Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 4 down vote accepted

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.

share|cite|improve this answer
Thanks a lot! A lot of new things to study! – Martin Gales Jun 25 '11 at 10:14

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.

share|cite|improve this answer
Very impressive! Thanks! – Martin Gales Jun 25 '11 at 10:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.