When proving that the log-canonical threshold is minus the largest root of the Bernstein-Sato polynomial one considers the integral

$$\int |f^2(z)|^{t+1}(P\bar{P}(\phi(z))) dz,$$ where $P$ is a linear partial differential operator and $\phi$ is a compactly support smooth function. In order for $t$ to be a root of the Bernstein-Sato polynomial of $f$ which is larger than $-lct(f)$ it would have to be the case that \begin{equation}\int |f^2(z)|^{t+1}(P\bar{P}(\phi(z))) dz=0\quad \text{for all }\phi\quad (1).\end{equation} Kollar in this document says that this seems strange, but cannot be excluded without knowing more about $P$. This makes me wonder, is $(1)$ really possible if $P\not=0$? Or does $P\not=0$ imply that $(1)$ is not possible?

When proving that the log-canonical threshold is minus the largest root of the Bernstein-Sato polynomial one considers the integral

$$\int |f^2(z)|^{t+1}(P\bar{P}(\phi(z))) dz,$$ where $P$ is a linear partial differential operator and $\phi$ is a compactly support smooth function. In order for $t$ to be a root of the Bernstein-Sato polynomial of $f$ which is larger than $-lct(f)$ it would have to be the case that \begin{equation}\int |f^2(z)|^{t+1}(P\bar{P}(\phi(z))) dz=0\quad \text{for all }\phi.\label{1}\tag{1}\end{equation} Kollar in this document says that this seems strange, but cannot be excluded without knowing more about $P$. This makes me wonder, is \eqref{1} really possible if $P\not=0$? Or does $P\not=0$ imply that \eqref{1} is not possible?