Necessary condition for an improper integral of two variables to converge

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 $f(s)= \cos(s^2)$.
• Thank you for your reply! to ensure my claim, the assumption that $\lim_{s\rightarrow \infty} f(s)$ exists is needed. – user155214 Jun 5 '18 at 11:36