Kinetic energy Hamiltonian in second quantized form and the delta function

I’m trying to transform the kinetic energy Hamiltonian to momentum basis. Starting with:
$$hat{T} = -frac{hbar^2}{2m}int d^3r hat{Psi}^{dagger}(vec{r}) nabla^2 hat{Psi}(vec{r})$$
I can expand the field operators in terms of momentum eigenstates, since the system is translationally invariant:
$$hat{Psi}(vec{r}) = frac{1}{sqrt{V}}sumlimits_{vec{k}}e^{ivec{k}cdotvec{r}}hat{a}_{vec{k}}.$$
I get:
begin{align}
hat{T} = &-frac{hbar^2}{2m}frac{1}{V}sumlimits_{vec{k}_1,vec{k}_2}hat{a}_{vec{k}_1}^{dagger}hat{a}_{vec{k}_2}int d^3r e^{-i{vec{k}_1}cdotvec{r}} nabla^2 e^{ivec{k_2}cdotvec{r}}
= &frac{hbar^2}{2m}frac{1}{V}sumlimits_{vec{k}_1,vec{k}_2}hat{a}_{vec{k}_1}^{dagger}hat{a}_{vec{k}_2}vec{k}_2^2int d^3r e^{-i({vec{k}_1}-vec{k}_2)cdotvec{r}}
= &frac{hbar^2}{2m}frac{1}{V}sumlimits_{vec{k}_1,vec{k}_2}hat{a}_{vec{k}_1}^{dagger}hat{a}_{vec{k}_2}vec{k}_2^2 Vdelta(vec{k}_1 – vec{k}_2).
end{align}
Here I don’t see how this expression should simplify to
$$sumlimits_{vec{k}}frac{hbar^2vec{k}^2 }{2m}hat{a}_{vec{k}}^{dagger}hat{a}_{vec{k}}, ,$$
given that the sum over $vec{k}_1=vec{k}_2$ gives $delta (0)=+infty$. Did I make a mistake somewhere? Could this be due to the delta function switching interpretation between a Dirac and a Kronecker delta?