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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

The closely related integral

$\Gamma(1/2)=\int_0^\infty\dfrac{e^{-x}dx}{\sqrt x}=\sqrt\pi$

can also be solved through a double integration ... with a different sort of coordinate transformation.

Let $I$ be the target integral and multiply it by itself to get the double integral

$I^2=\int_0^\infty\int_0^\infty\dfrac{e^{-(x+y)}(dx)(dy)}{\sqrt {xy}}$

And then rotate the coordinate system by 45°:

$\xi=(x+y)/\sqrt2,\eta=(y-x)/\sqrt2$

Therefore, using the algebraic identity $xy=(1/2)(\xi^2-\eta^2)$ with variables defined as above, we may render

$I^2=\int_0^\infty\int_{-\xi}^\xi\dfrac{\sqrt2e^{-\xi\sqrt2}(d\eta)(d\xi)}{\sqrt{\xi^2-\eta^2}}$

Where is the Jacobian conversion factor above? That's the beauty part. Unlike conversion to polar coordinates, the rotation of the Cartesian coordinates is area-preserving (both magnitude and sense of rotation about the boundary of the area), so the conversion factor is simply $+1$. We do not have to jump through hoops to identify that "extra" factor of $r$ we get with the polar conversion, and yet the exponential function integration will remain elementary.

So we separate terms dependent only on $\xi$ to get

$I^2=\int_0^\infty\sqrt2e^{-\xi\sqrt2}[\int_{-\xi}^{\xi}\dfrac{d\eta}{\sqrt{\xi^2-\eta^2}}]d\xi$

Plugging in $u=\eta/\xi$ converts the $\eta$ integral to

$\int_{-1}^1\dfrac{du}{\sqrt{1-u^2}}=\sin^{-1}u| _{-1}^1=\pi$

The $\xi$ integral gives directly

$\int_0^{\infty}\sqrt2e^{-\xi\sqrt2}d\xi=e^{-\xi\sqrt 2}|_0^{\infty}=1$

So

$I^2=(1)(\pi)=\pi, I>0; \therefore I=\Gamma(1/2)=\sqrt{\pi}\approx 1.772.$

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The above result may be generalized to cover the $\Gamma$ function for all positive arguments. Consider the integral

$\Gamma(a)=\int_0^{\infty}x^{a-1}e^{-x}dx, a>0$

Square and convert to double integral form, then rotate the coordinates as above:

$[\Gamma(a)]^2=\int_0^\infty\int_0^\infty\dfrac{e^{-(x+y)}(dx)(dy)}{\sqrt {xy}}$

$=\int_0^\infty\int_{-\xi}^\xi2^{1-a}e^{-\xi\sqrt2}(\xi^2-\eta^2)^{a-1}(d\xi)(d\eta)$

The $\eta$ integration is done by putting in $u=\eta/\xi$. Note that with $a\ne1/2$ we get an extra factor entering the $\xi$ integration:

$[\Gamma(a)]^2=2^{1-a}\int_0^\infty\xi^{2a-1}e^{-\xi\sqrt2}\int_{-1}^1(1-u^2)^{a-1}du$

$=2^{1-2a}\Gamma(2a)\int_{-1}^1(1-u^2)^{a-1}du$

And thus

$\Gamma(a)=2^{(1/2)-a}\sqrt{\Gamma(2a)\int_{-1}^1(1-u^2)^{a-1}du}$

Thus $\Gamma(a)$ is rendered in terms of $\Gamma(2a)$ and an algebraic-function integral. If $a$ is half a natural number, tgen $\Gamma(2a)=(2a-1)!$ is a factorial and the algebraic-function integral is elementary, thus recovering the familiar function values. For instance:

$\Gamma(1/2)=2^0\sqrt{0!\int_{-1}^1(1-u^2)^{-1/2}du}=1\sqrt{(1)(\pi)}=\sqrt\pi\approx 1.772.$

$\Gamma(1)=2^{-1/2}\sqrt{1!\int_{-1}^1(1-u^2)^0du}=\sqrt{1/2}\sqrt{(1)(2)}=1.$

$\Gamma(3/2)=2^{-1}\sqrt{2!\int_{-1}^1(1-u^2)^{1/2}du}=(1/2)\sqrt{(2)(\pi/2)}=(1/2)\sqrt\pi\approx 0.886.$

For other values of $a$ the integral is nonelementary and the $\Gamma$ function will also be nonelementary, but for some rational arguments expressions are available in terms of elliptic integrals. Let us explore the simplest such case, $a=1/4$.

For this case we render

$\Gamma(1/4)=2^{1/4}\sqrt{(\sqrt\pi)\int_{-1}^1(1-u^2)^{-3/4}du}$

The integral may be converted to a form matching the complete elliptic integral of the first kind:

$\int_{-1}^1(1-u^2)^{-3/4}du=2\int_0^1(1-u^2)^{-3/4}du$

$=2\int_0^1\dfrac{2t^3(dt)}{t^3\sqrt{1-t^4}}, t=(1-u^2)^{1/4}$

$=4\int_0^1\dfrac{dt}{\sqrt{1-t^4}}=4\int_0^1\dfrac{dt}{\sqrt{1-t^2}\sqrt{1+t^2}}=4K(i)$

where the complete elliptic integral of the first kind is defined as

$K(k)=\int_0^1\dfrac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}}$

(The argument $i$ is a square root of $-1$; the function is real for pure imaginary arguments.) Thereby

$\Gamma(1/4)=(32\pi)^{1/4}\sqrt{K(i)}\approx 3.626$

where the elliptic integral may be rendered with high efficiency using (real) arithmetic and geometric means.