Oscillating Function Converges?

Is the following working correct?

$\int_0^\infty \sin x \, dx = \int_0^\pi \sin x \, dx +\int_\pi^\infty \sin x \, dx$

The first integral in RHS is equal to $[ -\cos x ]_0^\pi = \cos 0 - \cos \pi = 2$ .

For the second integral in RHS, we use a substitution $y = x - \pi$ , then $x = y + \pi$ . So $\sin x = \sin(y + \pi) = -\sin y$ . We have

$\begin{aligned} \int_0^\infty \sin x \, dx &=& 2 - \int_0^{\infty} \sin y \, dy \\ &=& 2 - \int_0^{\infty} \sin x \, dx \\ 2 \int_0^\infty \sin x \, dx &=& 2 \\ \int_0^\infty \sin x \, dx &=& 1. \\ \end{aligned}$