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I am faced with the following problem:

$$Q(t) = \int_0^t \int_0^1 \int_0^1 \exp\left(-\frac{(x-x_s)^2+(y-y_s)^2}{\omega^2}\right) \,dx \, dy \, d\tau.$$

I know the first trivial step, since the integrand is not dependent of $\tau$, then we get

$$Q(t) = t \int_0^1 \int_0^1 \exp\left(-\frac{(x-x_s)^2+(y-y_s)^2}{\omega^2}\right) \,dx\,dy,$$

but that just kicks the can a little bit further down the road. It reminds of a Gaussian integral, but this contains two variables $x$ and $y$, not just one. I have been trying with a variable transformation by changing it to polar coordinates. Then I end up with some unknown boundary. What is the trick to this? Could anybody give me hint/solution?

Best regards//

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    $\begingroup$ Hint: $$\exp\left(-\frac{(x-x_{s})^{2}+(y-y_{s})^{2}}{\omega^{2}}\right)=\exp\left(-\frac{(x-x_{s})^{2}}{\omega^{2}}\right)\exp\left(-\frac{(y-y_{s})^{2}}{\omega^{2}}\right)$$ And notice that one exponential depends on $x$ and the other one depends on $y$. $\endgroup$
    – Zacky
    Jan 30, 2020 at 18:58
  • $\begingroup$ Yes of course, why didn't I see that. Then I evaluate these as "Gaussians" I presume. Thanks, I'll try this! $\endgroup$
    – SimpleProgrammer
    Jan 30, 2020 at 19:03
  • $\begingroup$ You can separate variables here. $\endgroup$
    – Allawonder
    Jan 30, 2020 at 21:31

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