Find $\lim_{m \to \infty} \int_{-\infty}^{\infty} \sin(e^t) e^{-(t - m)^2} dt$

Question

I would like to know if it's the case that $$\lim_{m \to \infty} \int_{-\infty}^{\infty} \sin(e^t) e^{-(t - m)^2} dt = 0.$$ For each fixed $t$, the integrand approaches zero as $m \to \infty$, so one might expect to apply the dominated convergence theorem. However, if one replaces $\sin(e^t)$ by a constant, then the integrand still converges pointwise to zero as $m \to \infty$, yet the integral is independent of $m$. So the oscillations of $\sin(e^t)$ (or perhaps lack of oscillation as $t \to -\infty$) must be leveraged somehow. I also suspect that the well-known estimate $\int_a^\infty e^{-ax^2} \le (2a)^{-1} e^{-a^2}$, $a > 0$, may be needed.

Hints or solutions are greatly appreciated.

Answer 1

Changing integration variables and integrating by parts gives \begin{align*} \int_{ - \infty }^{ + \infty } {\sin (e^t )e^{ - (t - m)^2 } dt} &\mathop = \limits^{t = m + s} \int_{ - \infty }^{ + \infty } {\sin (e^m e^s )e^{ - s^2 } ds} \\ & \;\;\,= - \frac{1}{{e^m }}\int_{ - \infty }^{ + \infty } {(2s + 1)\cos (e^m e^s )e^{ - s^2 - s} ds} \\ &\mathop = \limits^{s =t-1/2} - 2e^{1/4} \frac{1}{{e^m }}\int_{ - \infty }^{ + \infty } {t\cos (e^{m - 1/2} e^t )e^{ - t^2 } dt}. \end{align*} Now $$ \left| {\int_{ - \infty }^{ + \infty } {t\cos (e^{m - 1/2} e^t )e^{ - t^2 } dt} } \right| \le \int_{ - \infty }^{ + \infty } {\left| t \right|e^{ - t^2 } dt} = 1. $$ Thus, $$ \left| {\int_{ - \infty }^{ + \infty } {\sin (e^t )e^{ - (t - m)^2 } dt} } \right| \le 2e^{1/4} \frac{1}{{e^m }}, $$ showing that the limit is $0$.

Answer 2

Start with some standard transformations $$\begin{align} \int_{-\infty}^{\infty} \sin(e^t) e^{-(t-m)^2}dt &=\int_{-\infty}^{\infty} \sin(e^m e^t) e^{-t^2}dt = \int_{0}^{\infty} \sin(e^m s)e^{ -\ln(s)^2 -\ln(s)}ds \\ & = \frac{1}{2}\int_{\mathbb{R}}\sin( e^ms)g(s)ds \end{align}$$ where $g(s) =\text{sgn}(s)e^{ -\ln(|s|)^2 -\ln(|s|)}$

Clearly, $g \in L^1$, hence a standard result in complex analysis tells us that $\hat{g} \in \mathcal{C}_0$. And this clearly implies that : $$\lim_{m \rightarrow +\infty}\int_{\mathbb{R}}\sin( e^ms)g(s) = 0$$