Number of ideals of a given norm Is there any analytic expression(those involving exp, sin, cos  etc...) which gives the number of ideals of a given norm?
of course we are lying in an algebraic number field.
 A: Not really. There are upper bounds on the number of ideals of a given norm in a given ideal class, but I don't believe that the expression you're looking for exists.
If you take an ideal $\mathfrak{p}\lhd\mathcal{O}_K$ of norm bounded by $n$ and consider its image in the class group $[\mathfrak{p}]$, then we can find an inverse say $[\mathfrak{q}]$ in the class group, and hence $\mathfrak{p}\mathfrak{q}=(\alpha)$ is principal with $\alpha\in\mathfrak{q}$ and $|\operatorname{Nm}(\alpha)|\le n\operatorname{Nm}(\mathfrak{q})$. And if $\alpha\in\mathfrak{q}$, and $|\operatorname{Nm}(\alpha)|\le n\operatorname{Nm}(\mathfrak{q})$, then $\mathfrak{p}=(\alpha)\mathfrak{q}^{-1}$ is an ideal of norm bounded by $n$.
So the number of ideals of norm bounded by $n$ in a given class, say $X$, is equal to the number of principal ideals $(\alpha)$, where $\alpha\in\mathfrak{q}$ (and $[\mathfrak{q}]=X^{-1}$ in the class group), and $|\operatorname{Nm}(\alpha)|\le n\operatorname{Nm}(\mathfrak{q})$.
Let $x_1,\dots,x_d$ be an integral basis for $\mathcal{O}_K$, and let $\alpha=r_1x_1+\cdots+r_dx_d$. Then we can try and turn the problem of finding principal ideals $(\alpha)$ into one of counting the lattice points $(r_1,\dots,r_d)\in\mathbb{Z}^d$. This is explained in considerably more detail here.
