A double integral (differentiation under the integral sign) While working on a physics problem, I got the following double integral that depends on the parameter $a$:
$$I(a)=\int_{0}^{L}\int_{0}^{L}\sqrt{a}e^{-a(x-y+b)^2}dxdy$$
where $L$ and $b$ are constants.
Now, this integral obviously has no closed form in terms of elementary functions. However, it follows from physical considerations that the derivative of this integral $\frac{dI}{da}$ has a closed form solution in terms of exponential functions. Unfortunately, my mathematical abilities are not good enough to get this result directly from the integral. So, how does a mathematician solve this problem?

Nowadays many mathematicians (including me -:)) would be content to use some program to have 
$$I'(a)=\frac{e^{-a (b+L)^2} \left(2 e^{a L (2 b+L)}-e^{4 a b L}-1\right)}{4 a^{3/2}}.$$
As for the proof, put $t=1/a$ and let $G(b,t)=e^{-b^2/t}/\sqrt{\pi t}\ $ be a fundamental solution of the heat equation $u_t-u_{bb}/4=0\ $. Then 
$$
u(b,t)=I(1/a)/\sqrt\pi =\int_{0}^{L}\int_{0}^{L}G(b+x-y,t)\,dxdy.
$$
If to tinker a bit about what happens then $t\to+0$ we'll have that $u$ is a solution of the Cauchy problem with initial condition $u(b,0)=\psi(b)$ where $\psi(b)=L-|b|$ then $|b|\le L$ and $\psi(b)=0$ otherwise. So $u(b,t)=\int_{-\infty}^\infty G(b-z,t)\psi(z)\,dz\,\,\,$. Taking Fourier transform with respect to b we have
$$
\tilde u(\xi,t)=\tilde \psi(\xi) \tilde G(\xi,t)=-\frac{e^{-i L \xi} \left(-1+e^{i L \xi}\right)^2}{\sqrt{2 \pi } \xi^2} \frac{e^{-\frac{\xi ^2 t}{4}}}{\sqrt{2 \pi }}=
$$
$$
-\frac{\left(-1+e^{i L \xi}\right)^2 e^{-\frac{\xi ^2 t}{4}-i L \xi}}{2 \pi  \xi^2},$$
$$
\tilde u_t(\xi,t)=\frac{\left(-1+e^{i L \xi }\right)^2 e^{-\frac{1}{4} \xi  (\xi  t+4 i L)}}{8 \pi }.
$$
Taking inverse Fourier transform etc. will give the answer above.
