TITLE: What is integration along the fibers in D-module theory?
QUESTION [9 upvotes]: In Hotta, Takeuchi, Tanisaki's book on "D-modules, Perverse Sheaves, and Representation theory", for a morphism of smooth algebraic varieties $f:X \to Y$, they use the notation
$$
\int_f:D^b(D_X^{op}) \to D^b(D_Y^{op})
$$
for the derived pushforward of right $D$-modules. What motivates this notation with the integral? I have hear this has something to do with "integrating along the fibers", but wikipedia gives an article about partially integrating differential forms.

REPLY [7 votes]: I think, for a locally constant $D$-module it basically cashes out to

Defining a formal solution to the differential equation.
Defining a formal integral along the fibers.
Writing down all the differential relations satisfied by the formal integral that follow by integration by parts from the differential equation applied to the formal solution.
Converting this to a differential equation satisfied by the formal solution.

More precisely we are taking the de Rham cohomology, so there may be more than one integral over more than one cycle, all of which satisfy the same differential equation. To calculatee this you take a formal ensemble of integrals of an arbitrary differential operator applied to $f$ (which is just the D-module as a module) and find all the formal differential relations between them.
Unfortunately in your example you have a single PDE in two variables, so the solutions are not a holonomic D-module. I don't know if this still works in that case. If it does, you can observe that $\int \partial_x^2f dx =0$ (by integration by parts) forces $\partial_y^2\int  f df dx = \int \partial_y^2 f dx =0$ (by the differential equation).
Let's do a holonomic example instead. If we have the equations $\partial f/ \partial x=2xy f$, $\partial f/\partial y=x^2 f$, which is satisfied by $e^{x^2 y}$, whose integral is proportional to $\sqrt{y}$, we can find the differntial equation satisfied by $\sqrt{y}$ by observing
$$ y \partial_y \int f dx = \int y \partial_y f dy = \int y x^2 f dx= \int x \partial_x f dx/2 = \int \partial_x ( xf)dx/2 - \int fdx/2 dx= 0 - \int f dx/2$$
which is the correct differential equation, and I think you can check that this algebra is exactly what the definition of D-module cohomology by de Rham cohomology is doing.