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](https://www.uni-due.de/%7Emat903/sem/ws0708/pairs.pdf) 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?



Is it possible that $\int |f^2(z)|^{t+1}(P\bar{P})\phi(z) dz=0$ for all compactly supported $\phi$?
**Edit notice:** The answer is completely rewritten due to user2520938's comment. My original answer was that the linear operator $P$ depends on $t$, and we have $(1)$ as long as $P(t) = 0$. But as user2520938 pointed out, we always have $P(t) \ne 0$, since otherwise it contradicts the minimality of the Bernstein-Sato polynomial. So we need an example with $P(t) \ne 0$.


Here are some examples with $t = \frac{1}{k} - 1$ where $k$ is a positive integer.
In the one-variable situation, consider $f(z) = z^k$ and define
$$P(\phi) := 2\phi + z\frac{\partial\phi}{\partial z}.$$
We have
$$
\begin{split}
\int\int |f(z)^2|^{1/k}(P \bar{P}(\phi(z))) dxdy 
= \int \int \frac{\partial^2}{\partial z\partial \bar{z}}\left(z^2\bar{z}^2\phi \right) dxdy \\
\int \int \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)\left( (x^2 + y^2)^2\phi\right) dxdy
\end{split}
$$
with $x + iy = z$. From this expression, we see that
if $\phi$ has compact support,
then the integral vanishes.


Obviously $P$ is not a differential operator which characterizes the Bernstein-Sato polynomial of $f$, and this example is constructed without assuming that $P$ is a priori related to $f$.


