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?

  • $\begingroup$ $\sin(x^2)$ 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 $\endgroup$ – tired Feb 26 '16 at 12:37
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    $\begingroup$ unf.edu/~dbell/Poisson.pdf this was the paper i was talking about $\endgroup$ – tired Feb 26 '16 at 12:55
  • $\begingroup$ 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. $\endgroup$ – Yuriy S Feb 26 '16 at 13:00
  • $\begingroup$ u are welcome. for a way to calculate the $\int_R dx sin(x^2)$ see here: math.stackexchange.com/questions/187729/… $\endgroup$ – 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.

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

From Nate Eldredge's answer here:

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.

In a 2005 note in the American Mathematical MONTHLY, R. Dawson has observed that this is a trick that only works once; there are no other integrals that can be evaluated by this method. Specifically:

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 $f(x)=ke^{ax^2}$.

See: Dawson, Robert J. MacG. On a "singular" integration technique of Poisson. American Mathematical Monthly 112 (2005), 270-272.

  • $\begingroup$ See tired's comment to my question - he provided a more general reference $\endgroup$ – Yuriy S Feb 26 '16 at 14:29

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