# A double integral (differentiation under the integral sign)

Working on a physics problem I got the following double integral that depends on the parameter $a$:

$$I(a)=\int_{0}^{L}\int_{0}^{L}\sqrt{a}e^{-a(x-y+b)^2}dxdy;L=const,b=const$$ Now, this integral obviously has no closed form in terms of elementary functions. However it follows from the physical considerations that the derivative of this integral $\frac{dI}{da}$ has a closed form solution in terms of exponential functions. But my mathematical abilities are not so good to get this result directly from the integral. So how does a mathematician solve this problem?

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Have you tried differentiating under the integral sign? It'll make $(x-y+b)^2$ drop down from your exponential, probably something good can come out of this. I have not tried it though, it's just a suggestion. – Patrick Da Silva Jun 25 '11 at 6:44
@Patrick Da Silva: Yes, but with no success. – Martin Gales Jun 25 '11 at 7:14

Introduce new variables $u$, $v$ by means of $$x={1\over2}(u+v+L)\ ,\quad y={1\over2}(-u+v+L)$$ and get $$I(a)=\int_{|u|+|v|\leq L}{\sqrt{a}\over2}\exp\bigl(-a(u+b)^2\bigr){\rm d}(u,v)\ .$$ Now the inner integral, with respect to $v$, running from $-(L-|u|)$ to $L-|u|$, is elementary, and the resulting outer integral can be written as a linear combination of integrals of the form $\int_\ldots^\ldots (u+b)\exp\bigl(-a(u+b)^2\bigr) du$ and $\int_\ldots^\ldots \exp\bigl(-a(u+b)^2\bigr)du$, the first of which are also elementary.
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.