Star product of two commuting spinors Ok so this might be a very stupid and trivial question but I have spent a couple of hours on this little problem.
I am trying to derive a simple formula in a paper. We have a real commuting spinorial variable $y_\alpha$ ($\alpha = 1,2$) ($y^\alpha = \epsilon^{\alpha\beta}y_\beta$, $\epsilon^{\alpha\beta}=-\epsilon^{\beta\alpha}$, $\epsilon^{12}=\epsilon_{12}=1$).
The linear space of functions of the variables $y_\alpha$ is endowed with the structure of the algebra with the following $*$ product law:
$$(f*g)(y) = \int_{\mathbb{R}^4} \! d^2u~d^2v~ 
\exp \left[i u_{\alpha}  \epsilon^{\alpha\beta}  v_{\beta}\right]
f(y+u)g(y+v).$$
I am trying to compute the $*$ commutator 
$$[y_\alpha,y_\beta]_{*} := y_\alpha * y_\beta - y_\beta * y_\alpha = 2i\epsilon_{\alpha\beta},$$
but when I try to use the formula for lets say computing $y_1*y_2$, I get a divergent integral
$$y_1*y_2 = \int d^2u\,d^2v \,\,\, (y_1+u_1)(y_2+v_2)\exp[i (u_1v_2-u_2v_1)].$$
I feel like I am applying the formula wrong but I can't understand where. Can someone help me?
 A: The integral formula for the Moyal/Groenewold $*$ product in Ref. 1 reads
$$\tag{1} (f * g)(y)~=~\int_{\mathbb{R}^4} \! d^2u~d^2v~ 
\exp \left[i u_{\alpha}  \epsilon^{\alpha\beta}  v_{\beta}\right]
f(y+u)g(y+v).$$
Remark: The argument $u_{\alpha}  \epsilon^{\alpha\beta}  v_{\beta}$ inside the exponential of the integral formula (1) has an interesting geometric interpretation as (twice) a signed area of a triangle in phase space $\mathbb{R}^2$, cf. Ref. 2.
Usually the Moyal/Groenewold $*$ product is defined as
$$\tag{2} (f * g)(y)~:=~ f(y) \exp \left[\stackrel{\leftarrow}{\frac{\partial}{\partial y_{\alpha}}}  i\epsilon_{\alpha\beta}  \stackrel{\rightarrow}{\frac{\partial}{\partial y_{\beta}}} \right] g(y). $$
Let us for completeness prove$^1$ the integral formula (1) from the definition (2) for a pair of sufficiently well-behaved functions $f$ and $g$:
$$(f * g)(y)~\stackrel{(2)}{:=}~ f(y) \exp \left[\stackrel{\leftarrow}{\frac{\partial}{\partial y_{\alpha}}}  i\epsilon_{\alpha\beta}  \stackrel{\rightarrow}{\frac{\partial}{\partial y_{\beta}}} \right] g(y) $$
$$ ~=~\int_{\mathbb{R}^4} \! d^2u~d^2v ~\delta^2(u)~\delta^2(v) \exp \left[\frac{\partial}{\partial u_{\alpha}}  i\epsilon_{\alpha\beta}  \frac{\partial}{\partial v_{\beta}}\right]
f(y+u)g(y+v)$$
$$ ~=~\int_{\mathbb{R}^8} \! d^2u~d^2v~\frac{d^2p}{(2\pi)^2}\frac{d^2q}{(2\pi)^2} 
\exp i\left[p\cdot u+q\cdot v+\frac{\partial}{\partial u_{\alpha}}  \epsilon_{\alpha\beta}  \frac{\partial}{\partial v_{\beta}}\right]
f(y+u)g(y+v)$$
$$ ~\stackrel{\text{int. by part}}{=}~\int_{\mathbb{R}^8} \! d^2u~d^2v~\frac{d^2p}{(2\pi)^2} \frac{d^2q}{(2\pi)^2} 
\exp i\left[p\cdot u+q\cdot v-p^{\alpha}  \epsilon_{\alpha\beta}  q^{\beta}\right]
f(y+u)g(y+v)$$
$$ ~=~\int_{\mathbb{R}^6} \! d^2u~d^2v~\frac{d^2q}{(2\pi)^2} ~\delta^2(u_{\alpha}-\epsilon_{\alpha\beta}  q^{\beta}) ~e^{i q\cdot v}
f(y+u)g(y+v)$$
$$ ~\stackrel{(6)}{=}~\int_{\mathbb{R}^6} \! d^2u~d^2v~\frac{d^2q}{(2\pi)^2} ~\delta^2(  q^{\beta}-u_{\alpha}\epsilon^{\alpha\beta} ) ~e^{i q\cdot v}
f(y+u)g(y+v)$$
$$\tag{3} ~=~\int_{\mathbb{R}^4} \! d^2u~d^2v~ 
\exp \left[i u_{\alpha}  \epsilon^{\alpha\beta}  v_{\beta}\right]
f(y+u)g(y+v).$$
The functions $f$ and $g$ should be sufficiently well-behaved in order for the above integrals and manipulations (3) to make mathematical sense. Let us 
here give a necessary and sufficient condition for the integrand 
$$\tag{4} h(u,v) ~:=~\exp \left[i u_{\alpha}  \epsilon^{\alpha\beta}  v_{\beta}\right]
f(y+u)g(y+v), \qquad u,v,y ~\in ~\mathbb{R}^2, $$
of the integral formula (1) to be integrable. In general an integrand $h$ is integrable (i.e belongs to $L^1$) if and only if (i) the integrand $h$ is Lebesgue measurable, and (ii) the absolute value $|h|$ of the integrand has a finite integral $\int |h| dm<\infty$. Note that $|h(u,v)|=|f(y+u)|~|g(y+v)|$ factorizes in a $u$- and a $v$-dependent factor. Assuming that both functions $f$, $g$ (and therefore $h$) are Lebesgue measurable, this means (via Tonelli's and Fubini's theorems) that the integrand (4) is integrable $h\in L^1(\mathbb{R}^4)$ if and only if (i) both $f,g\in L^1(\mathbb{R}^2)$ are integrable, or (ii) at least one of the two functions $f$ and $g$ vanishes almost everywhere.
For instance, inserting the two non-integrable first-order polynomials $f(y)=y_{\alpha}$ and $g(y)=y_{\beta}$ inside the integral formula (1) is ill-defined as OP also mentions. 
On the other hand, the definition (2) does not involve integrals. So we can put $f(y)=y_{\alpha}$ and $g(y)=y_{\beta}$ in eq. (2) to derive the sought-for formula
$$\tag{5} y_{\alpha} * y_{\beta}~=~y_{\alpha} y_{\beta}+ i\epsilon_{\alpha\beta}. $$
References:


*

*M.A. Vasiliev, Unfolded representation for relativistic equations in 2 + 1 anti-de Sitter space, Class. Quant. Grav. 11, (1994) 649. Note that there is an imaginary unit $i$ missing in the published version of formula (1). A preprint version from KEK Preprint Library has the $i$ factor so it seems that the $i$ was lost during the publishing phase.

*C. Zachos, Geometrical Evaluation of Star Products, J. Math. Phys.  41 (2000) 5129, arXiv:hep-th/9912238.
--
$^1$ Note that Ref. 1 uses the convention
$$\tag{6} \epsilon_{\alpha\beta} \epsilon^{\beta\gamma}~=~-\delta_{\alpha}^{\gamma}. $$
A: I worked a lot with this kind of formulas :) It would be useful to recover the imaginary unit in $\exp$ and think of integration variables as taking real values. Then, upon appropriate normalization $$ \int\int \exp (i u_\alpha v^\alpha) d^2u d^2v=\int \delta^2(u) d^2u=1$$
Once you agree with this formula, which is a standard representation for the delta function (that the variables have some spinorial meaning is irrelevant, they are usual commuting variables) the rest is easy.
For example, let us compute $y_\alpha \star g(y)$ where $g(y)$ is any function
$$\int (y+u)_\alpha\, g(y+v) \exp (iu_\alpha v^\alpha)=\\\int y_\alpha\, g(y+v) \exp (iu_\alpha v^\alpha) +\int u_\alpha\, g(y+v) \exp (iu_\alpha v^\alpha)=A+B$$
A is easy, one can integrate over $u$ to get delta function
$$A=\int \delta^2(v) y_\alpha g(y+v) d^2 v=y_\alpha g(y)$$
To compute B we integrate by parts
$$ B=\int g(y+v)(-i\frac{\partial}{\partial v^\alpha}\exp (iu_\alpha v^\alpha))=\\
=\int (i\frac{\partial}{\partial v^\alpha}g(y+v))\exp (iu_\alpha v^\alpha)=\\
=i\frac{\partial}{\partial y^\alpha}\int g(y+v)\exp (iu_\alpha v^\alpha)=\\
=i\frac{\partial}{\partial y^\alpha} g(y)$$
Alltogether we derived,
$$y_\alpha \star g(y)=(y_\alpha+i \partial_\alpha)g(y)$$
which for example gives
$$y_\alpha \star y_\beta=y_\alpha y_\beta+i \partial_\alpha y_\beta=y_\alpha y_\beta+i\epsilon_{\alpha\beta}$$
The integral formula for star product is a very useful one if you are to compute star products of something more complicated than polynomials. This is what happens with Vasiliev theory since the functions encode higher-spin fields together with all derivatives thereof, so only the anti-de Sitter solution (vacuum) is of type $yy$ while perturbations are analytic functions of $y$. For example, the bulk-to-boundary propagator is somethins like $\exp(yy+y)$.
That the integral formula is more useful can be seen in trying to compute the star product of two gaussians $\exp(yy)\star \exp(yy)$, which is just a gaussian integral with the help of the integral formula and I have no idea how to compute this using $\exp(\partial\partial)$ formula
