2 particles in a box, perturbative solution

Two particles in the ground state of a quantum box of size L with the
unperturbed wavefunctions
$Psi_{mathrm{nm}}^{(0)}left(mathrm{x}_{1}, mathrm{x}_{2}right)= sin left(mathrm{n} pi mathrm{x}_{1} / mathrm{L}right) sin left(mathrm{m} pi mathrm{x}_{2} / mathrm{L}right)$
for $0<x_{1}, x_{2}<L$ and zero otherwise. The perturbation is
$$
mathrm{H}_{1}=deltaleft(mathrm{x}_{1}-mathrm{x}_{2}right)
$$
I’m trying to find the first order correction to the ground state wavefunction.

What I have tried is: using the following general formula,

$left|n^{(1)}rightrangle=sum_{k neq n} frac{leftlangle k^{(0)}|V| n^{(0)}rightrangle}{E_{n}^{(0)}-E_{k}^{(0)}}left|k^{(0)}rightrangle$ wrote the following equation,

$$
|left.psi_{n m}^{(1)}rightrangle=sum_{L neq n} sum_{k neq m} frac{leftlanglepsi_{k ell}^{(0)}left|H^{prime}right| psi_{n m}^{(0)}rightrangle |psi_{n m}^{(0)}rangle}{left(E_{n m}^{(0)}-E_{k ell}^{(0)}right)}
$$
Then,

$$
leftlanglepsi_{k l}^{(0)}left|H^{prime}right| psi_{n m}^{(0)}rightrangle=int_{0}^{L} int_{0}^{L} sin left(frac{k pi x_{1}}{L}right) sin left(frac{ell pi x_{2}}{L}right) deltaleft(x_{1}-x_{2}right) sin left(frac{n pi x_{1}}{L}right) sin left(frac{m pi x_{2}}{L}right) d_{x_1}d_{x_2}
$$

But after this part, I’m unable to decide what to do, suspecting that maybe changing $l$ and $m$ terms in the integral to $k$ and $m$ respectively yields square sin terms but that gives me confusion about why shouldn’t I use directly $sin^4(kx)$ term by changing all the dummy variables.