is there an equivalent of this theorem $\int_a^b f(x) dx=F(b)-F(a)$ for improper intergrals, maybe limits appear on the RHS ? If so, how to prove it ?
My intuition is that in $\int_a^b f(x) dx, a<x<b$ so maybe we will have $\int_a^b f(x) dx=\lim_{x\to b^-}F(b)-\lim_{x\to a^+}F(a)$
Thank you
Tom
By definition, the improper integral is $$\int_a^b f(x) dx = \lim_{u\to b-} \lim_{l\to a+} \int_l^u f(x)dx$$ By the FTC, $$\int_l^u f(x)dx = F(u) - F(l)$$ where $F' = f$.
So $$\begin{align}\int_a^b f(x) dx &= \lim_{u\to b-} \lim_{l\to a+} \int_l^u f(x)dx \\&= \lim_{u\to b-} \lim_{l\to a+}[F(u) - F(l)]\\&=\lim_{u\to b-} F(u) - \lim_{l\to a+}F(l)\end{align}$$