the Kowalevski configuration if in addition to , we have $$\begin{array}{l}
  H^{(2)}={\displaystyle \frac{1}{2}} \mathbf{g} \mathbf{M} {\bdot} \mathbf{M}, \qquad \mathbf{g}=\mathop{\rm diag}\nolimits \{\frac{1}{2},\frac{1}{2},1\}.
\end{array}$$ Here the constant multiplier of $H^{(2)}$ is due only to the choice of measure units. The essential condition is the ratio $1{:}1{:}2$ of the eigenvalues of the matrix $\mathbf{g}$ inverse to the inertia tensor of the body.

The classical Kowalevski case corresponds to $H^{(1)}\equiv 0, {{\boldsymbol \beta}}\equiv 0$: $$\label{eq2_5}
  H_{\rm class}=\omega_1^2+\omega_2^2 +\frac{1}{2}\omega_3^2-\alpha_1.$$ The first integrals then are the area integral $$\label{eq2_6}
\begin{array}{l}
  L_{\rm class} = {\displaystyle \frac{1}{2}} \mathbf{M} {\bdot} {{\boldsymbol \alpha}}= \alpha_1 \omega_1 + \alpha_2 \omega_2+ {\displaystyle \frac{1}{2}} \alpha_3 \omega_3\\
\end{array}$$ and the famous Kowalevski integral $$\label{eq2_7}
\begin{array}{l}
  K_{\rm class} = (\omega_1^2-\omega_2^2+\alpha_1)^2+ (2\omega_1 \omega_2+\alpha_2)^2.
\end{array}$$

For the general Kowalevski configuration under certain conditions the first integrals additional to $H$ were found in [@Bogo; @Yeh; @ReySem]. O.I.Bogoyavlensky [@Bogo] considered the case $H^{(1)}\equiv 0, {{\boldsymbol \beta}}\ne 0$ and found the first integral $$\begin{array}{l}
  K_{\rm B} = (\omega_1^2-\omega_2^2+\alpha_1-\beta_2)^2+ (2\omega_1 \omega_2+\alpha_2+\beta_1)^2.
\end{array}$$ Since for ${{\boldsymbol \beta}}\ne 0$ there is no area integral, Bogoyavlensky pointed out the special case $K_{\rm B}=0$ and proved the integrability of the system on this 4-dimensional submanifold in ${\mathcal{P}}^6$. In more general case $$\label{eq2_8}
  H^{(1)} = -\lambda M_3, 