Fourier transform of Gaussian

Question

In the abundant materials regarding solving the Fourier transform of Gaussian using differentiation and integration by part (such as Fourier transform of Gaussian?), I'm struggling understanding why the term $\left(\frac{i}{2a\sqrt{2\pi}}e^{-i\alpha x}e^{-x^2/a}\right)_{-\infty}^{+\infty} = 0$. Can anyone provide an explanation or some background materials for me to read?

Answer 1

As hinted at in the comments, this identity is to be understood as (omitting the factor in front)

$$ \lim_{T\rightarrow \infty}{\left[ e^{-i\alpha x}e^{-x^2/a}dx\right]_{x=-T}^{x=T}}= \lim_{T\rightarrow \infty}{\left( e^{-i\alpha T}e^{-T^2/a} -e^{i\alpha T}e^{-T^2/a} \right)} \\= \lim_{T\rightarrow \infty}{\left( e^{-i\alpha T}e^{-T^2/a} \right)} -\lim_{T\rightarrow \infty}{\left( e^{i\alpha T}e^{-T^2/a} \right)} = 0 - 0 = 0 $$

To understand this in the context of the answer linked to in your question, you may want to read up on the fundamental theorem of calculus and absolutely convergent integrals or improper (Riemann-)Integrals.