Irregularity of $\{a^iba^j : i \neq j\}$ using pumping lemma
Here is the contrapositive form of the pumping lemma, which is what we use to prove that a language is not regular:


  Let $L$ be a language. Suppose that for all $p \geq 1$ there exists a word $w \in L$ of length at least $p$ such that for every decomposition $w=xyz$ in which $|xy| \leq p$ and $y \neq \epsilon$ there exists an integer $i \geq 0$ such that $xy^iz \notin L$. Then $L$ is not regular.


Using the pumping lemma directly for the language $L$ in the question is a bit tricky. Given $p \geq 1$, let $w = a^p b a^{p+p!} \in L$; note that $|w| \geq p$. We have to show that for every decomposition $w = xyz$ in which $|xy| \leq p$ and $y \neq \epsilon$ there exists an integer $i \geq 0$ such that $xy^iz \notin L$. Consider any such decomposition. It is easy to check that $y = a^\ell$ for some $1 \leq \ell \leq p$, and that $xy^ij = a^{p+(i-1)\ell}ba^{p+p!}$. For $i = p!/\ell+1$ (which is an integer!), $xy^ij = a^{p+p!}ba^{p+p!} \notin L$. Thus the condition in the pumping lemma is satisfied, and we conclude that $L$ is not regular.