Please refere to the attached image (source: Wikipedia). They have derived the formula (summation one) for discrete random variable as a special case. Can anybody please help me understand how to derive that formula from the integral formula? I already have understanding for the integral formula(continuous case).
Henry has provided rather ingenious hints which OP can use to formally jot down the proof. Unless OP adds anything further regarding their attempt, I am not proceeding on that front.
For broader purpose, here is another take and a commentary.
$\bullet$ Observe
\begin{align}\sum_{n~=~0}^\infty \mathbf P[X >n] &= \sum_{n~=~1}^\infty \mathbf P[X \geq n]\\ &=\sum_{n~=~1}^\infty \left\{ \sum_{x~=~n}^\infty\mathbf P[X = x] \right\}\\ &= \sum_{x~=~1}^\infty \left\{ \sum_{n~=~1}^x\mathbf P[X = x] \right\}~~~~~~~~~\because\textrm{(Fubini's Theorem)}\\ &= \sum_{x~=~1}^\infty \left\{\mathbf P[X = x] \sum_{n~=~1}^x1\right\}; \tag 1\label 1\end{align}
from $\eqref 1$ the rest follows.
$\bullet$ If $X$ is only a non-negative random variable, then
\begin{align}\sum_{n~=~1}^\infty \mathbf P[X \geq n] & = \sum_{n~=~1}^\infty \left\{ \mathbf P[n \leq X < n+1] + \mathbf P[n+1 \leq X < n+2]+ \ldots\right\}\\ &=\sum_{\ell~=~1}^\infty \ell\times \mathbf P[\underbrace{\ell \leq X < \ell+1}_{\lfloor X\rfloor ~=~ \ell}]\\ &= \mathbb E\lfloor X\rfloor.\tag 2\end{align}
$\rm [I]$ A First Look at Rigorous Probability Theory, Jeffrey S. Rosenthal, World Scientific Publishing, $2006,$ sec. $4.3,$ p. $49.$
