Dimension of \lambda constant in \delta potential

[LagrangeEuler]

Time independent Schroedinger equation in $\delta$ potential $V(x)=-\lambda \delta(x)$, where $\lambda >0$ is given by
$$-\frac{\hbar^2}{2m}\frac{d^2}{d x^2}\psi(x)-\lambda \delta(x)\psi(x)=E\psi(x)$$.
How to find dimension of $\lambda$? Dimension of $V(x)$ is
$$[V(x)]=ML^2T^{-2}$$.
Because it is one dimensional problem dimension of $\psi(x)$ is
$$[\psi(x)]=L^{-\frac{1}{2}}$$.
Is then also
$$[\delta(x)]=L^{-\frac{1}{2}}$$?

 

[Demystifier]

The dimension of $\psi$ is irrelevant. From $\int dx\,\delta(x)=1$ it follows that the dimension of $\delta$ is $L^{-1}$. From that, the dimension of $\lambda$ should be easy.

 

[LagrangeEuler]

The dimension of $\psi$ is irrelevant. From $\int dx\,\delta(x)=1$ it follows that the dimension of $\delta$ is $L^{-1}$. From that, the dimension of $\lambda$ should be easy.

Thank you very much.