Let $H \stackrel{\Delta}{=} \nabla^2f(x)$ with $\|H(x) -H(y) \| \leq L \|x-y \|$ and let $D \stackrel{\Delta}{=} \text{diag} (H_{11},H_{22}, \dots, H_{dd})$. Then, $H$ can be written as $H \stackrel{\Delta}{=} A - D$, where $A$ is $H$ with the diagonal entries set to zero. Now I know from Taylor's Theorem [Theorem 2.1, here] that
$$\int_{0}^{1}H(x+ty)y\:dt = \nabla f(x+y) - \nabla f(y) \tag{1}$$
and I want to compute
$$\int_{0}^{1}\|A(x+ty)y\|\:dt. \tag{2}$$
I convenient result for me is to be zero or upper bounded by some quantity $C\|y\|^2$ but maybe this is not the case. So can you please someone provide some details on how we compute or upper bound $(1)$?