Wilson-Sommerfeld Quantization of Dirac delta in Infinite Square Well (ISW) I am curious to find the energies of Dirac delta potential inside the ISW (walls at $x=0,L$)
$$
H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V_0\delta(x-L/2)
$$
using Wilson-Sommerfeld Quantization (WSQ), 
so I looked at this integral (Here, $p_1=\sqrt{2 m E}, p_2=\sqrt{2 m (E- V_0\delta(x-L/2))}~$)
 $$\oint p~ dx = 2 \int_0^{L/2-\epsilon} p_1~ dx  + 2  \int_{L/2-\epsilon}^{L/2+\epsilon}  p_2 ~dx+ 2 \int_{L/2+\epsilon}^{L}  p_1~ dx     $$
$$= 2* \sqrt{2 m E}*(L-2  \epsilon) + 2  \int_{L/2-\epsilon}^{L/2+\epsilon}  p_2 (=\sqrt{2m (E-V_0 \delta(x-L/2))})~dx $$
The Integral term is zero E.g.- this integral (a particular case) , and WSQ  gives the energies of ISW Only.
If you recall that momentum is discontinuous for a Dirac delta potential, so integral term should not be zero!!!.
Now the Ques is: can we treat given Hamiltonian $H$ using WSQ ?, If yes, then how to tackle the integral (we have tackled this integral in Schrodinger's formalism)?
 A: Somehow your expression doesn't make quite sense, see: Don't understand the integral over the square of the Dirac delta function , What is the square root of the Dirac Delta Function? ,and (https://www.wolframalpha.com/input/?i=Integrate+sqrt(f(x)*(DiracDelta%5Bt%5D))+dx  )  Your input of Dirac function under square root doesn't make quite sense in standard mathematics, it automatically resolved into complex domain and then you almost automatically exited the WSQ.
A: At the physical (as opposed to rigorous mathematical) level, one can at least try to explore the weak and strong coupling limit of the Dirac delta potential. 


*

*On one hand, the quantum mechanically exact quantization condition is given by
$$ \sin\frac{kL}{2}~=~0\qquad \qquad\vee\qquad \tan\frac{kL}{2}~=~-\frac{2k}{\kappa},\tag{1} $$ 
where
$$ k~>~0, \qquad \kappa~:=~\frac{2m V_0}{\hbar^2},  \tag{2} $$
cf. e.g. Emilio Pisanty's answer here. The even-number modes
$$k~\in\frac{\pi}{L}2\mathbb{N} \tag{3}$$
can not "feel" the Dirac delta potential (because their wave functions vanish at the Dirac delta singularity). The odd-number modes has the following expansions


*

*Weak coupling limit ($V_0$ small):
$$ k~=~\frac{2}{L}\left(\frac{\pi}{2}+\arctan\frac{\kappa}{2k}\right)
~\approx~\frac{\pi}{L}(2j+1)+\frac{\kappa}{kL}, \qquad j~\in~\mathbb{N}.\tag{4}$$

*Strong coupling limit ($V_0$ large):
$$ \frac{kL}{2}+\frac{2k}{\kappa}~\approx~\pi j \qquad\Leftrightarrow\qquad k~\approx~\frac{\pi j}{\frac{L}{2}+\frac{2}{\kappa}}~\approx~
\frac{\pi}{L} 2j\left(1- \frac{4}{\kappa L}\right) .\tag{5}$$


*On the other hand, semiclassical WKB approximation yields the following.


*

*Weak coupling limit ($V_0$ small): The Bohr-Sommerfeld quantization rule is
$$ 2\pi n ~=~\oint \!\mathrm{d}x~ k(x)  ~=~ \oint \!\mathrm{d}x~\sqrt{2m(E_n-V(x))} $$
$$ ~\approx~ k_n\oint \!\mathrm{d}x~ \left(1- \frac{V(x)}{2E_n}\right)~=~2k_n\left(L-\frac{V_0}{2E_n}\right) , \qquad E_n \equiv \frac{(\hbar k_n)^2}{2m}, \qquad n\in\mathbb{N},\tag{6}$$
which leads to
$$ k^{(0)}_n~=~\frac{n\pi}{L}, \tag{7}$$
$$ k^{(1)}_n~=~\frac{n\pi}{L- \frac{V_0}{2E^{(0)}_n}}
~=~k^{(0)}_n\left(1- \frac{\kappa}{(k^{(0)}_n)^2L}\right)^{-1}~\approx~k^{(0)}_n+ \frac{\kappa}{k^{(0)}_n L},\tag{8}$$
perturbatively. In this limit products of Dirac delta distributions [from expanding the square root (6)] are effectively ignored. 


*Discussion. The zero-order approximation (7) correctly finds the even- & odd-number modes (3) & (4) when there is no Dirac delta potential. The first-order approximation (8) is correct for the odd-number modes, but this approach fails to take into account that the even-number modes are not affected by the Dirac delta potential.
