Does there exists a function $ \ f(z) \in \mathbb{C} \ $ such that $ \ \lim_{z \to 0^{+}} f(z)=\infty \ $ but $ \ \int_{0}^{1} f(z) dz < \infty \ $ ?
If exists then give an example otherwise state why not exists.
Answer:
I think there does not exists such functions.
Because ,
$ \ \lim_{z \to 0^{+}} f(z)=\infty \ \Rightarrow f(z) \ \ is \ undefined \ \ as \ \ z \to 0^{+} $
Thus the integral $ \ \ \int_{0}^{1} f(z) dz \to \infty \ $
But I am not sure.
Help me out.
Let $f(x)=\frac{1}{\sqrt{x}}$ for $x \in (0,1]$. Then $\lim_{x \to 0^{+}} f(x)=\infty$, but the improper integral $ \int_{0}^{1} f(x) dx $ is convergent.