An Infinite Summation Involving Sinc Function

The given integral can be written as $I=\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{\infty}\frac{\sin z}{z+2\pi n}dz\\=\sum_{n=1}^{\infty}\frac{1}{n}\int_0^{\infty}\frac{\sin nz}{z+2\pi}dz\\=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\int_{2(m-1)\pi}^{2m\pi}\frac{\sin nz}{z+2\pi}dz\\=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\int_{0}^{2\pi}\frac{\sin nz}{z+2m\pi}dz\\=\sum_{m=1}^{\infty}\int_{0}^{2\pi}\frac{1}{z+2m\pi}\sum_{n=1}^{\infty}\frac{\sin nz}{n}dz\\=\sum_{m=1}^{\infty}\int_{0}^{2\pi}\frac{\pi-z}{2(z+2m\pi)}dz\\=\pi\sum_{m=1}^{\infty}\left[\left(m+1\right)\ln\left(1+\frac{1}{m}\right)-1\right]=\pi\cdot S$ Now, $e^S=\lim_{M\to \infty}e^{-M}\prod_{m=1}^M \left(1+1/m\right)^{m+1/2}\\=\lim_{M\to \infty}e^{-M}\frac{(M+1)^M\sqrt{M+1}}{M!}\\=\lim_{M\to \infty}\frac{e^{-M}(M+1)^M\sqrt{M+1}}{M^Me^{-M}\sqrt{2\pi M}}\quad (\mbox{Using Stirling})\\=\frac{1}{\sqrt{2\pi}}\lim_{M\to \infty}\left(1+\frac{1}{M}\right)^M=\frac{e}{\sqrt{2\pi}}$ Thus, $I=\pi\ln\left(\frac{e}{\sqrt{2\pi}}\right)=\pi(1-\ln\sqrt{2\pi})$ Which gives the answer $\boxed{3}$ .