Let $\mathcal{H}$ be a finite dimensional Hopf algebra. A nonzero element $\Omega\in \mathcal{H}$ is called an integral in $\mathcal{H}$ if $$x~\Omega=\epsilon(x)\Omega,~~\forall x\in \mathcal{H}.\tag{1}$$ Sweedler has proved (see here) that $\Omega$ exists and is unique (up to a constant multiplier) for finite dimensional Hopf algebras. However, I can't find an explicit expression for $\Omega$ written in terms of the basis elements of $\mathcal{H}$ (which I need for specific calculations). I have the following guess, but can't prove it: $$\Omega= \sum_{x\in B(\mathcal{H})}x_{(1)}\delta_x(S^2 (x_{(2)})),\tag{2}$$ where $B(\mathcal{H})$ is a basis of $\mathcal{H}$, $\Delta(x)=x_{(1)}\otimes x_{(2)}$ is the coproduct written in Sweedler's notation and $\delta_x$ is the dual of $x$ in the dual Hopf algebra $\mathcal{H}^*$, i.e. we have $\delta_x(y)=\delta_{x,y},\forall x,y\in B(\mathcal{H})$.
For finite dimensional semisimple Hopf algebras, we have $S^2=\mathrm{id}$ by a theorem of Larson and Radford. In this case we have a even simple formula $$\Omega= \sum_{x\in B(\mathcal{H})}x_{(1)}\delta_x( x_{(2)}),\tag{3}$$ in which case $\Omega$ is actually the trace of the regular corepresentation of $\mathcal{H}$, and Eq.(3) gives back the correct expression $\Omega=\sum_g g$ when $\mathcal{H}$ is a finite group algebra. Can someone help me prove that $\Omega$ defined in Eq.(2) satisfies Eq.(1)?
This answer is based on the paper "THE HAAR MEASURE ON FINITE QUANTUM GROUPS".
We write $$\Omega= \sum_{x\in B(\mathcal{H})}x_{(1)}\langle\delta_x,S^2 (x_{(2)})\rangle,$$ where $\langle,\rangle$ denotes the dual pairing between $\mathcal{H}^*$ and $\mathcal{H}$. Then for any $y\in\mathcal{H}$, we have \begin{eqnarray} \epsilon(y)~\Omega &=& \sum_{x\in B(\mathcal{H})} x_{(1)}\langle \delta_x,y_{(1)}S(y_{(2)})S^2(x_{(2)})\rangle\\ &=&\sum_{x,z\in B(\mathcal{H})} x_{(1)}\langle \delta_x,y_{(1)}z\rangle\langle\delta_z,S(y_{(2)})S^2(x_{(2)})\rangle\\ &=&\sum_{z\in B(\mathcal{H})} (y_{(1)}z)_{(1)}\langle\delta_z,S(y_{(2)})S^2[(y_{(1)}z)_{(2)}]\rangle\\ &=&\sum_{z\in B(\mathcal{H})} y_{(1)}z_{(1)}\langle\delta_z,S(y_{(3)})S^2(y_{(2)}z_{(2)})\rangle\\ &=&\sum_{z\in B(\mathcal{H})} y_{(1)}z_{(1)}\langle\delta_z,S(y_{(3)})S^2(y_{(2)})S^2(z_{(2)})\rangle\\ &=&\sum_{z\in B(\mathcal{H})} y_{(1)}z_{(1)}\langle\delta_z,S[S(y_{(2)})y_{(3)}]S^2(z_{(2)})\rangle\\ &=&\sum_{z\in B(\mathcal{H})} y_{(1)}\epsilon(y_{(2)})z_{(1)}\langle\delta_z,S(1)S^2(z_{(2)})\rangle\\ &=&y\sum_{z\in B(\mathcal{H})} z_{(1)}\langle\delta_z,S^2(z_{(2)})\rangle\\ &=&y\Omega. \end{eqnarray} This proves that $\Omega$ is a left integral in $\mathcal{H}$.