Should the complex conjugate of a derivative of a Grassmann number include a sign? Take a real Grassmann variable, by which I mean $\theta=\theta^*$. We have
$$\int d\theta~ \theta =1,\qquad  \frac{\partial}{\partial\theta}\theta=1$$
If I define the conjugation of Grassmann variables to invert their order, $$(\eta\theta)^*= \theta^*\eta^*,$$ should I then have $$(d\theta\theta)^*=\theta^*d\theta^*~?$$ But this means
$$(\int d\theta~\theta)^*=1 \quad \Rightarrow \quad\int \theta^* d\theta^*=-\int d\theta^* ~\theta^*=1,$$
so if $\theta=\theta^*$, I should have $$d\theta^*=-d\theta.$$ The same can be found for the derivative of $\theta$.
Is this the usual convention to take, or should I instead choose $$(d\theta \theta)^*=d\theta^* \theta^*~?$$
 A: Yes. OP is right. There is a minus. Since by convention the complex conjugation obeys 
$$ (z w)^{\ast} ~=~ w^{\ast}z^{\ast}~=~(-1)^{|z|~|w|} z^{\ast}w^{\ast} \tag{1}$$
for any two supernumbers $z$, $w$ (of definite Grassmann parities $|z|$,$|w|$), we should also have 
$$ (A f)^{\ast} ~=~(-1)^{|A| ~|f|} A^{\ast}f^{\ast} \tag{2}$$ 
for complex conjugation of an operator $A$ and a function $f$, cf. e.g. Refs. 1 & 2. Eq. (2) reduces to eq. (1) if $A$ is a left multiplication operator. It is easy to check that eq. (2) implies that$^{1}$
$$   \left(\frac{\partial_L}{\partial z}\right)^{\ast}~\stackrel{(2)}{=}~ (-1)^{|z|} \frac{\partial_L}{\partial (z^{\ast})}.\tag{3}$$ 
Since Berezin integration is the same as left differentiation
$$ \int \!d\theta ~=~\frac{\partial_L}{\partial \theta}, \qquad 
\int \!d\theta^{\ast} ~=~\frac{\partial_L}{\partial (\theta^{\ast})},\tag{4} $$
we derive that complex conjugation of Grassmann-odd differentiation produces a minus
$$ \left( \int \!d\theta \right)^{\ast} 
~\stackrel{(4)}{=}~\left(\frac{\partial_L}{\partial \theta}\right)^{\ast}
~\stackrel{(3)}{=}~-\frac{\partial_L}{\partial (\theta^{\ast})} 
~\stackrel{(4)}{=}- \int \!d\theta^{\ast}.\tag{5} $$
References:


*

*B. DeWitt, Supermanifolds, Cambridge Univ. Press, 1992; eq. (2.2.19).

*S.J. Gates, M.T. Grisaru, M. Rocek & W. Siegel, Superspace, or One thousand and one lessons in supersymmetry, arXiv:hep-th/0108200; eq. (3.1.9).
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$^{1}$ The subscript $L$ ($R$) denotes left (right) differentiation, i.e. acting from left (right), respectively. For completeness, let us mention that 
left and right differentiation are connected via the formula
$$ \frac{\partial_L f}{\partial z}~=~(-1)^{(|f|+1)|z|}\frac{\partial_R f}{\partial z},\tag{6}  $$
so that complex conjugation satisfies
$$\left(\frac{\partial_L}{\partial z}\right)^{\ast}f~\stackrel{(3)+(6)}{=}~ (-1)^{|z||f|} \frac{\partial_R f}{\partial (z^{\ast})}, \qquad \left(\frac{\partial_R}{\partial z}\right)^{\ast}f~\stackrel{(3)+(6)}{=}~ (-1)^{|z||f|} \frac{\partial_L f}{\partial (z^{\ast})}.\tag{7}  $$
