I saw that physicists use the following integral $$ \int_{0}^{\infty} \sin x dx = 1 $$ in a distribution sense, i.e. $$ \int_{0}^{\infty} \sin x dx := \lim_{a\to 0} \int_{0}^{\infty} e^{-ax}\sin x dx = \lim_{a\to 0} \frac{1}{a^{2}+1} = 1 $$ However, I wonder we get the same value when we change the distribution part. For example, we have the following result (according to Wolfram alpha) $$ \lim_{a\to 0} \int_{0}^{\infty} e^{-ax^{2}}\sin x dx = \lim_{a\to 0}\frac{F\left(\frac{1}{2\sqrt{a}}\right)}{\sqrt{a}} = \lim_{a\to 0 }\frac{e^{-1/4a}}{\sqrt{a}}\int_{0}^{\frac{1}{2\sqrt{a}}}e^{y^{2}}dy =1 $$ where $F(x):=e^{-x^{2}}\int_{0}^{x}e^{y^{2}}dy$ is a Dawson's integral.
To show this, we have to show the following : suppose a function $f:\mathbb{R}_{\geq 0}\times \mathbb{R}_{\geq 0}\to \mathbb{R}$ is continuous and the integral $$ \int_{0}^{\infty} f(a, x)\sin x dx $$ converges for any $a\geq 0$. If $f(0, x)=0$ for all $x\geq 0$, we have $$ \lim_{a\to 0} \int_{0}^{\infty} f(a, x)\sin x dx = 0 $$ It seems that we cannot use dominated convergence theorem since $\sin x\not\in L^{1}$. If the above statement is true, is there any other function than $\sin x$ which satisfies the property and can define the improper integral in this way?