Quantum Mechanics

We have the following potential:
V\left (x\right ) = \frac{\hbar^2}{2m}\left [g_1\delta\left (x+a\right )+g_2\delta\left (x-a\right )\right ]
Assume a particle with an energy E > 0 comes from the left: determine the transmission coefficient.

How should I start with this? I solved the Schrödinger equation in the three regions, which gives me the following solution for the wave equation:
\psi\left (x\right )= \left\{
\begin{array}{l l}
  Ae^{-ik\left (x+a\right )} + Be^{ik\left (x+a\right )} & \quad \text{for } -\infty < x \leq -a\\
  Ce^{-ik\left (x-a\right )} + De^{ik\left (x-a\right )} & \quad \text{for } -a < x \leq a\\
  Fe^{ik\left (x-a\right )} & \quad \text{for } a < x < \infty\\
\end{array}
This is where I'm stuck. I can't apply the rules for continuity in the derivatives of the wave function, because the potential goes to \infty at -a and a. Any ideas?

Oops, I made a boo-boo:
\psi\left (x\right )= \left\{
\begin{array}{l l}
  Ae^{-ik\left (x+a\right )} + Be^{ik\left (x+a\right )} & \quad \text{for } -\infty < x \leq -a\\
  Ce^{-ik\left (x-a\right )} + De^{ik\left (x-a\right )} & \quad \text{for } -a < x \leq a\\
  Fe^{ik\left (x-a\right )} & \quad \text{for } a < x < \infty\\
\end{array}\right.

Try to establish a continuity rule for delta distributions. You should find that Phi(x) is continuous, while the derivate of Phi isn't.

>>3
That leaves me with:
A+B=C and C+D=F
I'm looking for [math]\frac{|F|^2}{|A|^2}[\math], so I seem to miss some boundary conditions in order to work it out.

>>4
You haven't used the rule for delta distributions that >>3 mentioned.  There must be jump discontinuities in the first derivative at x = \pm{a} to account for the two delta functions that show up in the potential.  The size of these jumps is given by the coefficients in front of the delta functions, g_1 and g_2.