
Then we have that
DΦ=(∂∂θ−iθ∂∂t)(x(t)+iθψ(t))=iψ(t)−iθddtx(t)
and
D2Φ=(∂∂θ−iθ∂∂t)(iψ(t)−iθddtx(t))=−iddtx(t)+θddtψ(t)
Now we have that
DΦD2Φ=(iψ(t)−iθddtx(t))(−iddtx(t)+θddtψ(t))
it is to say
DΦD2Φ=(ddtx(t))ψ(t)−θ(ddtx(t))2−iθψ(t)ddtψ(t)
Finally we have that
∫dθDΦD2Φ=∫((ddtx(t))ψ(t)−θ(ddtx(t))2−iθψ(t)ddtψ(t))dθ
it is to say
∫dθDΦD2Φ=ddθ((ddtx(t))ψ(t)−θ(ddtx(t))2−iθψ(t)ddtψ(t))
which is reduced to
∫dθDΦD2Φ=−(ddtx(t))2−iψ(t)ddtψ(t)
From the last equation we deduce that

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