Then we have that

$${\it D \Phi}= \left( {\frac {\partial}{\partial \theta}} -i \theta {\frac {\partial}{\partial t}} \right) \left(

x \left( t \right) +i\theta\,\psi \left( t \right) \right)= i\psi \left( t \right) -i\theta {\frac {d}{dt}}x \left( t \right) $$

and

$${\it D^2 \Phi}= \left( {\frac {\partial}{\partial \theta}} -i \theta {\frac {\partial}{\partial t}} \right) (i\psi \left( t \right) -i\theta {\frac {d}{dt}}x \left( t \right))= -i{\frac {d}{dt}}x \left( t \right) +\theta\,{\frac {d}{dt}}\psi

\left( t \right)$$

Now we have that

$${\it D \Phi}{\it D^2 \Phi}=( i\psi \left( t \right) -i\theta {\frac {d}{dt}}x \left( t \right) )(-i{\frac {d}{dt}}x \left( t \right) +\theta\,{\frac {d}{dt}}\psi \left( t \right))$$

it is to say

$${\it D \Phi}{\it D^2 \Phi}=\left( {\frac {d}{dt}}x \left( t \right) \right) \psi \left( t \right) - \theta( {\frac {d}{dt}}x \left( t \right) ) ^{2}

-i\theta\,\psi \left( t \right) {\frac {d}{dt}}\psi \left( t \right)$$

Finally we have that

$$\int {d\theta {\it D \Phi}{\it D^2 \Phi}}=\int (\left( {\frac {d}{dt}}x \left( t \right) \right) \psi \left( t \right) - \theta( {\frac {d}{dt}}x \left( t \right) ) ^{2}

-i\theta\,\psi \left( t \right) {\frac {d}{dt}}\psi \left( t \right)) {d\theta }$$

it is to say

$$\int {d\theta {\it D \Phi}{\it D^2 \Phi}}={\frac {d}{d\theta}}(\left( {\frac {d}{dt}}x \left( t \right) \right) \psi \left( t \right) - \theta( {\frac {d}{dt}}x \left( t \right) ) ^{2}

-i\theta\,\psi \left( t \right) {\frac {d}{dt}}\psi \left( t \right)) $$

which is reduced to

$$\int {d\theta {\it D \Phi}{\it D^2 \Phi}}=- ( {\frac {d}{dt}}x \left( t \right) ) ^{2}-i\psi \left(

t \right) {\frac {d}{dt}}\psi \left( t \right)$$

From the last equation we deduce that

and then we conclude that there is a missing minus sign in the equation (3.2):