In this post, I want to ask the community about an issue regarding an analytical approach to dealing with the integer factorization problem.
The problem definition is as follows:
1. Sets
Let $C=\{x^2:x\in\mathbb{Z}\}$.
We define the sets $A_n=C\cap \{x-n:x\in C\}$ and $B_n=C\cap\{x+n:x\in C\}$ to represent the congruences of squares modulo n (n is a non-square odd number), where $\forall a\in A_n, \exists b\in B_n:a\equiv b\pmod n$.
2. Indicator functions
Let $\mathbf{1}_C(x)=\begin{cases}1, x\in C\\0, x\notin C\end{cases}$
This function can be expressed as: $\mathbf{1}_C(x)=1-\lceil\sqrt x\rceil +\lfloor\sqrt x\rfloor$
We define the functions $\mathbf{1}_{A_n}(x)=\mathbf{1}_C(x)\cdot\mathbf{1}_C(x-n)$ and $\mathbf{1}_{B_n}(x)=\mathbf{1}_C(x)\cdot\mathbf{1}_C(x+n)$
3. Approximation
Let $\mathbf{1}_C(x)=\frac{\sin((2k+1)\pi x)}{2(2k+1)\sqrt x \sin(\pi\sqrt x)}$, where $k\in\mathbb{N}$.
We define the functions
$$\mathbf{1}_{A_n}(x)=\frac{\sin(\pi x)}{2\sqrt x \sin(\pi\sqrt x)}\frac{\sin(\pi(2k+1)(x-n))}{2(2k+1)\sqrt{x-n}\sin(\pi\sqrt{x-n})}$$
$$\mathbf{1}_{B_n}(x)=\frac{\sin(\pi x)}{2\sqrt x \sin(\pi\sqrt x)}\frac{\sin(\pi(2k+1)(x+n))}{2(2k+1)\sqrt{x+n}\sin(\pi\sqrt{x+n})}$$
To find ${\displaystyle {\underset {x}{\operatorname {arg\,max} }}\,\mathbf{1}_{A_n}(x)}$, we need to solve the following integral:
$$\int_n^x\frac{\sin(\pi t)}{2\sqrt t \sin(\pi\sqrt t)}\frac{\sin(\pi(2k+1)(t-n))}{2\sqrt{t-n}\sin(\pi\sqrt{t-n})}dt$$
To find ${\displaystyle {\underset {x}{\operatorname {arg\,max} }}\,\mathbf{1}_{B_n}(x)}$, we need to solve the following integral:
$$\int_0^x\frac{\sin(\pi t)}{2\sqrt t \sin(\pi\sqrt t)}\frac{\sin(\pi(2k+1)(t+n))}{2\sqrt{t+n}\sin(\pi\sqrt{t+n})}dt$$
I am trying to answer the following question: Is it possible to solve these integrals using analytical integration methods?