calculus - How to prove $operatorname{si}(0) = -pi/2$ without contour

How to prove $\operatorname{si}(0) = -\pi/2$ without contour integration ?
Where $\operatorname{si}(x)$ is the sine integral.

Answer

HINT:

Note that our integral may be rewritten as
$$\int_{0}^{\infty} \int_{0}^{\infty} e^{-xy} \sin x \ dy \ dx = \int_{0}^{\infty} \frac{\sin x}{x} \ dx$$
but integrating with respect to x we get that
$$\int_{0}^{\infty} \int_{0}^{\infty} e^{-xy} \sin x \ dx \ dy = \int_{0}^{\infty} \frac{1}{1+y^2} \ dy$$

Hence I hope you can handle it on your own.