definite integral

[man111]

$\displaystyle \int_{0}^{\infty}\frac{\sin x}{x}dx$

[ewert]

$\dfrac{\pi}2$

[alcoholist]

По теореме Фруллани

$$ \int_0^{+\infty}\frac{\sin x}{x}\,dx=\int_0^{+\infty}\frac{e^{ix}-e^{-ix}}{2ix}\,dx=\frac{\ln(-1)}{2i}=\frac{\pi}{2} $$