Hints for evaluating $ \lim_{y \to +\infty}y \int_0^{+\infty}{e^{-x^2}\sin(2xy) dx}$

Question

Please give me some hints for this limit.

$ \lim_{y \to +\infty} y\int_0^{+\infty}{e^{-x^2}\sin(2xy) dx}$

Answer 1

dIntegrate by parts twice. \begin{align} I &:=y\int_0^\infty e^{-x^2}\sin(2xy) dx \\ &=-\frac{1}{2}\int_{x=0}^\infty e^{-x^2}d\cos(2xy) \\ &= \frac{1}{2}-\int_{x=0}^\infty \cos(2xy)xe^{-x^2}dx \\ &= \frac{1}{2}-\frac{1}{2y}\int_{x=0}^\infty xe^{-x^2}d\sin(2xy) \\ &= \frac{1}{2}+\frac{1}{2y}\int_{x=0}^\infty \sin(2xy)(1-2x^2)e^{-x^2} dx \tag{1} \end{align} We see that $$\bigg|\int_{x=0}^\infty \sin(2xy)(1-2x^2)e^{-x^2} dx\bigg|<\int_{x=0}^\infty (1-2x^2)e^{-x^2} dx,$$ as $|\sin(u)|\le 1$. The right hand side is convergent, so the left hand side is bounded (actually convergent by examining the right tail integrals) uniformly for all $y$.

Therefore the second term of Eq. (1) vanishes as $y\to\infty$ and $$I=\frac{1}{2}+\mathcal O\Big(\frac{1}{y}\Big).$$