Calculus problem

These integrals are Cesaro Fourier sums for the function $f(x) \,=\, \big(1 + 2^{\sin\frac12x}\big)^{-1}$ , without a simple scale factor. Since $\sum_{n=0}^{N-1} \sum_{m=-n}^n e^{imx} \; = \; \left(\frac{\sin \frac12Nx}{\sin\frac12x}\right)^2 \;,$ with the variable substitution $x \mapsto -x$ , we see that (since $f(x) + f(-x) = 1$ ): $\begin{array}{rcl} \displaystyle I_N \; = \; \int_{-\pi}^\pi f(x) \left(\frac{\sin \frac12Nx}{\sin\frac12x}\right)^2\,dx & = & \displaystyle \int_{-\pi}^\pi f(-x) \left(\frac{\sin \frac12Nx}{\sin\frac12x}\right)^2\,dx \\ & =& \displaystyle \tfrac12\int_{-\pi}^\pi\big[f(x) + f(-x)\big] \left(\frac{\sin \frac12Nx}{\sin\frac12x}\right)^2\,dx \\ & = & \displaystyle \tfrac12 \int_{-\pi}^\pi \left(\frac{\sin \frac12Nx}{\sin\frac12x}\right)^2\,dx \\ & = & \displaystyle \tfrac12 \sum_{n=0}^{N-1} \sum_{m=-n}^n \int_{-\pi}^\pi e^{imx}\,dx \\ & = & \displaystyle \tfrac12 \sum_{n=0}^{N-1} \sum_{m=-n}^n 2\pi \delta_{m,0} \\ & = & N\pi \end{array}$ for all integers $N \ge 1$ , where $\delta_{m,n}$ is the Kronecker delta. From this it is clear that statement $1$ is false (should the inequality be the other way around?), but that statements $2,3,4$ are true, making the answer $2+3+4 = \boxed{9}$ .