If $I(t)$ is an improper integral type I, we already know that:
$$I(t)=\int\limits_{a}^{\infty} f(x,t)dx \text{ converges uniformly}$$ $$\Leftrightarrow \forall \epsilon>0, \exists A_0\in(a,\infty), \forall A \in (A_0,\infty), |\int\limits_{A}^{\infty} f(x,t)dx| <\epsilon$$ $$\Leftrightarrow \lim_{A\rightarrow \infty} \sup|\int\limits_{A}^{\infty} f(x,t)dx|=0 $$
My question is if $I(t)=\int\limits_{a}^{b} f(x,t)dx$ is an improper integral type II with $a$ is a singular point, then we have
$$I(t)=\int\limits_{a}^{b} f(x,t)dx \text{ converges uniformly}$$ $$\Leftrightarrow \forall \epsilon>0, \exists A_0\in(a,b), \forall A \in (a,A_0), |\int\limits_{A}^{b} f(x,t)dx| <\epsilon$$ $$\Leftrightarrow \lim_{A\rightarrow a^{+}} \sup|\int\limits_{A}^{b} f(x,t)dx|=0 $$
or we have $$I(t)=\int\limits_{a}^{b} f(x,t)dx \text{ converges uniformly}$$ $$\Leftrightarrow \forall \epsilon>0, \exists A_0\in(a,b), \forall A \in (a,A_0), |\int\limits_{a}^{A} f(x,t)dx| <\epsilon$$ $$\Leftrightarrow \lim_{A\rightarrow a^{+}} \sup|\int\limits_{a}^{A} f(x,t)dx|=0 $$
Which one is true? Anyone help me please? Can you give me a complete definition of uniform convergence in this way?