In the article: Residues Of Codimension one Singular Holomorphic Distribution, the author : Takeshi Izawa, states that:
Theorem: Let $X$ an $n$-dimensional compact complex manifold and $\mathcal{G}$ a rank-one locally-free subsheaf of $\Omega_{X}$. We assume that the singular supports of $\Omega_{X}/ \mathcal{G}$ are all isolated. Then we have:
$$ \int_{X}c_{n}(\Omega_{X} \otimes \mathcal{G}^{\vee}) = \displaystyle \sum_{j = 1}^{k}\text{Res}_{p_{j}} \left[ \begin{array}{cccc} df_{1}^{(j)} \wedge & \cdots & \wedge \,\, df_{n}^{(j)}\\ f_{1}^{(j)} \wedge & \cdots & \wedge f_{n}^{(j)} \end{array} \right] $$
where above there is the Grothendieck residue. In this article, the author provides no example of how to calculate this residue.
The Distribution $\mathscr{F}$of codimension one on $\mathbb{P}^{3}$ induced by: $$\omega = (z_{0}^{2} + z_{1}^{2} + z_{2}^{2})dz_{3} - (z_{3}z_{0} + z_{2}z_{1})dz_{0} + (z_{2}z_{0} - z_{3}z_{1})dz_{1} - z_{3}z_{2}dz_{2}$$
It has as singular scheme : $\lbrace 2[i : -1 : 0 : 0], 2[i : 1 : 0 : 0], [0 : 0 : 0 : 1] \rbrace$.
Frankly, I don't know how to calculate the Grothendieck residue for singular points for the above distribution. References on this subject and examples will be greatly appreciated.
Thanks a lot.
From the very definition of Grothendieck Residue: Let $\omega = \sum_{j=1}^n f_jdz_j$ be a a local one form defining the distribution around a point $p$ (which we choose to be the origin in the coordinates $z$). Then, for a small polydisc $P$ with essential boundary $\Gamma$ we have $$ \text{Res}_{p} \left[ \begin{array}{cccc} df_{1}^{(j)} \wedge & \cdots & \wedge \,\, df_{n}^{(j)}\\ f_{1}^{(j)} \wedge & \cdots & \wedge f_{n}^{(j)} \end{array} \right] = \frac{1}{(2\pi i)^n} \int_\Gamma \frac{\det(Jf)}{f_1\dots f_n}dz_1 \wedge \dots \wedge dz_n $$ where $Jf$ is the jacobian matrix of $(f_1 , \dots, f_n)$. For a reference see Soares - Lectures on Point Residues.
Now for the example. I will go for the last point $(0:0:0:1)$. Then $z_3=1$ and in coordinates $(z_0,z_1,z_2)$ we have $$ \omega = -(z_{0} + z_{2}z_{1})dz_{0} + (z_{2}z_{0} - z_{1})dz_{1} - z_{2}dz_{2} $$ i.e. $f_0 = -z_{0} -z_{2}z_{1}$, $f_1 = z_{2}z_{0} - z_{1}$ and $f_2 = - z_{2}$. $$ J(f_0,f_1,f_2) = \begin{pmatrix} -1 & -z_2 & -z_1 \\ z_2 & -1 & z_0 \\ 0 & 0 & -1 \end{pmatrix} $$ whose determinant is $\det J(f_0,f_1,f_2) = -(1+z_2^2)$. Give the an appropriate polydisc we have that the residue is $$ \frac{1}{(2\pi i)^3} \int_\Gamma \frac{-(1+z_2^2)dz_0 \wedge dz_1 \wedge dz_2}{(z_0+z_1z_2)(z_0z_2-z_1)z_2} = \frac{1}{(2\pi i)^3} \int_\Gamma\left[\frac{dz_0 \wedge dz_1 \wedge dz_2}{z_0z_1z_2}+ \dots\right] =1 $$ You'll certainly find the technical steps I have not gone through in Soares' notes.
ADDED: The other singularities $(i:\pm 1:0:0)$ lie on the open subset $\{z_0 \neq 0 \}$ where the form is $$ \omega = (1 + z_{1}^{2} + z_{2}^{2})dz_{3} + (z_{2} - z_{3}z_{1})dz_{1} - z_{3}z_{2}dz_{2} $$
then $f_1 = (z_{2} - z_{3}z_{1})$, $f_2 = - z_{3}z_{2}$ and $f_3 = (1 + z_{1}^{2} + z_{2}^{2})$. $$ J(f_1,f_2,f_3) = \begin{pmatrix} -z_3 & 1 & -z_1 \\ 0 & -z_3 & -z_2 \\ 2z_1 & 2z_2 & 0 \end{pmatrix} $$ whose determinant is $\det J(f_0,f_1,f_2) = -2(z_1^2z_3 +z_2^2z_3+z_2z_1)$. Then the residue is $$ \frac{1}{(2\pi i)^3} \int_\Gamma \frac{2(z_1^2z_3 +z_2^2z_3+z_2z_1)dz_1 \wedge dz_2 \wedge dz_3}{(z_{2} - z_{3}z_{1})(1 + z_{1}^{2} + z_{2}^{2})z_{3}z_{2}} $$ where $\Gamma$ is around one of the singularities, that in our chart are $(\pm i ,0,0)$. Lets compute the residue for $(i,0,0)$.
Then we make a translation $z_1 \mapsto z_1+i$ and compute the residue ate the origin (just to make computations easier). Our integrand becomes $$ \frac{2((z_1+i)^2z_3 +z_2^2z_3+z_2(z_1+i))}{(z_{2} - z_{3}(z_{1}+i))(1 + (z_{1}+i)^{2} + z_{2}^{2})z_{3}z_{2}} = \frac{2(-z_3+iz_2+2iz_1z_3+z_1^2z_3+z_2^2z_3+z_1z_2)}{(z_2-iz_3-z_1z_3)(2iz_1+z_1^2+z_2^2)z_2z_3} = \\ = \frac{2}{(z_2-iz_3-z_1z_3)z_2} + \frac{-2}{(z_2-iz_3-z_1z_3)(2iz_1+z_1^2+z_2^2)z_2}+ \\+ \frac{2i}{(z_2-iz_3-z_1z_3)(2iz_1+z_1^2+z_2^2)z_3}+ \frac{2z_1}{(z_2-iz_3-z_1z_3)(2iz_1+z_1^2+z_2^2)z_3} $$ The contributions of each of the four summands to the residue are, respectively $0,1,1,0$. Hence the residue is $2$.
Alternative approach: This residue is just the Milnor number of the map $(f_1,f_2,f_3)$ at $(i,0,0)$ defined by the dimension of the algebra $$\frac{\mathbb{C}[z_1,z_2,z_3]_{(i,0,0)}}{(f_1,f_2,f_3)} \simeq \mathbb{C}^2 $$ generated by the classes of $1$ and $z_2$.
Also note that if we change $i$ to $-i$ every thing goes the same way.