Calculate the improper double integral: integral
Calculate the improper double integral: $$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy. $$
Explanation of Method
Because the integrand factors as $e^{-x^2}\cdot e^{-y^2}$ and the region is the Cartesian product $\mathbb R\times\mathbb R$, the double integral factors into a product of identical one-dimensional Gaussian integrals. Independently, converting directly to polar coordinates evaluates the double integral without any prior knowledge of the one-dimensional Gaussian value — and the two routes can be combined to derive that classical value.
- Convert to polar coordinates.
- Evaluate the resulting elementary radial integral.
- Combine with the angular integral.
Step 1 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$, $x^2+y^2=r^2$, $dx\,dy=r\,dr\,d\theta$:
$$ I = \int_0^{2\pi}\int_0^\infty e^{-r^2}\cdot r\,dr\,d\theta = 2\pi\int_0^\infty r e^{-r^2}\,dr. $$
Step 2 · Evaluate the radial integral
With $u=r^2$, $du=2r\,dr$:
$$ \int_0^\infty r e^{-r^2}\,dr = \frac12\int_0^\infty e^{-u}\,du = \frac12\Big[-e^{-u}\Big]_0^\infty = \frac12. $$
Step 3 · Combine
$$ I = 2\pi\cdot\frac12 = \pi. $$
Additional Notes
- Verified numerically: direct evaluation over a large truncated square converges to $\approx3.14159265$, matching $\pi$.
- This is the standard route to deriving the one-dimensional Gaussian integral: writing $J=\int_{-\infty}^\infty e^{-x^2}\,dx$, factoring gives $J^2=\iint_{\mathbb R^2}e^{-(x^2+y^2)}\,dx\,dy=\pi$ (by the computation above), so $J=\sqrt\pi$.
- More generally, for $a>0$, the substitution $x\to x/\sqrt a$, $y\to y/\sqrt a$ gives $\displaystyle\iint_{\mathbb R^2}e^{-a(x^2+y^2)}\,dx\,dy=\dfrac\pi a$.
Result
The solution to the integral is: