Beautiful Constant

Using Laplace transformation

$\large{L \left(\frac{\sin x}{x}\right)=\displaystyle \int^{\infty}_{s}\frac{1}{s^2+1}ds}$

$\large{L \left(\frac{\sin x}{x}\right)=\arctan |^{\infty}_{s}}$

$s=\pi$

$\large{\Rightarrow L \left(\frac{\sin x}{x}\right)=\text{arccot} (\pi)}$

$\large{\lfloor 100 arccot (\pi) \rfloor =30}$

$\displaystyle\int_0^\infty e^{-\pi x}\frac{\sin{x}}{x}dx = \frac1{2i}\int_0^\infty \frac{e^{-(\pi-i)x}}{x}-\frac{e^{-(\pi+i)x}}{x}dx$

Individually each integral cannot be evaluated like a Gaussian because $\Gamma(0)=\infty$ (below)

$\displaystyle\int_0^\infty x^be^{-ax^n}dx = \frac1{a^{\frac{b+1}{n}}n}\Gamma(\frac{b+1}{n})$

Using Feynman's trick: $\displaystyle I(a) = \int_0^\infty\frac{e^{-ax}-1}{x} = -\ln{a}+C$ , the original integral becomes:

$\displaystyle\Re(\frac{\ln(\frac{\pi+i}{\pi-i})}{2i}) = \Re(\frac{\ln(\pi+i)}{i}) = \boxed{\arctan{\frac1{\pi}}}$