I need to show that:

$\displaystyle \displaystyle{\int_{0}^{\infty}\frac{\sin{x}}{x}\, dt=\frac{1}{2i}\int_{-\infty}^{\infty}\frac{e^{ix}-1}{x}}$

It seems obvious that i need to use

$\displaystyle \sin{x}=\frac{e^{ix}-e^{-ix}}{2i}$

but I can't get it to work.

I have tried thinking it of as a complex curve integral

$\displaystyle \displaystyle{\int_{\gamma}^{}f(z)\,dz \ \,\ \ f(z)=\frac{\sin{z}}{z}$

with

$\displaystyle \gamma(t)=t, 0<t<\infty$

and trying to use a different parametrization for $\displaystyle \gamma(t)$