$$F(X)=\frac{99}{1-X}\quad\text{and}\quad X\equiv Uniform[0,1]$$ What is the expected value of above distribution. I tried calculating but I find the expected value integral doesn't converge.
Is my calculation correct? What is the implication of expected value of integral not converging? Is the variance also not convergent?
Indeed, it does not converge: $$ \mathbb{E}[F(X)] = \int_0^1 F(x)dx = \int_0^1 \frac{99}{1-x}dx = +\infty $$
This is alright, and can happen: not every random variable has a finite expectation! (For instance, look at the famous Cauchy distribution.)
What this means is exactly that: the expectation of the random variable $F(X)$ is infinite. (It is, at least, defined, since $F(X)\geq 0$: it just is infinite. The expectation of a Cauchy r.v. is not even defined, since a Cauchy r.v. takes both negative and positive values.)
A consequence? Indeed, then it does not have a variance either. This is because "having a variance implies having an expectation" (or, in fancier terms, $L_1$-integrable implies $L_2$-integrable for random variables). To show why, you can e.g. use Cauchy--Schwarz: $$ 0 \leq \mathbb{E}[\lvert Y\rvert ] = \mathbb{E}[\lvert Y\rvert\cdot 1] \leq \sqrt{\mathbb{E}[Y^2]\mathbb{E}[1]} = \sqrt{\mathbb{E}[Y^2]} $$ Thus, if $\mathbb{E}[Y^2] <\infty$, then $\mathbb{E}[\lvert Y\rvert ]<\infty$ as well, meaning that $Y$ is integrable (has a finite expectation).