Using symmetries to evaluate integrals is not something that I have ever been shown or taught and I have come across a problem that I am very unsure of my answers to. Any help would be appreciated!
I have shown that $$I=\iint_A x^2e^{-(x^2+y^2)}dxdy=\frac{\pi}{4}\left(\frac{1}{2}-\frac{1}{e}\right)$$ where $A$ is the intersection of the interior of the unit circle with the first quadrant.
Now, with $D$ as the interior of the unit circle, I would like to find the following using just symmetry arguments:
$$I_1=\iint_D x^2e^{-(x^2+y^2)}dxdy,\;I_2=\iint_D y^2e^{-(x^2+y^2)}dxdy,\;I_3=\iint_D xye^{-(x^2+y^2)}dxdy$$
$I_1$: The integrand of $I$ is invariant to both $x\mapsto-x$, $y\mapsto-y$, and any composition of the two. The first mapping would change $I$ to the integral over the second quadrant, the second mapping gives the fourth quadrant, and applying both gives the third quadrant. Since $I_1$ is the sum of the integral over each quadrant, $$I_1=4\left(\frac{\pi}{4}\left(\frac{1}{2}-\frac{1}{e}\right)\right)=\pi\left(\frac{1}{2}-\frac{1}{e}\right)$$
$I_2$: This integral has the same integrand with a reflection in the line $y=x$. I know that noting the integrals of $sin^2$ and $cos^2$ are equal over $[0,\frac{\pi}{2}]$, combined with the above justification for $I_1$, gives $I_2=I_1$ (?). I cannot think of a way to justify this part without comparing the calculation of the integral instead of using the $y=x$ reflection.
$I_3$: For this I feel like there is no choice but to calculate the integral explicitly for the first quadrant and then the mappings mentioned for $I_1$ show that the four quadrants cancel each other for this integral and so $I_3=0$.
I would like to know to what extent my method is correct and rigorous, how is best to justify a symmetry argument like for $I_1$, and if there is a way to evaluate the second/third integrals without calculating any integrals. Thank you!
You would have $I_3=0$ because the integrand is odd. In particular,
$$\iint_A (-x)ye^{-(-x)^2-y^2}\,\mathrm dx\,\mathrm dy = -I \\\\ \iint_A x(-y)e^{-x^2-(-y)^2}\,\mathrm dx\,\mathrm dy = -I \\\\ \iint_A (-x)(-y)e^{-(-x)^2-(-y)^2}\,\mathrm dx\,\mathrm dy = I$$
so that $\iint_Dxye^{-x^2-y^2}\,\mathrm dx\,\mathrm dy=I-I-I+I=0$. You don't even need to know the value of $I$ to arrive at this.