Show that the Singular Value Decomposition of the operator
$$ A\colon L^2([0,1])\to L^2([0,1]), x\mapsto\int\limits_0^t x(s)\, ds $$
is given by
$$ \sigma_j=\frac{1}{(j-1/2)\pi},~~~~~v_j(x)=\sqrt{2}\cos((j-1/2)\pi x),~~~~~u_j(x)=\sqrt{2}\sin((j-1/2)\pi x). $$
My first question is: Is this operator compact at all so that it makes sense to talk of a SVD?
Then: What do I have to do to solve the task?
First, I determined the adjoint operator $A^*$; it is given by $x\mapsto\int\limits_t^1 x(s)\, ds$. Then I attested that $Av_j=\sigma_ju_j$ and $A^*u_j=\sigma_jv_j$.
Moreover, I showed with substitution $\omega=(j-1/2)\pi x$ that
$\langle v_j,v_j\rangle_{L^2([0,1])}=\frac{2}{(j-1/2)\pi}\int\limits_0^{(j-1/2)\pi}\lvert\cos(\omega)\rvert^2\, d\omega=1$ and similarly $\langle u_j,u_j\rangle_{L^2([0,1])}=1$.
Furthermore, I calculated for $k\neq j$ that
$\langle v_j,v_k\rangle_{L^2([0,1])}=\int\limits_0^1 v_j(x)\overline{v_k(x)}\, dx=\int\limits_0^1 v_j(x)v_k(\overline{x})\, dx=\int\limits_0^1 v_j(x)v_k(x)\, dx$ because $x\in [0,1]$. This integral is 0.
Similarly $\langle u_j,u_k\rangle_{L^2([0,1])}=0$.
So far so good. But is this the whole task? Is there something i have to show additionally?