The commutativity of minimal extension and direct image by blowing-down Let $X$ be a sooth algebraic variety over $\mathbb{C}$.
Let us assume that there exists the commutative diagram
$\require{AMScd}$
\begin{CD}
U @>{i}>> \hat{X}\\
@| @VV{\phi}V\\
U @>{j}>> X
\end{CD}
where $\phi : \hat{X} \rightarrow X$ is the blowing-up, and assume that the inclusion maps $i : U\hookrightarrow \hat{X}$,  $j : U\hookrightarrow X$ are affine.
Let $M$ be a $D_U$-module on $U$.
Then we have the minimal extension functor $j_{!*}$ (resp. $i_{!*}$) from the (derived) category of $D_U$-modules to the (derived) category of $D_X$-modules (resp. $D_{\hat{X}}$-modules),
and direct image functor $\phi_+(\bullet) := R\phi_*(D_{X \leftarrow \hat{X}}\otimes ^L_{D_{\hat{X}}}\bullet)$ (there are various notations in the literature for the direct image , e.g. $\int_{\phi}$).
My question is
is it true that
$$ j_{!*}(M) \cong \phi_+(i_{!*}(M)) $$
as $D_X$-modules?
Or, are there any reference for them?
 A: No. Here's an example. Let $X=\Bbb C^2$, $\hat{X}$ the blow-up of $X$ at the origin, $U=(\Bbb C^*)^2$, and $M=\mathcal{O}_U$. Then $j_{!*}M=\mathcal{O}_X$ and $i_{!*}M=\mathcal{O}_{\hat{X}}$.

Claim: $\phi_+\mathcal{O}_{\hat{X}}\ncong \mathcal{O}_X$.
Proof. Let $E=\phi^{-1}(0)$, and let $i_E\colon E\hookrightarrow \hat{X}$ and $i_0\colon \{0\} \hookrightarrow X$ be inclusion. It suffices to show that $i_0^+\mathcal{O}_X\ncong i_0^+\phi_+\mathcal{O}_{\hat{X}}$.[1] First, we have
$$i_0^+\mathcal{O}_X = Li_0^*\mathcal{O}_X[-2] = \Bbb{C}[-2].$$
Next, denoting by $a_E$ the map from $E$ to a point, we have
$$i_0^+\phi_+\mathcal{O}_{\hat{X}} = (a_E)_+ i_E^+\mathcal{O}_{\hat{X}} = (a_E)_+ \mathcal{O}_E[-1],$$
where the first equality is via base change. But $(a_E)_+\mathcal{O}_E[-1]$ is (up to shift) the de Rham cohomology of $E\cong \Bbb P^1$. In particular, it has cohomology in more than one degree. This finishes the proof of the claim.

[1] Given a morphism $f\colon X\to Y$ of smooth varieties, I denote by $f^+$ the (shifted) $D$-module inverse image functor. I.e.
$$f^+:=Lf^*[\dim X - \dim Y].$$
