# Linear probing and tabulation hashing

I'm currently reading the paper "The Power of Simple Tabulation Hashing" by Mihai Patrascu and Mikkel Thorup [1] because I want to adapt the proof of the constant time complexity of linear probing for modified tabulation hashing. I would like to ask a question about Corollary 11 (page 17). In this corollary is shown that $$Pr[R \ge l] \le 2^{-\Omega(l\epsilon^2)} + (l/m)^\gamma$$ for $$\alpha \ge 1/2$$ and $$Pr[R \ge l] \le \alpha^{\Omega(l)} + (l/m)^\gamma$$ for $$\alpha \le 1/2$$ when $$\gamma = O(1)$$ and $$l \le n^{1/(3c)}/\alpha$$. Here R is the length of the run during linear probing and $$c \ge 1$$.

Then as I understand, we can prove that $$E[R] = O(1/\epsilon^2)$$ using the formula $$E[X] = \sum_{l=1}^{n} Pr[X \ge l]$$. Could you please explain how do we apply the last formula if $$l > n^{1/(3c)}/\alpha$$? Intuitively, I understand that we have a tight concentration around $$O(1/\epsilon^2)$$ and, thus, the expectation is around it. But how do we take into account that $$l$$ can be greater than $$n^{1/(3c)}/\alpha$$ when showing that $$E[R] = O(1/\epsilon^2)$$? Moreover, what if $$1/\epsilon^2 > n^{1/(3c)}/\alpha$$? Then how do we prove that $$E[R] = O(1/\epsilon^2)$$?

Thank you very much for you help.