Evaluate the Improper Integral(help)

Question

I encountered the following integral while solving a log-normal distribution question. Initially, I thought since its a odd function, it evaluates to zero. But I think, since its a improper integral, we cannot to do simply. Upon further inspection, I found that neither does the indefinite integral exist for it. How to evaluate it then?
$$\int_{-\infty}^{+\infty}(\sin{2\pi x}) e^{-x^{2}/2}dx $$

Answer 1

Here, you can use the fact the integrand is odd, because the improper integral exists, so in calculating the improper integral, you can calculate $\int_{-r}^r (\cdots)$ and then take $r\to \infty$.

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More explicitly, note that for all $x$, $\left|\sin(2\pi x) e^{-x^2/2}\right| \leq e^{-x^2/2}$, and $\int_{-\infty}^{\infty} e^{-x^2/2}\, dx < \infty$. Therefore $\int_{-\infty}^{\infty}\left|\sin(2\pi x) e^{-x^2/2}\right|\,dx < \infty$. Now recall that absolute convergence of integral implies regular convergence. This means the following limit (which is the definition of improper integral) exists: \begin{align} \int_{-\infty}^{\infty}\sin(2\pi x) e^{-x^2/2}:= \lim_{r\to \infty}\int_{0}^{r}\sin(2\pi x) e^{-x^2/2}\, dx + \lim_{\alpha\to \infty}\int_{-\alpha}^{0}\sin(2\pi x) e^{-x^2/2}\, dx \end{align}

Because this limit exists, we can show that \begin{align} \int_{-\infty}^{\infty}\sin(2\pi x) e^{-x^2/2}\, dx &= \lim_{r \to \infty} \int_{-r}^r \sin(2\pi x)e^{-x^2}\, dx \\ &= 0, \end{align} where the second equality is because for each $r$, oddness of integrand implies the integral is $0$, so the result is $0$ even after the limit.