Solution
= Solution
The <Gaussian function> $e^{-x^2}$ is even and $\sin x$ is odd, so their product is an <odd function>. Moreover $|e^{-x^2}\sin x|\leq e^{-x^2}$, whose integral on the real <line> is finite. Thus the <improper integral> is absolutely convergent, not merely a symmetric principal value, and the contributions of its two half-lines cancel:
$$
\boxed{\int_{-\infty}^{\infty}e^{-x^2}\sin x\,dx=0.}
$$