Gaussian Integral

The Gaussian integral, also called the Probability Integral, is the integral of the 1-D Gaussian over $(-\infty,\infty)$. It can be computed using the trick of combining two 1-D Gaussians

$$\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\left({\int_{-\infty}^\infty e^{-y^2}\,dy}\right)\left({\int_{-\infty}^\infty e^{-x^2}\,dx}\right)}$$

$$= \sqrt{\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)}\,dy\,dx}$$ (1)

and switching to Polar Coordinates,

$$\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{ \int^{2\pi}_0 \int^\infty_0 e^{-r^2}r\,dr\,d\theta}$$

$$= \sqrt{ 2\pi\left[{-{\textstyle{1\over 2}}e^{-r^2}}\right]_0^\infty} = \sqrt{\pi}.$$ (2)

However, a simple proof can also be given which does not require transformation to Polar Coordinates (Nicholas and Yates 1950).

The integral from 0 to a finite upper limit $a$ can be given by the Continued Fraction

$$\int_0^a e^{-x^2}\,dx = {\sqrt{\pi}\over 2} {1\over a+} {2\over 2a+} {3\over a+} {4\over 2a+\ldots}.$$ (3)

The general class of integrals of the form

$$I_n(a)\equiv \int_0^\infty e^{-ax^2}x^n\,dx$$ (4)

can be solved analytically by setting

$$x \equiv a^{-1/2}y$$ (5)

$$dx = a^{-1/2}\,dy$$ (6)

$$y^2 = ax^2.$$ (7)

Then

$$I_n(a) = a^{-1/2}\int_0^\infty e^{-y^2}(a^{-1/2})^n\,dy$$

$$= a^{-(1+n)/2} \int_0^\infty e^{-y^2} y^n\,dy.$$ (8)

For $n=0$, this is just the usual Gaussian integral, so

$$I_0(a)= {\sqrt{\pi}\over 2} a^{-1/2} = {1\over 2}\sqrt{\pi\over a}.$$ (9)

For $n=1$, the integrand is integrable by quadrature,

$$I_1(a)= a^{-1} \int_0^\infty e^{-y^2}y\,dy = a^{-1}[-{\textstyle{1\over 2}}e^{-y^2}]^\infty_0 = {\textstyle{1\over 2}}a^{-1}.$$ (10)

To compute $I_n(a)$ for $n>1$, use the identity

$$-{\partial\over\partial a}I_{n-2}(a) = -{\partial\over\partial a} \int_0^\infty e^{-ax^2}x^{n-2}\,dx$$

$$= -\int_0^\infty -x^2e^{-ax^2}x^{n-2}\,dx$$

$$= \int_0^\infty e^{-ax^2}x^n\,dx = I_n(a).$$ (11)

For $n=2s$ Even,

$$I_n(a) = \left({-{\partial\over\partial a}}\right)I_{n-2}(a) = \left({-{\partial\over\partial a}}\right)^2 I_{n-4}$$

$$= \ldots = \left({-{\partial\over\partial a}}\right)^{n/2} I_0(a)$$

$$= {\partial^{n/2}\over\partial a^{n/2}} I_0(a) = {\sqrt{\pi}\over 2} {\partial^{n/2}\over\partial a^{n/2}} a^{-1/2},$$ (12)

so

$$\int^\infty_0 x^{2s}e^{-ax^2}\,dx = {(s-{\textstyle{1\over 2... ...ver 2a^{s+1/2}}= {(2s-1)!!\over 2^{s+1}a^s} \sqrt{\pi\over a}.$$ (13)

If $n=2s+1$ is Odd, then

$$I_n(a) = \left({-{\partial\over\partial a}}\right)I_{n-2}(a) = \left({-{\partial\over\partial a}}\right)^2 I_{n-4}(a)$$

$$= \ldots = \left({-{\partial\over\partial a}}\right)^{(n-1)/2} I_1(a)$$

$$= {\partial^{(n-1)/2}\over\partial a^{(n-1)/2}} I_1(a) = {1\over 2} {\partial^{(n-1)/2}\over\partial a^{(n-1)/2}} a^{-1},$$ (14)

so

$$\int^\infty_0 x^{2s+1}e^{-ax^2}\,dx = { s!\over 2a^{s+1}}.$$ (15)

The solution is therefore

$$\int_0^\infty e^{-ax^2}x^n\,dx =\cases{ {(n-1)!!\over 2^{n/... ...$\ even\cr {[(n+1)/2]!\over 2a^{(n+1)/2}} & for $n$\ odd.\cr}$$ (16)

The first few values are therefore

$$I_0(a) = {1\over 2}\sqrt{\pi\over a}$$ (17)

$$I_1(a) = {1\over 2a}$$ (18)

$$I_2(a) = {1\over 4a}\sqrt{\pi\over a}$$ (19)

$$I_3(a) = {1\over 2a^2}$$ (20)

$$I_4(a) = {3\over 8a^2}\sqrt{\pi\over a}$$ (21)

$$I_5(a) = {1\over a^3}$$ (22)

$$I_6(a) = {15\over 16a^3}\sqrt{\pi\over a}.$$ (23)

A related, often useful integral is

$$H_n(a)\equiv {1\over\sqrt{\pi}}\int_{-\infty}^\infty e^{-ax^2}x^n\,dx,$$ (24)

which is simply given by

$$H_n=\cases{ {2I_n(a)\over\sqrt{\pi}} & for $n$\ even\cr 0 & for $n$\ odd.\cr}$$ (25)

References

Nicholas, C. B. and Yates, R. C. ``The Probability Integral.'' Amer. Math. Monthly 57, 412-413, 1950.