In Moyal's 1949 paper, he derives the quantum characteristic function as $$ M(\tau,\theta)=\frac{1}{\sqrt{\hbar}}\int_{\mathbb{R}}\int_{\mathbb{R}}[\psi^{*}(q)\phi(p)e^{\frac{ipq}{\hbar}}]e^{i(\tau p+\theta q)}e^{-\frac{i}{2} \hbar \tau \theta}dq\:dp $$ This function is the Fourier inverse of the Wigner function, where $\tau$ and $\theta$ represent the Fourier space variables of $q$ anf $p$ respectively.
Supposedly, integrating this by parts should return $$ M(\tau,\theta)=\frac{1}{\sqrt{\hbar}}\int_{\mathbb{R}}\int_{\mathbb{R}}e^{\frac{\hbar}{2i}\frac{\partial^{2}}{\partial p \partial q}}[\psi^{*}(q)\phi(p)e^{\frac{ipq}{\hbar}}]e^{i(\tau p+\theta q)}dq\:dp $$ where the partial derivatives are only acting within the square brackets.
After this, it is easy to obtain his final expression by performing the Fourier transform, but I can't see how integrating by parts would return this expression at all. Is there an alternative form of the integration by parts formula which returns exponentiated partial differentials in this fashions? If not how would you show the equivalence?
Integrating this by parts a necessary number of times will net you the celebrated expression. First, convince yourself that $$ \int_{\mathbb{R}}\int_{\mathbb{R}} e^{i(\tau p+\theta q)}{\frac{\hbar}{2i}\frac{\partial^{2}}{\partial p \partial q}}L(q,p) ~~ dq\:dp =\int_{\mathbb{R}}\int_{\mathbb{R}} e^{i(\tau p+\theta q)} L(q,p) ~\left({-\frac{\hbar}{2i} ~ \tau \theta}\right ) dq\:dp, $$ which you may then repeat n times to obtain the analogous result for the n-th power of the differential operator.
Then collect all such powers to the exponential pseudodifferential operator involved.
Your paper may have muffed the over-all sign of the exponent.