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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.

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up vote 7 down vote accepted

$$I = \frac{e^{n^2/2}}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\frac{(x-n)^2}{2}} \sin (2\pi x) \, dx \stackrel{x = x-n}{=} \frac{e^{n^2/2}}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\frac{x^2}{2}} \sin (2\pi x) \, dx$$

Now divide the integral into two parts:

$$\int_{-\infty}^{+\infty} e^{-\frac{x^2}{2}} \sin (2\pi x) \, dx = \int_{-\infty}^{0} e^{-\frac{x^2}{2}} \sin (2\pi x) \, dx + \int_{0}^{+\infty} e^{-\frac{x^2}{2}} \sin (2\pi x) \, dx$$ Take one of them and substitute $t=-x$: $$\int_{-\infty}^{0} e^{-\frac{x^2}{2}} \sin (2\pi x) \, dx = -\int_0^{+\infty}e^{-\frac{t^2}{2}} \sin (2\pi t) \, dt$$ Because these integrals are finite, i.e.: $$\int_0^{+\infty} \left| e^{-\frac{t^2}{2}} \sin (2\pi t) \right| \, dt \le \int_0^{+\infty}e^{-\frac{t^2}{2}} \, dt = \sqrt{\frac{\pi}{2}}$$ We can write $I = 0$ and we are not dealing with any kind of indeterminate form like $\infty - \infty$.

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Thanks. I always forget these kind of ideas... – AndreasS Jun 15 '12 at 17:04
The first line is essentially enough since the integrand on the right is odd. There's no convergence issue since $e^{-{x^2 \over 2}} \leq e^{-|x|}$ for $|x| > 2$ for example and the exponential function has finite integral. – Zarrax Jun 15 '12 at 17:05
@AndreasS I'm pinging you on behalf of Zarrax in case you haven't seen his comment. – Rudy the Reindeer Jul 4 '12 at 16:48
@MattN. Thanks. I saw the comment. However, I think qogosz made the steps more explicit (However, it works well with any odd function). And Zarrax also cares about the convergence - otherwise he wouldn't have to argue about it. – AndreasS Jul 4 '12 at 20:22
@AndreasS Good then : ) Just wanted to make sure since he didn't ping you. – Rudy the Reindeer Jul 4 '12 at 20:24

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