let now $T>0$. Since 
$$u(t)=\sum_{n=1}^\infty u_n(t)$$ 
converges uniformly on $[-T,T]$, we have - M-test - that 
$$g(x):=\sum_{n=1}^\infty g_n(x)$$
converges uniformly on $[-T,T]$, for every $T>0$.
Now, since $g(x)$ converges uniformly, we have that:
$$\begin{align*}\int_0^tg(x)dx=&\int_0^t\sum_{n=1}^\infty g_n(x) dx=\sum_{n=1}^\infty\int_0^tg_n(x)dx\Leftrightarrow\\
\Leftrightarrow&\int_0^t\sum_{n=1}^\infty\sin(t-x)u_n(x)dx=\sum_{n=1}^\infty\int_0^t\sin(t-x)u_n(x)dx\tag{$\star$}
\end{align*}$$
Now, we have:
$$\begin{align*}
u(t)=&\sum_{n=1}^\infty u_n(x)=u_1(t)+\sum_{n=2}^\infty u_n(t)=f(t)+\sum_{n=1}^\infty u_{n+1}(t)=\\
=&f(t)+\sum_{n=1}^\infty\int_0^t\sin(t-s)u_n(s)ds\overset{(\star)}{=}\\
=&f(t)+\int_0^t\sum_{n=1}^\infty\sin(t-s)u_n(s)ds=\\
=&f(t)+\int_0^t\sin(t-s)\sum_{n=1}^\infty u_n(s)ds=\\
=&f(t)+\int_0^t\sin(t-s)u(s)ds
\end{align*}$$
So, finally:
$$\boxed{u(t)=f(t)+\int_0^t\sin(t-s)u(s)ds}$$
For (3), setting $u(t)=1$, we have:
$$f(t)=1-\int_0^t\sin(t-s)ds\overset{u=t-s}{\underset{du=-ds}{=}}1-\int_t^0\sin u(-du)=1+\int_0^t(-\sin u)du=\cos t$$
So,
$$\boxed{f(t)=\cos t}$$
