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 $f(s)= \cos(s^2)$.


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

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