I want to evaluate the following integral ($n \in \mathbb{N}\setminus \{0\}$): $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp\left(n x-\frac{x^2}{2}\right) \sin(2 \pi x)\ dx$ Maple and WolframAlpha tell me that this is zero and I also hope it is zero, but I don't see how I can argue for it.
I thought of rewriting the sine via $\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ or using Euler's identity on $\exp(n x-\frac{x^2}{2})$. However, in both ways I am stuck...
Thanks for any hint.