Evaluate the following integral:
$$
I=\iint_D e^{-(x^2+y^2)}\,dx\,dy,\ \ \ D=\left\{(x,y)\Bigm|2\leqslant |x|+|y|\leqslant 3\right\}
$$
Well, to make things a bit easier, I can find an integral $J$ first:
$$
J=\int\limits_0^3dx\int\limits_0^{-x+3}e^{-x^2-y^2}dy
$$
And then
$
I=4J
$.
Afterwards, if I try to make a polar coordinate substitution, I get some problems:
$$
J=\int\limits_0^{\pi/2}d\phi\int\limits_0^{\frac{3}{\sin\phi+\cos\phi}}e^{-r^2}r\,dr=\frac{1}{2}\int\limits_0^{\pi/2}\left(e^{-\left(\frac{3}{\sin\phi+\cos\phi}\right)^2}-1\right)d\phi
$$
And it is rather difficult to evaluate the final integral. So, perhaps there is a better approach to this problem?
Best Answer
Let $u=\frac{x+y}{\sqrt2}$ and $v=\frac{y-x}{\sqrt2}$. Then the $(u,v)$ coordinates $(u,v)$ are obtained from $(x,y)$ by a $45^\circ$-rotation in the clockwise direction. (In my comment above, I mistakenly defined $v=\frac{x-y}{\sqrt2}$. Call this $v'$ instead to avoid confusion. While most of the idea remains unchanged, it should be noted that in the comment, the $(u,v')$ coordinates is obtained by a rotation and a reflection of the $(x,y)$ coordinates.)
Then in the new coordinates, the image $D'$ of $D$ looks like $$D'=\left\{(u,v)\in\Bbb{R}^2:\frac{2}{\sqrt2}\le |u|,|v|\le \frac{3}{\sqrt2}\right\}=A\setminus B,$$ where $A$ and $B$ are the squares $$A=\left\{(u,v)\in\Bbb{R}^2:|u|,|v|\le \frac{3}{\sqrt2}\right\}$$ and $$B=\left\{(u,v)\in\Bbb{R}^2:|u|,|v|< \frac{2}{\sqrt2}\right\}.$$ Since $u^2+v^2=x^2+y^2$, we get $$\iint_D e^{-x^2-y^2} dx\ dy=\iint_{D'}e^{-u^2-v^2}du\ dv =\iint_A e^{-u^2-v^2}du\ dv-\iint_B e^{-u^2-v^2}du\ dv.$$ Clearly \begin{align}\iint_A e^{-u^2-v^2}du\ dv&=\int_{-3/\sqrt2}^{3/\sqrt2} \int_{-3/\sqrt2}^{3/\sqrt2} e^{-u^2-v^2}du\ dv\\&=\int_{-3/\sqrt2}^{3/\sqrt2}e^{-u^2}du \int_{-3/\sqrt2}^{3/\sqrt2} e^{-v^2}dv\\&=\left(\int_{-3/\sqrt2}^{3/\sqrt2}e^{-t^2}dt\right)^2=\Biggl(\sqrt{\pi}\operatorname{erf}\left(\frac{3}{\sqrt2}\right)\Biggr)^2.\end{align} Likewise \begin{align}\iint_B e^{-u^2-v^2}du\ dv=\left(\int_{-2/\sqrt2}^{2/\sqrt2}e^{-t^2}dt\right)^2=\left(\int_{-\sqrt2}^{\sqrt2}e^{-t^2}dt\right)^2=\Big(\sqrt{\pi}\operatorname{erf}\left({\sqrt2}\right)\Big)^2.\end{align} Therefore \begin{align} \iint_D e^{-x^2-y^2} dx\ dy&=\left(\int_{-3/\sqrt2}^{3/\sqrt2}e^{-t^2}dt\right)^2-\left(\int_{-\sqrt2}^{\sqrt2}e^{-t^2}dt\right)^2 \\&=\Biggl(\sqrt{\pi}\operatorname{erf}\left(\frac{3}{\sqrt2}\right)\Biggr)^2-\Big(\sqrt{\pi}\operatorname{erf}\left({\sqrt2}\right)\Big)^2 \\&=\pi\Biggl(\left(\operatorname{erf}\left({\textstyle\frac{3}{\sqrt2}}^{\vphantom{a^2}}_{\vphantom{a_2}}\right)\right)^2-\big(\operatorname{erf}\left({\sqrt2}\right)\big)^2\Biggr)\approx0.26244. \end{align} I don't think you can get the answer in terms of elementary functions. You at least need the error function $\operatorname{erf}$.