divergence operator w.r.t. $D$, $\div^D Y=D_i Y^i$ for a vector field $Y$, $(DX)^t$ is the transpose of $D X$.

Under local coordinates $\{\P_i\}$, reads as $$\begin{aligned}
    D_i(X^jD_jX^i)=&D_iX^j D_jX^i+X^jD_jD_iX^i+R^{D}_{ji}X^iX^j.
    \label{Bochner1}\end{aligned}$$ Moreover, $$\begin{aligned}
    D_i(X^jD_jX^i-X^iD_jX^j) =&D_iX^j D_jX^i-(D_jX^j)^2+R^{D}_{ji}X^iX^j.
    \label{Bochner equ 1}\end{aligned}$$

Using tensor calculus, we have $$\begin{aligned}
 D_i(X^jD_jX^i)=&D_iX^j D_jX^i+X^jD_iD_jX^i. \label{tensor calc}\end{aligned}$$ Combining with the Ricci identity we get . We also have by using tensor calculus that $$\begin{aligned}
 X^jD_jD_iX^i=& D_j(X^jD_iX^i)- (D_j X^j)(D_i X^i)
\nonumber\\=&D_i(X^iD_jX^j)- (D_i X^i)^2.
\label{tensor calc 2}\end{aligned}$$ Inserting into , we get .

Our aim is to derive an integral formula from the local Bochner formula . From now on, let $(M^n, \bar g, V)$ be an $n$-dimensional Riemannian triple and $\a,\g\in \mathbb{R}$. Let $D=D^{\a,\g}$ be the affine connection defined as in (\[defi D\]). Note that $Ric^{D}$ is symmetric. In order to obtain a useful integral formula, we need the following two important ingredients:

1.  $DX$ is symmetric, i.e., $$\begin{aligned}
         D_iX^j=D_jX^i;\end{aligned}$$

2.  The left hand side of (\[Bochner equ 1\]) needs to be a “divergent form" with respect to some choice of volume form.

In the following two lemmata, we will find an appropriate vector field $X$ and also a compatible volume form.

\[sym vf\] Let $\phi$ be a smooth function on 