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If $\sin(x+iy)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)=0$ , $y=0$ . So $\sin(\pi x)$ can be expressed as an product of $x$ , $(x-n)$ s for $n\in\mathbb{N}$ , $(x+n)$ s for $n\in\mathbb{N}$ , and some non zero constant. For finite $a$ , $p(x)=cx\prod_{n=1}^a(x-n)(x+n)$ is a finite polynomial and goes to infinity with $x$ ; $c$ such that the polynomial best approximates $\sin(\pi x)$ near a finite $x$ will better apply to a wider range of $x$ s around zero as $a$ approaches infinity. The maximum value of $\sin(\pi x)$ is 1 at, arbitrarily, $x=0.5$ . So for a growing range around $x=0$ , $\frac{p(x)}{p(0.5)}$ will better approximate $\sin(\pi x)$ as $a$ approaches infinity. Thus we have $\sin(\pi x)=\lim_{a\to\infty}\frac{x\prod_{n=1}^a(x-n)(x+n)}{0.5\prod_{n=1}^a(0.5-n)(0.5+n)}=$ $\frac{ x \left( \frac{ (x+a)! } { x! } \right) \left( \frac{ (x-1)! } { (x-a-1)! } \right) } { \frac{ 0.5(0.5+a)!(-0.5)! } { 0.5!(-0.5-a)! } }$ . Knowing $\frac{1}{2}!=\frac{\sqrt{\pi}}{2}$ and $\left(-\frac{1}{2}\right)!=\sqrt{\pi}$ , the above can be simplified to $\lim_{a\to\infty}\frac{(x+a)!(-0.5-a)!}{(x-a-1)!(0.5+a)!}$ . Wallis’ formula gives $\frac{\pi}{2}=\lim_{x\to\infty}\prod^x_{n=1}\frac{4n^2}{4n^2-1} =\lim_{x\to\infty}\frac{4^x(x!)^2}{\prod_{n=1}^x\left(4n^2-1\right)}=\lim_{x\to\infty}\frac{x!^2}{\prod_{n=1}^x\left(n^2-0.25\right)}$ . We can multiply these forms of $\frac{\pi}{2}$ and $\sin(\pi x)$ to get the inner part of the problem’s expression, which between $x=0$ and $x=$ some positive odd integer, encloses a signed area of 1 with the x axis.