I'm being very confused with an improper integral of two variables.
When an improper integral converges, $$ \lim_{t\rightarrow \infty} \int_0^t f(s)ds $$ , it implies $\lim_{s\rightarrow \infty} f(s) = 0$.
But, what if the integrand $f(s)$ is changed to $f(t,s)$? In other words,
When $\lim_{t\rightarrow \infty} \int_0^t f(t,s) ds $ converges, what can we say about necessary condition for $f(t,s)$?
My intuitive answer is $$ \lim_{t\rightarrow \infty} \lim_{s\rightarrow t} f(t,s) = 0. $$
For example, if the folowing improper integral converges, $$ \lim_{t\rightarrow \infty} \int_0^t g(s)(1 + t-s) ds $$ , can we say $$ \lim_{t\rightarrow\infty}\lim_{s\rightarrow t} g(s)(1+t-s) = \lim_{t\rightarrow\infty} g(t) =0? $$
My guess is true? if so, how to prove it? Thanks in advance!
It is not true, that if the improper integral $\int_0^{\infty} f(s)ds$ converges , it implies that $\lim_{s\rightarrow \infty} f(s) = 0$ !!
Take for example $f(s)= \cos(s^2)$.
https://en.wikipedia.org/wiki/Fresnel_integral