Compute iterated kernels for symmetric kernel $K(x,t)=\sum_{n=1}^\infty \frac{\sin(n\pi x) \sin(n \pi t)}{n}.$ 
Compute iterated kernls for symmetric kernel $$K(x,t)=\sum_{n=1}^\infty \frac{\sin(n\pi x) \sin(n \pi t)}{n}.$$

I am studying integral equations with symmetric kernels and there is a given problem. This problem is very unclear to me since there is no interval given, just the expression of the kernel.
My attempt so far: the expression looks very familiar to the form to the Mercer's theorem's expression of the kernel: $$K(x,t)=\sum_{n=1}^\infty \frac{\phi_n(x)\phi_n(t)}{\lambda_n}$$ assuming the values of the functions are real. It wouldn't be quite right to say that we have $\phi(x)=\sin(n\pi x)$ and $\lambda_n=n$ but in this case we could say that $$\phi_n(x)=\lambda_n\int_a^b K(x,t) \phi_n(t)dt \implies \sin(n \pi x)=n\int_0^1 K(x,t)\sin(n \pi t)dt.$$ assuming our interval is $[0,1].$ Is it possible to get $K(x,t)$ out of this equation?
What's the method to generate 'proper' kernel $K(x,t)$ out of the sum above? Any help would be highly appreciated.
 A: Your intuition is correct, kernel $K$ can be written as
$$K(x,t)=\sum_{n=1}^\infty \frac{\phi_n(x)\phi_n(t)}{\lambda_n}$$
with $\phi_n(x)=\sin(n\pi x)$ and $\lambda_n=n$.
This is no special theorem, we're just writing the definition of $K$.
So, assuming the integration interval is $[0, 1]$, and using the following notation for the inner product in $L^2([0,1])$:
$$\langle f, g\rangle = \int_0^1 f(x)g(x)dx$$
the square of $K$ is
$$K^2(x,t)=\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{\phi_n(x)\phi_m(t)}{\lambda_n^2}\langle \phi_n, \phi_m\rangle$$
You can verify that $$\langle \phi_n, \phi_m\rangle=\int_0^1 \sin(n\pi u)\sin(m \pi u)du$$
is equal to $\frac 1 2$ if $n=m$, and is equal to $0$ otherwise.
Thus, $$K^2(x,t)=\frac 1 2 \sum_{n=0}^{+\infty}\frac{\phi_n(x)\phi_m(t)}{\lambda_n^2}$$
the $p$-th power of $K$ is given by
$$\boxed{K^p(x,t)=\frac 1 {2^{p-1}} \sum_{n=1}^\infty\frac{\phi_n(x)\phi_m(t)}{\lambda_n^p}=\frac 1 {2^{p-1}} \sum_{n=1}\frac{\sin(n\pi x) \sin(n \pi t)}{n^p}}$$
Edit: Following the comment from @Jean-Marie (merci!), an alternative form can be obtained by using the polylogarithm:
$$Li_p(z)=\sum_{n\geq 1}\frac {z^p}{n^p}$$
With this,
$$\begin{split}
K^p(x,t)&=\frac 1 {2^{p-1}} \sum_{n=1}\frac{\sin(n\pi x) \sin(n \pi t)}{n^p}\\
&= \frac 1 {2^p}\sum_{n=1}\frac{\cos(n\pi(x-t))-\cos(n\pi(x+t))}{n^p}\\
&= \frac 1 {2^p}\Re \left( \sum_{n=1}\frac{e^{in\pi(x-t)}-e^{in\pi(x+t)}}{n^p} \right)
\end{split}$$
$$\boxed{K^p(x,t) = \frac 1 {2^p} \Re \left( Li_p(e^{i\pi(x-t)}) - Li_p(e^{i\pi(x+t)}) \right)}$$
I'm not sure if that can be simplified further.
With $p=1$, $L_1(z) = -\ln(1-z)$, you get $$K(x,t)=\dfrac12\operatorname{ln}\left|\dfrac{\sin(\pi(x+t)/2)}{\sin(\pi(x-t)/2)}\right|$$
