# Is Gaussian integral the only one that can be easily solved by this double integral trick?

For a lot of people the favorite way of solving Gaussian integral $I=\int^{\infty}_{-\infty} e^{-x^2} dx$ is to find $I^2$ in polar coordinates and then take a root.

The trick may be useful in this case, but I struggle to find any other integral it can be applied to. The obvious condition for the integrated function is:

$$f(x) \cdot f(y)=g(x^2+y^2)=h(|r|)$$

I don't know any other function aside from $e^{bx^2}$ that meets this condition.

Moreover, the limits for the argument should be infinite. Otherwise we can't equate integration in the square $x,y \in (-a,a)$ with integration in the cirlce $r \in (0,a)$.

But maybe this method can be generalized? For example, there may be some functions that give elementary integrals in polar form when multiplied $f(x)f(y)$ even if their product depends on the angle too?

• $\sin(x^2)$ and $\cos(x^2)$ is also possible. it also exists a paper which rigerously proves why this trick is useful only for this very restricted class of functions – tired Feb 26 '16 at 12:37
• unf.edu/~dbell/Poisson.pdf this was the paper i was talking about – tired Feb 26 '16 at 12:55
• Thank you! This is exactly what I was asking. By the way, the general condition in this paper is $f(x)f(y)=g(x^2+y^2)h(y/x)$, so the angle dependence is also considered. – Yuriy S Feb 26 '16 at 13:00
• u are welcome. for a way to calculate the $\int_R dx sin(x^2)$ see here: math.stackexchange.com/questions/187729/… – tired Feb 26 '16 at 13:01

See also my article, "Poisson's remarkable calculation --a method or a trick?", Elem. Math. 65 (2010), where a more general version of Dawson's result is proved.

• Thank you, the link was provided in a comment by @tired and I've read you excellent paper – Yuriy S Jun 19 '16 at 8:33

One well-known trick is a way to evaluate the Gaussian integral $G = \int_\mathbb{R} e^{-x^2}dx = \sqrt{\pi}$ by writing $$G^2 = \left(\int_\mathbb{R} e^{-x^2}dx\right)\left(\int_\mathbb{R} e^{-y^2}dy\right) = \int_{\mathbb{R}^2} e^{-(x^2+y^2)}dxdy$$ which when transformed to polar coordinates becomes $$G^2 = 2\pi \int_0^\infty e^{-r^2} r dr = \pi \int_0^\infty e^{-u} du = \pi$$ via the substitution $u=r^2$. It appears this idea is due to Poisson.
Theorem. Any Riemann-integrable function $f$ on $\mathbb{R}$, such that $f(x)f(y) = g(\sqrt{x^2+y^2})$ for some $g$, is of the form $f(x)=ke^{ax^2}$.